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PHP - All Image Information Gets Stored In The Database Except For The Last 2
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 Similar TutorialsI 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. Okay, so I have a database with user log on info, and unique ID's. How to I allow the user to save info from a form, and be able to log out and come back and log on to see/edit that info. Thanks! 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 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 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 $row['word_document']; any ideas would be really appreciated. I am writting a php function that uses mysql to get user data - pretty common, right Well, my issue is that I need to run a check in my file system. Users profile pictures are stored in my image directory as .png's. I need to have my function check that directory and if an image matches their id, then return their information. I only want the user data if they have an image uploaded. Here is my current function: Code: [Select] function fetch_users_login($limit) { $limit = $limit(int); $sql = "SELECT `users`.`id`, `users`.`firstname`, `users`.`lastname`, `users`.`username`, `user_privacy`.`avatar` FROM `users` LEFT JOIN `user_privacy` ON `users`.`id` = `user_privacy`.`uid` WHERE `users`.`status` > 2 AND `user_privacy`.`avatar` = 1 ORDER BY `users`.`id` DESC LIMIT 0, {$limit}"; $result = mysql_query($sql) or die(mysql_error()); $users = array(); $i = 0; while(($row = mysql_fetch_assoc($result)) !== false) { $users[$i] = array( 'id' => $row['id'], 'firstname' => $row['firstname'], 'lastname' => $row['lastname'], ); $users[$i]['avatar'] = getUserAvatar($row['username']); $i++; } return $users; } Hi
I am winning I think
I have got the records displayed for the current user logged in so basically they can only see their submitted listings and just working on the edit of them so the current user logged in can update their listing and is all working apart from the update of the images
the images are stored by the file name on the database and then gets moved onto the server so the actual images are not stored on the database only the file names are and the images are moved onto the server, hope that makes sense
what I can't do at the mo is work out how to update the file names of the images on the database and update on the server
I get the following error
Notice: Undefined index: photo on line 209 and line 211
the coding for them lines are below
$pic1= basename($_FILES['photo']['name'][0]); $pic2= basename($_FILES['photo']['name'][1]);The rest of the coding below that is
if(!empty($_FILES['photo']['tmp_name']))
{
// Number of uploaded files
$num_files = count($_FILES['photo']['tmp_name']);
/** loop through the array of files ***/
for($i=0; $i < $num_files;$i++)
{
// check if there is a file in the array
if(!is_uploaded_file($_FILES['photo']['tmp_name'][$i]))
{
$messages[] = 'No file uploaded';
}
else
{
// move the file to the specified dir
if(move_uploaded_file($_FILES['photo']['tmp_name'][$i],$target.'/'.$_FILES['photo']['name'][$i]))
{
$messages[] = $_FILES['photo']['name'][$i].' uploaded';
}
else
{
// an error message
$messages[] = 'Uploading '.$_FILES['photo']['name'][$i].' Failed';
}
}
}
The update query is below
// save the data to the database
mysql_query("UPDATE privatelistings SET listingtitle='$listingtitle', make='$model', model='$model', exteriorcolour='$exteriorcolour', enginesize='$enginesize', fueltype='$fueltype', yearregistered='$yearregistered', transmission='$transmission', mileage='$mileage', nodoors='$nodoors', bodystyle='$bodystyle', price='$price', photo='$pic1', photo1='$pic2' WHERE id='$id'")
or die(mysql_error());
I am going to be updating to mysqli just want to get it working first
My HTML form coding is below
<form action="" method="post" enctype="multipart/form-data"> <input type="hidden" name="id" value="<?php echo $id; ?>"/> <div> <strong>Listing Title: *</strong> <input type="text" name="listingtitle" value="<?php echo $listingtitle; ?>"/> <br/> <strong>Make: *</strong> <input type="text" name="make" value="<?php echo $make; ?>"/> <br/> <strong>Model: *</strong> <input type="text" name="model" value="<?php echo $model; ?>"/> <br/> <strong>Exterior Colour: *</strong> <input type="text" name="exteriorcolour" value="<?php echo $exteriorcolour; ?>"/> <br/> <strong>Engine Size: *</strong> <input type="text" name="enginesize" value="<?php echo $enginesize; ?>"/> <br/> <strong>Fuel Type: *</strong> <input type="text" name="fueltype" value="<?php echo $fueltype; ?>"/> <br/> <strong>Year Registered: *</strong> <input type="text" name="yearregistered" value="<?php echo $yearregistered; ?>"/> <br/> <strong>Transmission: *</strong> <input type="text" name="transmission" value="<?php echo $transmission; ?>"/> <br/> <strong>Mileage: *</strong> <input type="text" name="mileage" value="<?php echo $mileage; ?>"/> <br/> <strong>Number of Doors: *</strong> <input type="text" name="nodoors" value="<?php echo $nodoors; ?>"/> <br/> <strong>Body Style: *</strong> <input type="text" name="bodystyle" value="<?php echo $bodystyle; ?>"/> <br/> <strong>Price: *</strong> <input type="text" name="price" value="<?php echo $price; ?>"/> <br/> <strong>Photo One:</strong> <input type='hidden' name='size' value='350000'><input type='file' name='photo[]'> <br> <strong>Photo Two:</strong> <input type='hidden' name='size' value='350000'><input type='file' name='photo[]'> <br> <input type="submit" name="submit" value="Submit"> </div> </form>sorry was not sure what other I need to show, so you guys get the idea? I am having trouble pulling a youtube embedded code from my database. Everything else comes out fine however it just doesnt pull anything out where the embedded code is supposed to be. Any ideas I put the code below. All help would be greatly appreciated, also if anyone is feeling generous and would like to help me some more please message me I have a couple other small questions. Code: [Select] <?php //open database $connect = mysql_connect("Database","name","password") or die("Not connected"); mysql_select_db("database") or die("could not log in"); $query = "SELECT * FROM boox ORDER BY date DESC"; $result = mysql_query($query); // Get the page number, if none is set - it is 0 if( isset($_GET['page']) ) { $page =$_GET['page']; } else { $page = 0; } $resultsPerPage = 15; $num = mysql_num_rows($result); // amount of rows $loops = $page*$resultsPerPage; // starting loops at.. while ($loops < $num && $loops < ($page+1)*$resultsPerPage ) { $link = mysql_result($result,$loops,"link"); // get result from the 'Title' field in the table $username = mysql_result($result,$loops,"username"); // get result from the 'Content' field in the table $messsage = mysql_result($result,$loops,"message"); $date = mysql_result($result,$loops,"date"); if ($pagelimit == 0) { $pagelimit == 1; } if ($pagelimit <= 15) // echo stuff here $loopz = $loops + 1; echo "   </br><align='left'><table width='297' height='900' border='1' align='center' bgcolor='#111'> <tr> <td>$loopz. </br> $message </br> Posted By: $username $date </td> </tr> </table></br><br>"; $count++ ; $pagelimit++; $loops++; } if ( $page!=0 ) // Show 'Previous' link { $page--; $prevpage = ($page + 1); echo "<br><br><br><a href='index.php?page=$page'>Previous $prevpage </a>"; $page++; } if ($loops > 5&&($page+1)*$resultPerPage < $num ) // Show 'next' link { $page++; $nextpage = ($page + 1); echo "<a href='index.php?page=$page'> Next $nextpage</a>"; } ?> I'm not sure I know how to explain what I want to allow you to understand. I'm working on a ventilation app that has different fan sizes. In the input php file I have a dropdown list that's populated by what's in the database. I'd like a results page that lists the fan size and number of each size and the cfm's each fan kicks out. the problem is that I don't know how to do anything with it other than hard code it. What I'd like, is for it to also pull all of the fan sizes and list them, that way the database can be fluid. This is the code I'd like to replace with some sort of loop. //9" fan $results = mysql_query("SELECT * FROM fan WHERE size=9", $link); while ($row = mysql_fetch_array($results)) { $capacity9 = $row['capacity']; }; //--------------------------------------------------------- //10" fan $results = mysql_query("SELECT * FROM fan WHERE size=10", $link); while ($row = mysql_fetch_array($results)) { $capacity10 = $row['capacity']; }; //--------------------------------------------------------- //14" fan $results = mysql_query("SELECT * FROM fan WHERE size=14", $link); while ($row = mysql_fetch_array($results)) { $capacity14 = $row['capacity']; }; //--------------------------------------------------------- //16" fan $results = mysql_query("SELECT * FROM fan WHERE size=16", $link); while ($row = mysql_fetch_array($results)) { $capacity16 = $row['capacity']; }; //--------------------------------------------------------- //18" fan $results = mysql_query("SELECT * FROM fan WHERE size=18", $link); while ($row = mysql_fetch_array($results)) { $capacity18 = $row['capacity']; } //--------------------------------------------------------- //20" fan $results = mysql_query("SELECT * FROM fan WHERE size=20", $link); while ($row = mysql_fetch_array($results)) { $capacity20 = $row['capacity']; }; //--------------------------------------------------------- //24" fan $results = mysql_query("SELECT * FROM fan WHERE size=24", $link); while ($row = mysql_fetch_array($results)) { $capacity24 = $row['capacity']; }; //--------------------------------------------------------- //36" fan $results = mysql_query("SELECT * FROM fan WHERE size=36", $link); while ($row = mysql_fetch_array($results)) { $capacity36 = $row['capacity']; }; Hello, I recently came upon this website and I want to implement something similar. The 'remove' box is what I would like to do with PHP. I have no idea where to start. I would like to be able to search through my database, select one of its components and add it to that window. Within that window I would also like to be able to remove it. Does anyone know where I should go from there? here is the site: http://www.sealandserpent.org/schedgen/schedulegenerator.php I have these two files and I need to solve the problem that I need the user data in the role column to be taken when logging in. I have written it so that the data can be extracted from the database, but it does not work for me that it is “forwarded” from process.php to admin.php . Please don’t know what the error is that the admin.php file doesn’t want to load $ _SESSION['Role'] from process.php ? Thanks to everyone for helping and here is the code: process.php
<?php
require_once('connect.php');
session_start();
if(isset($_POST['Login']))
{
if(empty($_POST['Username']) || empty($_POST['Password']))
{
header("location:index.php?Empty= Please Fill in the Blanks");
}
else
{
$query="select * from role_test where Username='".$_POST['Username']."' and Password='".md5($_POST['Password'])."'";
$result=mysqli_query($con,$query);
if(mysqli_fetch_assoc($result))
{
$_SESSION['User']=$_POST['Username'];
while($row = mysqli_fetch_array($result) ){
$_SESSION['Role']=$row['role'];
}
header("location:admin.php");
}
else
{
header("location:index.php?Invalid= Please Enter Correct User Name and Password ");
}
}
}
else
{
echo 'Not Working Now Guys';
}
?>
admin.php
<?php
session_start();
if(!(isset($_SESSION['User'])))
{
header("Location: index.php");
exit(0);
}
?>
<!DOCTYPE html>
<html>
<head>
<title>Role</title>
<meta charset="UTF-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
</head>
<body>
<?php
$_SESSION['Role']=$role;
echo $role;
?>
</body>
</html>
I am creating a music blogging site however the main page will only show one video the code is below anyone have any ideas? " <?php //open database $connect = mysql_connect("******","username","password") or die("Not connected"); mysql_select_db("collegebooxboox") or die("could not log in"); $query = "SELECT * FROM boox ORDER BY date DESC"; $result = mysql_query($query); // Get the page number, if none is set - it is 0 if( isset($_GET['page']) ) { $page =$_GET['page']; } else { $page = 0; } $resultsPerPage = 15; $num = mysql_num_rows($result); // amount of rows $loops = $page*$resultsPerPage; // starting loops at.. while ($loops < $num && $loops < ($page+1)*$resultsPerPage ) { $link = mysql_result($result,$loops,"link"); // get result from the 'Title' field in the table $username = mysql_result($result,$loops,"username"); // get result from the 'Content' field in the table $messsage = mysql_result($result,$loops,"message"); $date = mysql_result($result,$loops,"date"); if ($pagelimit == 0) { $pagelimit == 1; } if ($pagelimit <= 15) // echo stuff here $loopz = $loops + 1; echo "   </br><align='left'><table width='297' height='900' border='1' align='center' bgcolor='#111'> <tr> <td>$loopz. $link </br> $message </br> Posted By: $username $date </td> </tr> </table></br><br>"; $count++ ; $pagelimit++; $loops++; } if ( $page!=0 ) // Show 'Previous' link { $page--; $prevpage = ($page + 1); echo "<br><br><br><a href='index.php?page=$page'>Previous $prevpage </a>"; $page++; } if ($loops > 5&&($page+1)*$resultPerPage < $num ) // Show 'next' link { $page++; $nextpage = ($page + 1); echo "<a href='index.php?page=$page'> Next $nextpage</a>"; } ?> " |
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