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PHP - Using A Sql Query Thats Stored In A Table On The Database.
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. Similar TutorialsHi, In some cases I need to execute a default query , so I am storing the SELECT statement in a variable like this and executing it: Code: [Select] $default_query="SELECT * from user where userid='$userid'"; $user_query=mysql_query($default_query); The above code returns an empty results. But if I execute it without the variable it works fine. Code: [Select] $user_query=mysql_query("SELECT * from user where userid='$userid'"); Am I syntatically wrong somewhere? i want the name of a picture stored in my db after i upload it the data is not stored in the db after i run this script, but i dont get errors either i print the two vars before sending them, and they get printed fine any help on this would be greatly appreciated thanks ! <?php error_reporting(E_ALL); ini_set("display_errors", 1); // INCLUDE THE CLASS FILE include('ImageLib.Class.php'); include("./includes/egl_inc.php"); $displayMessage = ''; if($_POST){ if(isset($_FILES['image_file'])){ // SEE THE MAGIC HAPPEN $destination_path = 'uploads/'; $post_file_name = 'image_file'; $width = 600; $height = 400; $scale = false; $trim = true; $uniqueName = true; $img = ImageLib::getInstance()->upload($post_file_name, $destination_path, $uniqueName)->resize($width, $height, $scale, $trim)->save(); $imgstr = mysql_real_escape_string ($img); $fileName = $_FILES['image_file']['name']; $displayMessage = '<div class="image"><img src="'.$destination_path.$fileName.'" /><br />Uploaded And Resized...With new file name : "'.$img.'"</div><br /><br />'; $playerid=$_SESSION['tid']; $matchdetails = mysql_fetch_array(mysql_query("SELECT id FROM ffa_matches WHERE status=2 and admin=$playerid")); $id = $matchdetails[id]; print $img;print $imgstr; print $id; mysql_query(" INSERT INTO ffa_screens (imgname,match) VALUES( '" . mysql_real_escape_string($imgstr) . "', '" . mysql_real_escape_string($id) . "' )"); }} ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html> <head> <title>ImageLib Samples By Rahul Kate</title> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <style> body{font-family: arial; font-size:12px; color:#444444; padding:20px;} li{margin-top:10px;} .image{color:green;} .image img{margin-bottom:5px;} </style> </head> <body> <h1>ImageLib | Upload Image, move it to Uploads folder and Resize it and Trim</h1> <?php echo $displayMessage; ?> <form method="post" enctype="multipart/form-data"> Select Image<br /> <input type="file" name="image_file" id="image_file" /> <br /> <br /> <input type="submit" name="submit" value="Submit" /> <br /> <br /> <a href="index.html">Back TO Home</a> </form> </body> </html> 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 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 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 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. 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 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! 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 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 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 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. the second database found on the cloud
i try to get JSON data but how to insert and update them to another online database with the same table my php script to return json data <?php include_once('db.php'); $users = array(); $users_data = $db -> prepare('SELECT id, username FROM users'); $users_data -> execute(); while($fetched = $users_data->fetch()) { $users[$fetchedt['id']] = array ( 'id' => $fetched['id'], 'username' => $fetched['name'] ); } echo json_encode($leaders);
i get
{"1":{"id":1,"username":"jeremia"},"2":{"id":2,"username":"Ernest"}} Edited March 24 by mahenda hello I want query from one table and insert in another table on another domain . each database on one domain name. for example http://www.site.com $con1 and http://www.site1.com $con. can anyone help me? my code is : <?php $dbuser1 = "insert in this database"; $dbpass1 = "insert in this database"; $dbhost1 = "localhost"; $dbname1 = "insert in this database"; // Connecting, selecting database $con1 = mysql_connect($dbhost1, $dbuser1, $dbpass1) or die('Could not connect: ' . mysql_error()); mysql_select_db($dbname1) or die('Could not select database'); $dbuser = "query from this database"; $dbpass = "query from this database"; $dbhost = "localhost"; $dbname = "query from this database"; // Connecting, selecting database $con = mysql_connect($dbhost, $dbuser, $dbpass) or die('Could not connect: ' . mysql_error()); mysql_select_db($dbname) or die('Could not select database'); //query from database $query = mysql_query("SELECT * FROM `second_content` WHERE CHANGED =0 limit 0,1"); while($row=mysql_fetch_array($query)){ $result=$row[0]; $text=$row[1]."</br>Size:(".$row[4].")"; $alias=$row[2]; $link = '<a target="_blank" href='.$row[3].'>Download</a>'; echo $result; } //insert into database mysql_query("SET NAMES 'utf8'", $con1); $query3= " INSERT INTO `jos_content` (`id`, `title`, `alias`, `) VALUES (NULL, '".$result."', '".$alias."', '')"; if (!mysql_query($query3,$con1)) { die('Error: text add' . mysql_error()); } mysql_close($con); mysql_close($con1); ?> wrote a stored procedure this morning and i don’t know how to get the values out of it through a class function in php or phpmyadmin. here is what i wrote :
public function totalProcedures($friend_name,$session_id)
{ /* *query to fetch stored procedure */
try {
//executing the stored procedure
$sql_sp="CALL timeline (:friend, :session,@updates, @group_posts)";
$stmt_sp= $this->_db->prepare($sql_sp);
$stmt_sp->bindValue(":friend",$friend_name);
$stmt_sp->bindValue(":session",$session_id);
$stmt_sp->execute();
$rows=$stmt_sp->fetch(PDO::FETCH_ASSOC);
$stmt_sp->closeCursor(); // closing the stored procedure
//trying to get values from OUT parameters.
$stmt_sp_2=$this->_db->prepare("select @updates,@group_posts");
$stmt_sp_2->execute();
return $stmt_sp_2->fetch(PDO::FETCH_ASSOC);
} catch (PDOException $ei)
{
echo $ei->getMessage(); }
}
can someone helpme how to get results. here is the storedprocedu DELIMITER $$ CREATE DEFINER=`root`@`localhost` PROCEDURE `timeline`(IN `friend` VARCHAR(255), IN `session_id` VARCHAR(255), OUT `updates` VARCHAR(62555), OUT `group_posts` VARCHAR(62555)) BEGIN select * FROM updates where author in (friend,session_id) order by time desc limit 5; select * FROM group_posts where author_gp in (friend,session_id) order by pdate desc limit 5; END$$ DELIMITER ; i get the result in php myadmin as follows:
how do i do this inside a php class function.
CALL timeline('shan2batman','aboutthecreator', @updates, @group_posts);
Hello All, Trying to write a log-in script using PDO. But have one question. With the below when I query the database and even run a print_r on $RES nothing is outputted? Any ideas? Thanks if (isset($_POST['Submit'])) { $email = $_POST['email']; $password = $_POST['password']; $QUERY = $dbc->prepare("SELECT email, password FROM tbl WHERE email = :email"); $QUERY->bindParam(':email', $email); $QUERY->execute(); $RES = $QUERY->fetchColumn(); echo "Email is:".$RES['email']; echo "PW is: ".$RES['password']; print_r($RES); } Is it possible to query my database and display results from a varible that changes each time from a users search? I know how to get results from a variable like a field name and display them but i was wondering if for example; the user searches for criteria and gets the result there looking for (postcode); they then click that result in the browser and are redirected to another page with lots more information.(company_name and location). There result from the original search would be different each time but i would like to know if i could take that result and display all other fields in my table associated with that one result. ***Freelistings*** id company_name location postcode 1 Dave's Storage London PO Box1 2 Steve's Trucks Birmingham WS12 3 3 Sue's Tyres Manchester M14 6 4 Paul's Birmingham WS11 7 ***Basicpackge*** id company_name location postcode 1 Storage Ltd London LN12 5TW 2 Fran's Grit Birmingham WS5 37 3 Raj Walls Manchester M14 60 4 Paul's Light's Birmingham WS12 17 ***Premiumuser*** id company_name location postcode 1 Name1 Location1 PC1 2 Name2 Location2 PC2 3 Name3 Location3 PC3 4 Name4 Location4 PC4 I'm sure this can be done, i'm just getting stuck on how to treat the user's search result. Currently it's called '%$var%'. Can i use: $query = "(SELECT company_name, location FROM freelistings WHERE company_name, location like '%$var%') UNION (SELECT company_name, location FROM basicpackage WHERE company_name, location like '%$var%') UNION (SELECT company_name, location FROM premiumuser WHERE company_name, location like '%$var%') ORDER BY postcode"; So that would be looking at all three tables, getting the details from each field that is associated with that (result) postcode. I hope i have given enough information to explain my problem i'm having. for me quite a tough one? any help would be appreciated, thanks guys in advance! I just did a massive hardware/software upgrade - first new computer (a Mac) in 6 1/2 years, first MAMP upgrade in several years, etc. I'm now using these software versions: Apache 2.2.21, PHP 5.3.6, MySQL 5.5.9 Just about everything seems to be working fine except my database connections. I appear to be making a simple mistake, but I can't figure out the solution. I've checked out all sorts of online help pages, but I'm only getting more confused. Rather than ask what the problem is, it might be better to go back to square one and ask how you would write this query from scratch. I pasted my current, somewhat convoluted code below, to give you an idea of how I was doing it. But here's what I'm trying to do: Imagine a website with a static page at World: MySite/World. It displays dynamic pages, like these: MySite/World/France, MySite/World/Spain...just as long as "France" or "Spain" match values stored in a database table (gw_geog). In my query, I represent France, Spain and other place names with the variable $MyURL. So I want to query that database table. If $MyURL = 'Japan' (e.g. MySite/World/Japan), then $result should = 1, fetching a web page. If $result = 0 (e.g. a misspelled URL, like MySite/World/Japax), then it fetches a 404 error page. If $result = 2 or more (e.g. MySite/World/Georgia - the name of a country and a U.S. state), then it fetches a duplicate note. I wrote this script years ago, when I didn't know much about PHP/MySQL (I still don't), and I've upgraded my software, so I need to start fresh. Can anyone tell me how YOU would a script to accomplish this task? Thanks. * * * * * Code: [Select] <?php // DATABASE CONNECTION $conn = mysql_connect("localhost", "Username", "Password"); if (!$conn) { echo "Unable to connect to DB: " . mysql_error(); exit; } if (!mysql_select_db("db_general")) { echo "Unable to select db_general: " . mysql_error(); exit; } // DATABASE QUERY $sql = " SELECT COUNT(URL) FROM gw_geog WHERE URL = '$MyURL' "; $sql_result = mysql_query($sql,$conn); $result = $sql_result; // DO SOMETHING WITH THE RESULTS switch ($result) { case 1: echo "\n"; include_once($BaseINC."/$MyPHP/inc/B/DB2/Child/World.php"); include_once($BaseINC."/$MyPHP/inc/B/Child2.php"); include_once($BaseINC."/$MyPHP/inc/D/Child.php"); echo "\n"; break; case 0: include_once($BaseINC."/404.php"); break; default: // More than one results, as in Georgia (state and republic) include_once($_SERVER['DOCUMENT_ROOT']."/Dupe.php"); echo "\n"; break; } ?> |
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