|
PHP - How To Compare The Radio Button Value Stored In Database.
i am developing online test ,after succesful completion of the test i want to display the correct answe,in databse i have stored corect answer as radio button values(ex 1,2,3,4,) i want to display the coreect answer with a right mark,how can i achive this?could anybody help?
here is my code
<?php $testid=$_GET['testid'];
include_once("header.php");
?>
<html>
<head>
<body>
<?php
include_once('connect.php');
$sqltest="select testname from test where testid='$testid'";
$sqltestresult=mysql_query($sqltest) or die(mysql_error());
while($result=mysql_fetch_array($sqltestresult))
{
$testname=$result['testname'];
}
?>
<!--pass test name and time left -->
<table width="100%" cellpadding="0" cellspacing="0" border="0"><tr><td width="50%" valign="middle">
<h1 style="margin-left:0;"><?php
if(isset($testname))
{
echo $testname ;
}
?></h1>
</td>
<td width="50%" align="right" valign="middle">
<ul class="tabs" id="tabs">
<li id="showtime" style="font-weight:bold; color:#FF0000; bottom:5px; font-size:16px; float:right"></li>
</ul>
</td>
</tr></table>
<div class="div-tabs-container" class="div-tabs-container" style="width:79%;margin-left:3%" >
<table cellpadding="0" cellspacing="0" border="0" id="questiontable" >
<tr>
<?php
include_once('connect.php');
$sql="select * from testquestions where testid='$testid'";
$result=mysql_query($sql) or die(mysql_error());
if (mysql_num_rows($result)>0)
{
$data = array(); // create a variable to hold the information
while (($row = mysql_fetch_array($result, MYSQL_ASSOC)) !== false)
{
$data[] = $row; // add the row in to the results (data) array
}
//shift off the first value
array_shift($data[0]);
$merge = call_user_func_array('array_merge', $data);
$size=sizeof($merge);
$newdata=array();
$num=1;
$correctanswerarray=array();
$selctedanswers=array();
/* echo "<form action='fetchquestion.php' method='post' id='formsubmit' onsubmit='return out()' name='test'>"; */
foreach ($merge as $key => $value )
{
$sql2="select qid, question,option1,option2,option3,option4,correct_answer from question where qid='$value' ";
$result2=mysql_query($sql2) or die(mysql_error());
while($row2=mysql_fetch_assoc($result2))
{
$questionid=$row2['qid'];
$question=$row2['question'];
$option1=$row2['option1'];
$option2=$row2['option2'];
$option3=$row2['option3'];
$option4=$row2['option4'];
$correctanswer=$row2['correct_answer'];
#array to store the correct answer.
$correctanswerarray[] = $questionid.$correctanswer;
#$newdata[]=$row2;
echo "
<table class='bix-tbl-container' cellspacing='0' cellpadding='0' border='0' width='100%' id='questiontable'><tr>
<td class='bix-td-qno jq-qno-2413' rowspan='2' valign='top' align='left'>$num</td>
<td class='bix-td-qtxt' valign='top'><p>$question</p></td>
</tr>
<tr>
<td class='bix-td-miscell' valign='top'><table class='bix-tbl-options' id='tblOption_2413' border='0' cellpadding='0' cellspacing='0' width='100%'><tr><td nowrap='nowrap' class='bix-td-option bix-td-radio' width='1%' id='tdOptionNo_A_2413'>
<input type='radio' class='result-option cls_2413' name='option_$questionid' value='1' id='disable'/>
</td><td class='bix-td-option' width='1%'>A.</td>
<td class='bix-td-option' width='48%' id='tdOptionDt_A_2413'><table border='0' cellpadding='0' cellspacing='0'>
<tr>
<td class='bix-inner-td-option'>$option1</td>
<td id='tdAnswerIMG_A_2413' class='jq-img-answer' valign='middle'style='display:none;padding-left:10px;'>
<img src='/_files/images/website/accept.png' alt='' />
</td>
</tr>
</table></td></tr><tr><td nowrap='nowrap' class='bix-td-option bix-td-radio' width='1%' id='tdOptionNo_B_2413'>
<input type='radio' class='result-option cls_2413' name='option_$questionid' value='2' id='disabble'/>
</td><td class='bix-td-option' width='1%'>B.</td>
<td class='bix-td-option' width='48%'id='tdOptionDt_B_2413'><table border='0' cellpadding='0' cellspacing='0'>
<tr>
<td class='bix-inner-td-option'>$option2</td>
<td id='tdAnswerIMG_B_2413' class='jq-img-answer' valign='middle' style='display:none;padding-left:10px;'>
<img src='/_files/images/website/wrong.gif' alt=''/>
</td>
</tr>
</table></td></tr><tr><td nowrap='nowrap' class='bix-td-option bix-td-radio' width='1%' id='tdOptionNo_C_2413'>
<input type='radio' class='result-option cls_2413' name='option_$questionid' value='3' id='disable'/>
</td><td class='bix-td-option' width='1%'>C.</td>
<td class='bix-td-option' width='48%' id='tdOptionDt_C_2413'><table border='0' cellpadding='0' cellspacing='0'>
<tr>
<td class='bix-inner-td-option'>$option3</td>
<td id='tdAnswerIMG_C_2413' class='jq-img-answer' valign='middle' style='display:none;padding-left:10px;'>
<img src='/_files/images/website/wrong.gif' alt='' />
</td>
</tr>
</table></td></tr><tr><td nowrap='nowrap' class='bix-td-option bix-td-radio' width='1%' id='tdOptionNo_D_2413'>
<input type='radio' class='result-option cls_2413' name='option_$questionid' value='4' id='disable'/>
</td><td class='bix-td-option' width='1%'>D.</td>
<td class='bix-td-option' width='48%' id='tdOptionDt_D_2413'><table border='0' cellpadding='0' cellspacing='0'>
<tr>
<td class='bix-inner-td-option'>$option4</td>
<td id='tdAnswerIMG_D_2413' class='jq-img-answer' valign='middle' style='display:none;padding-left:10px;'>
<img src='/_files/images/website/wrong.gif' alt='' />
</td>
</tr>
</table></td></tr></table>
\n
<input type='hidden' name='count' value='$num' id='count'>
<input type='hidden' name='queId[]' value='$questionid' id='queId'>
</td>
</tr>
</table>
";
$num++;
}
}
echo "
</tr>
</table>
";
#array of correct answrer
#echo "array of correct answer<br/>";
#print_r($correctanswerarray);
#echo "size of the coreect answer array";
$size1=sizeof($correctanswerarray);
echo "<br/>";
}
else
{
echo " no questions available for selected test";
}
?>
</div>
<!-- add div right side -->
<div style="height:500px;width:150px;float:right;margin-top:-500px;border:1px solid #99CCFF">
</div>
<div style="height:30px">
</div>
<!-- add div bottom -->
<div style="height:125px;border:1px solid #99CCFF">
</div>
<?php
include('footer.html');
?>
</body>
</html>
Edited by mac_gyver, 25 October 2014 - 10:05 AM. code tags around code please Similar Tutorials
Query About How To Retrieve A Password From The Database And Compare To The One The User Has Entered
When I display the data in my database I have included a radio button in each row. When the button is clicked, the intention is to delete all the information in that row. But it won't delete anything... Here's the form: Code: [Select] <FORM Method = "POST" action ="update_tee_times.php"> <?php include("opendatabase.php"); $dateID = $_POST[teetimedate]; $result = mysql_query(" SELECT Date, Player1,Player2,Player3,Player4,Player5,Time, IF( TIME(Time) BETWEEN '17:30:00' AND '19:30:00','FFFF66','FFFFFF') AS color FROM Daysheets WHERE Date ='$dateID'"); echo "<h1>"; echo date("l, F d, Y", strtotime($dateID)); echo "</h1>"; echo "<table align=center width=700 border=1>\n"; echo "<th>Clear</th><th>Time</th><th>Player 1</th><th>Player 2</th><th>Player 3</th><th>Player 4</th><th>Player 5</th> <th>M</th><th>W</th><th>G</th>"; $i=0; while($row = mysql_fetch_array( $result )) { $dateID = $row['Date']; $timeID = $row['Time']; echo "<tr>"; echo "<td>"; echo "<input type='radio' name='time1[$i]' value='$timeID'>"; //HERE'S THE RADIO BUTTON echo "</td><td>"; echo date("g:i ", strtotime($row[Time])); echo "</td><td>"; echo "<input type='text' size='17' value= '$row[Player1]' name='player1[$i]' style ='background-color:#{$row['color']};'>"; echo "</td><td>"; echo "<input type='text' size='17' value= '$row[Player2]' name='player2[$i]' style ='background-color:#{$row['color']};'>"; echo "</td><td>"; echo "<input type='text' size='17' value= '$row[Player3]' name='player3[$i]' style ='background-color:#{$row['color']};'>"; echo "</td><td>"; echo "<input type='text' size='17' value= '$row[Player4]' name='player4[$i]' style ='background-color:#{$row['color']};'>"; echo "</td><td>"; echo "<input type='text' size='17' value= '$row[Player5]' name='player5[$i]' style ='background-color:#{$row['color']};'>"; echo "</td><td>"; echo "<input type='text' size='2' value= '$row[Men]' name='women' style ='background-color:#{$row['color']};'>"; echo "</td><td>"; echo "<input type='text' size='2' value= '$row[Women]' name='women' style ='background-color:#{$row['color']};'>"; echo "</td><td>"; echo "<input type='text' size='2' value= '$row[Guest]' name='guests' style ='background-color:#{$row['color']};'>"; echo "<input type='hidden' name='date[$i]' value='$dateID'>"; echo "<input type='hidden' name='time2[$i]' value='$timeID'>"; echo "</td></tr>"; $i++; } echo "</table>"; mysql_close($con); ?> <br /><br /> <input type="Submit" name="Submit" value="Update Tee Times"> </FORM> And here's the form action: Code: [Select] <?php header("Location: /teetimes/preteetimesadmin.php"); include("opendatabase.php"); $size = count($_POST['player1']); $timeID1 = $_POST['time1'][$i]; //HERE'S WHERE I'M TRYING TO DELETE THE DATA. mysql_query(" UPDATE Daysheets SET Player1='', Player2='', Player3='', Player4='', Player5='' WHERE Time ='$timeID1' "); //HERE'S THE UPDATE, WHICH WORKS FINE. $i=0; while ($i < $size) { $dateID = $_POST['date'][$i]; $timeID2 = $_POST['time2'][$i]; $player1 = ($_POST['player1'][$i]); $player2 = ($_POST['player2'][$i]); $player3 = ($_POST['player3'][$i]); $player4 = ($_POST['player4'][$i]); $player5 = ($_POST['player5'][$i]); mysql_query(" UPDATE Daysheets SET Player1='$player1', Player2='$player2', Player3='$player3', Player4='$player4', Player5='$player5' WHERE Date ='$dateID' AND Time ='$timeID2' "); $i++; } mysql_close($con) ?>When I add or update any information it works fine. It just won't delete anything where the radio button is clicked. Any help appreciated! Thanks. I have page with male female radio button Here's the code Code: [Select] $qry=mysql_query("SELECT * FROM reg_table where id=$id "); $res=mysql_fetch_array($qry); $radio = $res['gender']; switch($radio) { case "male": $mal = "checked"; break; case "female": $fem = "checked"; break; } Code: [Select] Gender: <br /> <input type="radio" name="colour" value="male" checked="<?php $mal; ?>" />Male <input type="radio" name="colour" value="female" checked="<?php $fem; ?>" />Female Whats the use of checked in <input type="radio" name="colour" value="male" checked="<?php $mal; ?>" /> Hi there. I have this course registering website whereby i will need to : 1. pull out all course info and display (done that) 2. when user choose the course they are interested in using radio button and click submit , it will display you're registered and update the database to decrease the head count by 1. i m stuck on the radio button part... after i click submit nothing happens... Please advice... thanks a lot Code: [Select] <html> <head> <title>TimeTable</title> </head> <body> <h1> Here is the timetable</h1> <?php @ $db = new mysqli('localhost', 'root', '', 'mylifestyle'); if (mysqli_connect_errno()) { echo 'Error: Could not connect to database. Please try again later.'; exit; } $query = "SELECT course.CourseName , course.Instructor , course.ClassSize , courseschedule.Date ,courseschedule.Time , courseschedule.Location FROM course INNER JOIN courseschedule ON course.CourseID = courseschedule.CourseID"; $result = $db->query($query); $num_results = $result->num_rows; echo "<table border='1'> <tr> <th>Course Name</th> <th>Instructor</th> <th>Class size</th> <th>Date</th> <th>Time</th> <th>Location</th> <th>Register</th> </tr>"; ?> <?php for ($i=0; $i <$num_results; $i++) { $row = $result->fetch_assoc();?> <tr> <form action = "" method = "post"> <td><? echo $row['CourseName']; ?></td> <td><? echo $row['Instructor']; ?></td> <td><? echo $row['ClassSize']; ?></td> <td><? echo $row['Date']; ?></td> <td><? echo $row['Time']; ?></td> <td><? echo $row['Location']; ?></td> <td align="center"><input name="radio" type="radio" id="radio[]" value="<? echo $row['ClassSize']; ?>"></td> </tr> <?php } ?> <td colspan="5" align="center" bgcolor="#FFFFFF"><input name="submit" type="submit" id="submit" value="submit"></td> <?php if($submit){ for($i=0;$i<$num_results;$i++){ $del_id = $radio[$i]; $query = "UPDATE course SET ClassSize = ClassSize -1 Where id='$del_id'"; $result = mysql_query($query); } // if successful redirect to delete_multiple.php if($result){ echo "$CourseName was deleted from database - myliftstyle!!<br />"; // Compute the item costs and total cost } } ?> </div> </body> </html> [attachment deleted by admin] Hey guys, I have been banging my head against a wall here with this. I am saving my sessions in my database via session_set_save_handler(). So let me walk you through what I have here, what works, and what the issue seems to be. basically, the problem is the $_SESSION array is empty on page load. common.php: I have the open / close / read / write / destroy / and gc functions. These all work properly as when i use a session variable it stores into the database and i can see all of the information in there.. inside of common.php i have session_start().. I am positive that the session_start() is running becasue common.php is included into index.php and i tried adding session_start() to index.php again and i got an E_NOTICE saying the session already began. (yes, for the read function i am returning a string... i included that below). for the table itself, i have set the session_data as both text and blob, same issue with both. index.php: includes common.php. I know it's included as other aspects of the file are included and work properly. if i call var_dump($_SESSION) i get an empty array. and i prited out the session_id() and it matched my session id in the database. and the values that are stored in there are correct. i have for example: count|i:0 now with that value actually in the database and when calling session_id() and i get the ID that matches in the table. so i will get an E_NOTICE of an undefined index.. i was trying something like: if(!isset($_SESSION['count'])) $_SESSION['count'] = 0; else $_SESSION['count']++; echo $_SESSION['count']; Everytime the page is reloaded, count is reset to 0.. I can tell that it is reset to 0 as the expired time changes in the database after every load of the page (which is part of the write function). I double checked that it wasn't a problem for some reason with the ++ by adding in a variable that sets to 1 when in the if, and 0 if in the else, and it always outputs a 1. i have included here my read function since that apparently is the issue.. function sess_read($sess_id) { global $DB; $sql = "SELECT `session_data` FROM `sessions` WHERE session_id = '".$sess_id."' AND session_agent = '".$_SERVER["HTTP_USER_AGENT"]."' AND session_expire > '".time()."';"; $query = $DB->query($sql); if($DB->num_rows($query) > 0) { $r = $DB->fetch($query); return settype($r['session_data'], 'string'); } return ''; } I tried also with removing the agent and expire check to see if it was an issue there but same problem. I probably have missed something really dumb but i can't for the life of me figure it out and I have googled around for a similar issue. The actual output of the page is correct.. all of the HTML and CSS information loads properly.. no errors are reported (and i have E_ALL on). Thanks How can I run php code that is stored in a database? I have my pages created by grabbing the contents of a pageBody mySQL table cell, but some pages require further php to be able to display correctly. For example, I have a players page, which will list all players, and some information about them. The players and information is all saved in a players table in my database (separate to the pages table where pageBody is stored). I use simple php to grab all the players and format all of their information so it can be displayed on the page. This is the flow of information that I would like: User clicks on players page browser loads content page (this is a template page), and grabs the pageid from $_GET browser then reads pages table in database to get the pageBody associated with the pageid The pageBody will contain more php, which will run the players script to retrieve the list, and display Doing it the above way will make it much easier to extend the website to include more types of pages that has to run additional php scripts. The only way I can think of this working is by doing this: user clicks on players page browser loads content page, grabs pageid from $_GET browser checks the pageid for the one associated with players (hard coded into the php script) browser then loads the players.php script instead of going to the database This above way means that I will need to edit the index.php page everytime I add a new list type (example, coaches, volunteers etc). Anyone have any suggestions? I know my question is long, but I was finding it hard to explain it in an easy to understand way, but I'm still not sure if I have :S. Cheers Denno Hello - I need a way to execute a stored procedure from a wordpress site with an external database where the stored procedure is housed. I am very limited in php code. All I know is from what little I have used My Custom Functions plugin. Please be very detailed. This stored procedure I run truncates and re-populates a number of tables I use for various reports. I am pulling data from the wp db and creating tables in the external db which make the data easier to read than through the wp meta data. The tables are housed in a separate db so as not to slow down the wp db. The hosting company does not allow 'Events' to run on the mySql db so I am left with figuring out how to create a cron job using a php file which would be the best. Aside from that, I thought a button on an admin page for execution would make it easier if I have to do it be hand. Thanks for your help. Hi Guys, I have a webpage which has subsequent pages stored in a database e.g. index.php?id=1 The problem being, is that id=1 has it's data pulled from a database. This was fine & dandy until I started to insert PHP...I am trying to get the below to executre <?php $rater_id=1; $rater_item_name='Rate This'; include("rater.php");?> However nothing shows & nothing happens, I know eval can be used but have not been succesfull in implementing this, can someone please help! I put together the following blocs of code for uploading pictures into a database and displaying them on a webpage. The pictures are supposed to be displayed on the member's only page of a website I'm working on, upon logging in, and they are supposed to be the member's uploaded picture. I created several members and and used one of my existing member accounts to test the uploading process. The picture upload process appeared to have been successful when I checked on myphpadmin. Yet, when I login with this account, no picture is displayed, instead, a tiny jpg icon is displayed at the top left corner of the box in which the picture was supposed to be displayed. Same thing when I login with the other accounts with which I haven't yet uploaded a picture. I'll start with the code that installs the table in the database $query = "CREATE TABLE images ( image_id INT UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY , member_id INT UNSIGNED, like_id INT UNSIGNED, image LONGBLOB NOT NULL, image_name varchar(255) NOT NULL, image_type varchar(4) NOT NULL, image_size int(8) NOT NULL, image_cartegory VARCHAR(20) NOT NULL, image_path VARCHAR(300), image_date DATE )"; Then here is the code which allows the member to upload his pictu <form enctype="multipart/form-data" action="insert_image.php" method="post" name="changer"> <input name="MAX_FILE_SIZE" value="102400" type="hidden"> <input name="image" accept="image/jpeg" type="file"> <input value="Submit" type="submit"> </form> And here is the insertimage.php which inserts the image into our database: Note that I have to authenticate the user in order to register his session id which is used later on in the select query to identify him and select the right image that corresponds to him. <?php //This file inserts the main image into the images table. //address error handling ini_set ('display_errors', 1); error_reporting (E_ALL & ~E_NOTICE); //authenticate user //Start session session_start(); //Connect to database require ('config.php'); //Check whether the session variable id is present or not. If not, deny access. if(!isset($_SESSION['id']) || (trim($_SESSION['id']) == '')) { header("location: access_denied.php"); exit(); } else{ // Make sure the user actually // selected and uploaded a file if (isset($_FILES['image']) && $_FILES['image']['size'] > 0) { // Temporary file name stored on the server $tmpName = $_FILES['image']['tmp_name']; // Read the file $fp = fopen($tmpName, 'r'); $data = fread($fp, filesize($tmpName)); $data = addslashes($data); fclose($fp); // Create the query and insert // into our database. $query = "INSERT INTO images (member_id, image_cartegory, image_date, image) VALUES ('{$_SESSION['id']}', 'main', NOW(), '$data')"; $results = mysql_query($query); // Print results print "Thank you, your file has been uploaded."; } else { print "No image selected/uploaded"; } // Close our MySQL Link mysql_close(); } //End of if statmemnt. ?> On the page which is supposed to display the image upon login in, I inserted the following html code in the div that's supposed to contain the image: <div id="image_box" style="float:left; background-color: #c0c0c0; height:150px; width:140px; border- color:#a0a0a0;border-style:outset;border-width:1px; margin:auto; "> <img src=picscript.php?imname=potwoods> </div> And finally, the picscript.php contained the select query: <?php //address error handling ini_set ('display_errors', 1); error_reporting (E_ALL & ~E_NOTICE); //include the config file require('config.php'); $image = stripslashes($_REQUEST[imname]); $rs = mysql_query("SELECT* FROM images WHERE member_id = '".$_SESSION['id']."' AND cartegoty = 'main' "); $row = mysql_fetch_assoc($rs); $imagebytes = $row[image]; header("Content-type: image/jpeg"); print $imagebytes; ?> Now the million dollar question is, "What is preventing the picture from getting displayed?" I know this is a very lengthy and laborious problem to follow but I'm sure there is someone out there who can point out where I'm not getting it. Thanks. Hi, I am trying to store a sql query in a database, but every time I try to read it, it shows all the code as text. eg just print ' . $parts_row['part_number'] . ' Here is the code on the page:
$parts_sql = "SELECT * FROM parts WHERE part_cat = " . $category_row['cat_id'] . " ORDER BY CAST(part_a AS DECIMAL(10,2))";
$parts_result = mysql_query($parts_sql);
if(!$parts_result)
{
echo '<tr><td>The parts could not be displayed, please try again later.</tr></td></table>';
}
else
{
while($parts_row = mysql_fetch_array($parts_result))
{
$pageformat_sql = "SELECT * FROM category_pages WHERE catpage_number = " . $category_row['cat_page'] . "";
$pageformat_result = mysql_query($pageformat_sql);
if(!$pageformat_result)
{
echo 'Error';
}
else
{
while($pageformat_row = mysql_fetch_assoc($pageformat_result))
{
$data = $pageformat_row['catpage_table'];
echo '<table class="table2 centre" style="width:750px">
' . $pageformat_row['catpage_tabletitle'] . '' . $data . '';
}
}
}}
And this is what is stored in the database table:
<tr> <td class="cell left">' . $parts_row['part_number'] . '</td> <td class="cell centre">' . $parts_row['part_a'] . ' mm</td> <td class="cell centre">' . $parts_row['part_b'] . ' mm</td> <td class="cell centre">' . $parts_row['part_c'] . ' mm</td> <td class="cell centre">' . $parts_row['part_d'] . ' mm</td> <td class="cell centre">' . $parts_row['part_e'] . ' mm</td> <td class="cell centre">' . $parts_row['part_imped100'] . '</td> <td class="cell centre"><a href="datasheets/' . $parts_row['part_datasheet'] . '.pdf"><img border="0" src="images/pdf.jpg" alt="Download"</a></td></tr>I would be very grateful to anyone who can help me with this. this is the line in my script that I have to show the image: Code: [Select] $output .= "<img>{$row['disp_pic']}</img></br>\n"; As you can see I added the image tag, but it wont show the actual image. IE shows it as a small square with another small square picture icon in the middle of it (i'm sure you guys know what i mean). Hi Everyone, I have a program that generates 200 unique images keeping the first image static in each run.The images keep scrolling on to the screen pause for 3 seconds and scroll off I'm able to generated all 200 unique images without repetition, everything is working well except for the lase two images the last two images are scrolling on to the screen but are not been displayed in the database, Moreover The last image is a duplicate of 197th image.I don't know what is happening..... Here is MY code.......... <?php session_start(); $sid = $_SESSION['id']; $_SESSION['imageDispCnt'] = 0; $myQuery = "SELECT * from image"; $conn = mysql_connect("localhost","User","Passwd"); mysql_select_db("database_Name",$conn); $result = mysql_query($myQuery); $img =Array(); $id =Array(); $i =0; $imagepath = 'http://localhost/images/'; while ($row = mysql_fetch_array($result)) { $img[$i] = $imagepath.$row['img_name']; $id[$i] = $row['imageid']; $i = $i + 1; } ?> </head> <script language="JavaScript1.2"> var scrollerwidth='800px'; var scrollerheight='600px'; var scrollerbgcolor='white'; var pausebetweenimages=3200; var s; var sec; var d; var j; var imgid; var milisec = 0; var seconds = 0; var flag = 1; var ses_id = '<?php echo $sid;?>'; var count = 0; var i = 0; var imgname; var imgid; var k =0; var slideimages=new Array(); var img_id = new Array(); var index; <?php $l =0; $count = array(); $j = rand(0,199); while($l < 200) { while(in_array($j, $count)) { $j = rand(0,199); } $count[$l] = $j; $l++; }?> <?php $k = 0; for($k = 0;$k<count($count);$k++){ ?> index = <?php echo $k;?>; <?php $indx = $count[$k];?> if(index == 0){ slideimages[0] = '<img src="http://localhost/images/hbag044.jpg" name="r_img" id="0"/>'; img_id[0] = '<input type="hidden" value="0" id="imgId" />'; } else if(index > 0) { slideimages[<?php echo $k?>] = '<img src="<?php echo $img[$indx]?>" name="r_img" id="<?php echo $id[$indx]?>"/>'; img_id[<?php echo $k?>] = '<input type="hidden" value="<?php echo $id[$indx]?>" id="imgId" />'; } <?php } ?> Can Any one plese help me Appreciate your help... Thanks Good afternoon all!! I have a very specific issue that, in my opinion, is quite complicated. In words, here is what I wish to achieve. I would like a page which displays users from the database. Each individual user has their own Div hidden below their name. Within this div, there is a form. The form is full of radio buttons. Once the user's name is clicked and form submitted, I wish to write the form data to the database for the specific user that was selected. See below example for better understanding: Dale Gibbs Chris Hansen Steve Jobs If I click Chris Hansen, the following happens Dale Gibbs ============== Chris Hansen HIDDEN DIV FORM CONTENT HIDDEN DIV FORM CONTENT HIDDEN DIV FORM CONTENT HIDDEN DIV FORM CONTENT HIDDEN DIV FORM CONTENT Submit Button ============== Steve Jobbs So as of now, the display is correct. I am seeing exactly what I want to see from my database. Also, my div IDs work just fine as well as the Javascript toggleSlidebox function. The only issue is for some reason, whenever I submit the form (no matter which user I select from my list), I can only write to the last inputted ID. So for example, if the last ID entered into the database was 6, then thats the only ID that will be returned when my form is submitted and the only place data will be written to, even if I select a user with ID 2. Please see below code for more information Code: [Select] <?php $staff_display_query = "SELECT staff_info.id, staff_info.fname, staff_info.lname FROM staff_info, staff_projects WHERE staff_info.id = staff_projects.staff_id AND staff_projects.proj_id = '$c_project_id'"; $staff_display_sql = mysql_query($staff_display_query) or die (mysql_error()); while ($row = mysql_fetch_array($staff_display_sql)) { $current_staff_id = $row['id']; $staff_fname = $row['fname']; $staff_lname = $row['lname']; $list_staff .= ' ' . $current_staff_id . '<br /> <a href="#" onclick="return false" onmousedown="javascript:toggleSlideBox(' . $current_staff_id . ');">' . $staff_fname . ' ' . $staff_lname . '</a> <div class="hiddenDiv" id="' . $current_staff_id . '" style="border:#FFF 2px solid; width:553px;"> <!--TASK 1--> <div id="task_1_permissions" class="task_permissions"> <input name="permissions_1" type="radio" value="1"/> <input name="permissions_1" type="radio" value="2" /> <input name="permissions_1" type="radio" value="3" /> <input name="permissions_1" type="radio" value="0" /> </div> <!--TASK 2--> <div id="task_2_permissions" class="task_permissions"> <input name="permissions_2" type="radio" value="1"/> <input name="permissions_2" type="radio" value="2" /> <input name="permissions_2" type="radio" value="3" /> <input name="permissions_2" type="radio" value="0" /> </div> <!--TASK 3--> <div id="task_3_permissions" class="task_permissions"> <input name="permissions_3" type="radio" value="1"/> <input name="permissions_3" type="radio" value="2" /> <input name="permissions_3" type="radio" value="3" /> <input name="permissions_3" type="radio" value="0" /> </div> <!--TASK 4--> <div id="task_4_permissions" class="task_permissions"> <input name="permissions_4" type="radio" value="1"/> <input name="permissions_4" type="radio" value="2" /> <input name="permissions_4" type="radio" value="3" /> <input name="permissions_4" type="radio" value="0" /> </div> <input name="submit_user_permissions" type="submit" value="Submit Permissions" /> </div> </div> <br /><br /> '; } if (isset($_POST['submit_user_permissions'])) { $permissions_1 = $_POST['permissions_1']; $permissions_2 = $_POST['permissions_2']; $permissions_3 = $_POST['permissions_3']; $permissions_4 = $_POST['permissions_4']; $query = "UPDATE staff_projects SET task_1='$permissions_1', task_2='$permissions_2', task_3='$permissions_3', task_4='$permissions_4' WHERE proj_id='$c_project_id' AND staff_id='$current_staff_id'"; $sql = mysql_query($query) or die (mysql_error()); echo 'Permissions set successfully.<br />'; } After this PHP I have my standard HTML. Below is the javascript function I use for the slide box: Code: [Select] <script src="js/jquery-1.5.js" type="text/javascript"></script> <script language="javascript" type="text/javascript"> function toggleSlideBox(x) { if ($('#'+x).is(":hidden")) { $(".hiddenDiv").slideUp(200); $('#'+x).slideDown(300); } else { $('#'+x).slideUp(300); } } </script> Javascript works just fine by the way. Below is the form I have in the HTML of the code. Code: [Select] <form action="" method="post" enctype="multipart/form-data"> <?php echo "$list_staff"; ?> </form> This is of course wrapped around body tags and all the other necessary HTML. The view of the form is working right, the functions within the form are also working correctly. I just cant seem to separate the variables for each individual user. If I am in Chris Hansen's div, it should be his ID number that is being referenced for the database, not the ID number of the last person entered into the system. Any ideas? As always, thanks in advance guys!!! Bl4ck Maj1k Hi! I was wondering if there is a way to execute php code which is stored in mysql database using php. At the minute I am using a echo to try and run php code stored in a mysql database but this just displays the code and does not run the php code. Thanks for any help! Hi there! On a page of mine, random tables get generated. The tables consist of 8 characters, so there's 5 . 10^13 possibilities. Yet ofcourse there is a possibility that 2 tables with the same name can be generated, which ofcourse is not desired. So I need an adjustment that can see whether there is already a name that's the same, and if so, make another random name. And if that one also exists, make a new one again. Untill the name doesn't exist yet and it remains. Here's the script I have right now: function createName($length) { $chars = "abcdefghijkmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"; $i = 0; $password = ""; while ($i <= $length) { $password .= $chars{mt_rand(0,strlen($chars))}; $i++; } return $password; } $name = createName(8); mysql_select_db($database, $con); $sql = "CREATE TABLE " . $name . "( id int NOT NULL AUTO_INCREMENT, PRIMARY KEY(id), devraag varchar(500), weergave varchar(500), naam varchar(500), inhoud varchar(500), tijd varchar(100))"; mysql_query($sql, $connection)or die(mysql_error()); I'm not goot with the combination of PHP and mysql, and really have no Idea how to do this. Any help? Hi all, I've got a website for an event, each team have their details on a page which are recalled from a SQl database. But I'm wanting to create a password input box for each team, so when they enter the correct password they are taken to a page containing forms where they can edit the team details. Here is the page with the users details on where they anter the password: http://www.wharncliffenetwork.co.uk/wrc/entered/team.php?id=8 I'm not sure how to code it, Can an IF statement be used? Anyone got any pointers? I'f been unsuccessful in finding a tutorial or something similar. Hope that makes sense :S Cheers. Hi guys, I have a file location stored in mysql. when i populate the table i need this file location to be a hyperlink to the file itself, so the visitors can click like a normal link and open the file in word and pdf (both formats stored). example of file location as in db "_private/Incident_Reports/Incident%20-%20Applecross%20-%2017%20December%202010%20-%20Website.doc" example of php code Code: [Select] echo $row['word_document']; any ideas would be really appreciated. Hello, Sorry for my english... Well, I make a smaill website. members has to login and then member can see pictures (saved in a folder). When the admin upload pictures a folder wil be created. To get the actual folders in the "album" folder I use this code. <?php //get the folders $handle = opendir(realpath('../album/Album/.')); while ($direc = readdir($handle)) { if($direc == '.' || $direc == '..') continue; else { echo "\t\t<input type='checkbox' name='map[]' id='map[]' value='" . $direc. "' />" . $direc . "<br />\n"; } } closedir($handle); ?> The admin has to assign the folders wich are allowed to see by a member. I use checkboxes for that. And now my problem: When a member can see folder 1 to 4 and the admin upload 2 new folders and want to give that member to see these 2 new folders I want the following: 6 checkboxes (checkbox 1 > album 1, checkbox 2 > album 2... etc). As is said, this member can see folder 1 to 4, so in the CMS album 1 to 4 has to be checked ans album 5 to 6 not. This is my code for that: <?php $sql2 = "SELECT url FROM url WHERE gebruiker_id =" . $_GET['id'] . " ORDER BY url ASC"; $res2 = mysql_query($sql2); while ($rij = mysql_fetch_array($res2)) { $url = $rij['url']; // mappen ophalen $handle = opendir(realpath('../album/Album/.')); while ($direc = readdir($handle)) { if($direc == '.' || $direc == '..' || $direc == $url) { continue; } echo "\t\t<input type='checkbox' name='map[]' id='map[]' value='" . $direc. "' />" . $direc . "<br />\n"; } closedir($handle); } ?> Now my output is: <input type='checkbox' name='map[]' id='map[]' value='album 2' />album 2<br /> <input type='checkbox' name='map[]' id='map[]' value='album 3' />album 3<br /> <input type='checkbox' name='map[]' id='map[]' value='album 4' />album 4<br /> <input type='checkbox' name='map[]' id='map[]' value='album 5' />album 5<br /> <input type='checkbox' name='map[]' id='map[]' value='album 6' />album 6<br /> <input type='checkbox' name='map[]' id='map[]' value='album 1' />album 1<br /> <input type='checkbox' name='map[]' id='map[]' value='album 2' />album 2<br /> <input type='checkbox' name='map[]' id='map[]' value='album 4' />album 4<br /> <input type='checkbox' name='map[]' id='map[]' value='album 5' />album 5<br /> <input type='checkbox' name='map[]' id='map[]' value='album 6' />album 6<br /> <input type='checkbox' name='map[]' id='map[]' value='album 1' />album 1<br /> <input type='checkbox' name='map[]' id='map[]' value='album 2' />album 2<br /> <input type='checkbox' name='map[]' id='map[]' value='album 3' />album 3<br /> <input type='checkbox' name='map[]' id='map[]' value='album 5' />album 5<br /> <input type='checkbox' name='map[]' id='map[]' value='album 6' />album 6<br /> What do I have to do to check the checkboxes (album 1-4) and not to check the album 5-6??? My guess it has to do with the While in While loop but... Greetings, Ferdi Has anyone got any script or refence to a tut where i can find a script that compares two mysql database (current) and outdated db and then takes the current db and updates the outdated one to match accordingly. Thanks Hi, after following lots of advice and changing to MySqli I am running into a few probs. This is me just probably missing something stupid, I know what I want, but can't figure out what query I should use and where I should place it. All the queries I have tried have failed.
I just need a query that gets the $current_stored_password from the password field on the database, to confirm the last check
elseif
($current_password !== $current_stored_password) {
include 'includes/overall/header.php';
echo $current_password . ' AND ' . $_POST['current_password'] . ' Password and password again do not match';
include 'includes/overall/header.php';
}
Here is the whole script.
<?php
session_start();
error_reporting(0);
//ini_set('display_errors', '1');
require( 'database.php' );
$username = $_SESSION['loggedinuser'];
$current_stored_password = $_SESSION['password'];
$current_password = $_POST['current_password'];
$password = mysqli_real_escape_string($con, md5( $_POST['password']));
$password_again = mysqli_real_escape_string($con, md5( $_POST['password_again']));
// Run checks
if (isset($_POST['current_password'], $_POST['password'], $_POST['password_again'])) {
if( strlen( $_POST['current_password'] ) < 8 )
{
include('includes/overall/header.php');
echo "Password Must Be 8 or More Characters.";
include('includes/overall/footer.php');
}
elseif( strlen( $_POST['password'] ) < 8 )
{
include('includes/overall/header.php');
echo "Password Must Be 8 or More Characters.";
include('includes/overall/footer.php');
}
elseif
( strlen( $_POST['password_again'] ) < 8 )
{
include('includes/overall/header.php');
echo "Password Must Be 8 or More Characters.";
include('includes/overall/footer.php');
}
elseif
($password !== $password_again) {
include 'includes/overall/header.php';
echo ' Password and password again do not match';
include 'includes/overall/header.php';
}
elseif
($current_password !== $current_stored_password) {
include 'includes/overall/header.php';
echo $current_password . ' AND ' . $_POST['current_password'] . ' Password and password again do not match';
include 'includes/overall/header.php';
} else {
// Define a query to run
$query = "UPDATE `user` SET `password` = '$password' WHERE `username` = '$username'";
// Query the database
$result = mysqli_query($con,$query);
// Check if the query failed
if( !$result )
{
die('There was a problem executing the query ('.$query.'):<br>('.mysqli_errno($con).') '.mysqli_error($con));
}
else {
include 'includes/overall/header.php';
echo 'Password has been changed';
include 'includes/overall/footer.php';
}
}
}
// Close the connection
mysqli_close($con);
?>
At the moment the message displayed when the form is submitted is
echo $current_password . ' AND ' . $_POST['current_password'] . ' Password and password again do not match';How do I retrieve the password from the database to compare against the current password entered by the user? Any help is much appreciated. PS. Yes I know I have repeated code and that md5 is not secure, but I am just building onto a template I got and will be making changes to shorten the code and secure the password soon |
|||||||