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PHP - Unable To Jump To Row Zero
I get the unable to jump to row zero mysql error.
Code: [Select] function is_admin($uid, $cid) { $uid = (int)$uid; $cid = (int)$cid; $sql = "SELECT `users`.`id` AS `uid`, `companies`.`companyid` AS `cid`, `companies`.`adminid` AS `aid` FROM `companies` LEFT JOIN `users` ON `users`.`id` = companies.adminid WHERE `users`.`id` = {$uid} AND `companies`.`companyid` = {$cid}"; $user = mysql_query($sql); return (mysql_result($user, 0) == '1') ? true : false; } Similar TutorialsSo I'm basically done an assignment, however I'm having the biggest brainfart at the moment. For the life of me I cannot remember how to "jump line" in a file and have to put "end" at the end of my txt files. Also I have to put a break in my code which messes up my formatting:
$coursedata = explode('
', rtrim($coursedata));
$studentdata = explode('
', rtrim($studentdata));
fwrite($fp, $number.'||'.$_POST['course'].' '.$content);Here is the main code below for the script:
<html>
<head>
<?php
$success = 0;
$number_of_registered = 0;
$course_array = getdatafile('courses.txt');
$registered_array = getdatafile('register.txt');
#______________global variables____________________
function getdatafile($file) {
# code...
$coursedata = file_get_contents($file);
$coursedata = explode('
', rtrim($coursedata));
unset($coursedata[count($coursedata) - 1]);
$course_array = array();
foreach($coursedata as $row) {
$course_array[] = explode('||', $row);
}
return $course_array;
}
#_____________functions_______________________
if (isset ($_POST['name']) and isset ($_POST['number']) and isset ($_POST['course'])) {
#________________test if the post exists
$name = $_POST['name'];
$number = $_POST['number'];
$found = 0;
$findregistered = 0;
for ($i = 0; $i < sizeof($registered_array); $i++) {
if ($registered_array[$i][0] == $_POST['number'] && $registered_array[$i][1] == $_POST['course']) {
# ___________test if the student is enrolled in this course
$findregistered = 1;
}
if($registered_array[$i][1] == $_POST['course']) {
#________counter of the number registered in this course
$number_of_registered++;
}
}
for ($i = 0; $i < sizeof($course_array); $i++) {
if ($course_array[$i][1] == $_POST['course']) {
# ______________this test if the number of registered is the max
$course_name = $course_array[$i][0];
$findcourse = 1;
if($number_of_registered < $course_array[$i][2]) {
$number_max = 0;
}
else {
$number_max = 1;
}
}
}
$studentdata = file_get_contents('students.txt');
$studentdata = explode('
', rtrim($studentdata));
for ($i = 0; $i < sizeof($studentdata); $i++) {
list($studentname, $studentnumber) = explode("||", rtrim($studentdata[$i]));
if (strtoupper($name) == strtoupper($studentname) && strtoupper($number) == strtoupper($studentnumber)) {
$found = 1;
if ($findcourse == 1 && $findregistered == 0 && $number_max == 0) {
# _____________________do all the tests to record the student in the course("register.txt") and do it if all the test are checked
$success = 1;
$content = file_get_contents('register.txt');
$fp = fopen('register.txt', 'w');
fwrite($fp, $number.'||'.$_POST['course'].'
'.$content);
fclose($fp);
echo "<title>".strtoupper($name).",".strtoupper($studentname)."</title>";
}
}
}
}
?>
<style>
body, p{padding: 0;margin: 0;}
</style>
</head>
<body>
<form action="index.php" method="POST">
<?php
if (isset ($_POST['course'])) {
$course = htmlentities ($_POST['course']);
# ___________________print the information of rgester in the screen for
if ($success == 1) {
echo "You have successfully enrolled in $course_name, $course!";
}
if($found == 0) {
echo 'The credentials you have entered are incorrect, please try again.';
}
else {
if($findregistered == 1) {
echo 'You have already enrolled in this course.';
}
else {
if($number_max == 1) {
echo "Sorry $course_name, $course is full, please select another course.";
}
}
}
}
?>
<p>Courses</p>
<p><input name = "name" placeholder = "Student Name" type = "text"></p>
<p><input name = "number" placeholder = "Student Number" type = "text"></p>
<p><select name = "course" size = "1">
<option id = "select">-- Select Course --</option>
<?php
for ($i = 0; $i <sizeof($course_array); $i++) {
$number_of_registered1 = 0;
for ($j = 0; $j < sizeof($registered_array); $j++) {
if($registered_array[$j][1] == $course_array[$i][1]) {
$number_of_registered1++;
}
}
$a = '<option value = "'.$course_array[$i][1].'">'.$course_array[$i][0].' registered:'.$number_of_registered1.'</option>';
echo $a;
}
?>
</select></p>
<p><input type = "submit" value = "Send"><input type = "reset" value = "Clear"></p>
</form>
</body>
</html>
Attached Files
courses.txt 94bytes
0 downloads
register.txt 3bytes
0 downloads
students.txt 103bytes
0 downloads
index.php 4.19KB
0 downloadsHow do I execute display.php if condition is true? I have only included relevant code. $query = "SELECT * FROM airplanes WHERE ama='$ama'"; if (mysql_num_rows($result) == 0) { //print error message if (mysql_num_rows($result) == 1) { // execute display.php. How do I do this? I do plan on changing the 2 ifs to if/else
For example with an IDE to run a program is there a way to put some code in the path and let it tell me where it came from like a referral?
$ref = $_SERVER['HTTP_REFERER'];
echo("$ref<br>";
Result: https://www.mydomain.com/wp-admin/admin.php?page=lambo As you can clearly see the page that we have entered the code in is referred by the URL here. It tells me that it was last at a word press page /wp-admin/admin-php and I can go to that page and it will be the last page it was at.. So bottom line then if i add this code to the admin page and click the same button i had before.. it will tell me the file that called out the admin page. Baslcaly I am looking for ideas on how to track the page i am now all the way back to the page with the button on it and all intermediate pages that it passes thru to get to the final destination. It seems that to succesfully troubleshoot a fault is to follow the path of the entire job and see if any breaks... Any thoughts or suggestions on how to best do this would greatly be appreciated.
Code: [Select] <?php function checking_out() { $conn = db_connect(); $nickname=$_SESSION['valid_user']; $query="select sum(price) from preorders where name='".$nickname."'"; $result = $conn->query($query); if ($result) { echo '<h1>'.$result.'</h1>'; } } ?> This is not working, there is no result in the browser, any idea ? Hi all,
I am not sure why my header is not displaying the header image after using the CSS
I have a png file that repeats horizotally.
Please help
<!DOCTYPE html> <html> <head> <meta charset="UTF-8"> <title>Wikigets The online store</title> <style type="text/css"> body { margin:0 px;} #pageTop { background: url(style/headerline1.png); height:110 px; } </style> </head> <body> <div id="pageTop"> </div> <div id="pageMiddle"></div> <div id="pageBottom"></div> </body> </html> headerline1.png 2.82KB 0 downloads headerline1.png 2.82KB 0 downloads Hello, I've never really used the update command before for mysql and I'm attempting to use it and struggling a little bit. I'm trying to use mysqli prepared statements.. here's the code that I have thus far: if($query = $database->connection->prepare("UPDATE videos SET comments=?, views=?, uploader=? WHERE title = ?")) { $query->bind_param('iiss', $comments, $views, $uploader, $title); $query->execute(); $result = $query->affected_rows; $query->close(); } For some reason I cannot get this working. I have created a modification page for the administrators to be able change any of the values and wanting to update the database to reflect the changes. When using the MySQL UPDATE command do all of the values have to get changed or modified, or am I able to pass back some of the same values? Like with the above code.. if I only wanted to update the views, would I still be able to just pass in the same values for comments and uploader and it would just replace the values? Unable to execute query (SELECT * FROM lb-players) in the database : You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '-players' at line 1 I don't know why i am getting this error. All i can think of that is taking away lb from "lb-players" when you add a " - "? Code: [Select] [m]<?php $conn = mysql_connect("23.23.23.23", "unknown", "itsAsecertxD");//ignore this xD if (!$conn) { echo "Unable to connect to the database : " . mysql_error(); exit; } if (!mysql_select_db("minecraft-rc")) { echo "Unable to select database mydbname : " . mysql_error(); exit; } $sql = 'SELECT * FROM lb-players'; $result = mysql_query($sql); if (!$result) { echo "Unable to execute query ($sql) in the database : " . mysql_error(); exit; } if (mysql_num_rows($result) == 0) { echo "No rows found, nothing to display."; exit; } while ($row = mysql_fetch_assoc($result)) { echo $row["playername"]; } mysql_free_result($result); ?>[/m] tell me if i posted this in the wrong place its been forever since i last been on this site :$ I have written my index.php script where after verifying the password the script displays the user's name. But the user name is not displayed by it. Here is my script Hi guys, Currently my result displayed a 'comma' at the end Lower Primary English, Math, Science, Chinese, How do I remove the comma which is bold in red? Thank you so much My code Code: [Select] while($row3 = mysqli_fetch_array($data3)) { if ($row3['level_id'] == 2) { echo ''.$row3['subject_name'].', '; } } Hello, I'm working with Zend Framework on Linux, and I'm trying to generate a CAPTCHA using Zend_Form_Element_Captcha. Whenever the CAPTCHA page loads I get this error: [12-Jan-2011 18:14:54] PHP Warning: imagepng() [<a href='function.imagepng'>function.imagepng</a>]: Unable to open '/var/www/square/application/../public/captcha/ebf44d292149b3ebda05571c54c463a8.png' for writing: Permission denied in /usr/local/zend/share/ZendFramework/library/Zend/Captcha/Image.php on line 563 Here's my code for generating the CAPTCHA: // create captcha $captcha = new Zend_Form_Element_Captcha('captcha', array( 'captcha' => array( 'captcha' => 'Image', 'wordLen' => 6, 'timeout' => 300, 'width' => 300, 'height' => 100, 'imgUrl' => '/captcha', 'imgDir' => APPLICATION_PATH . '/../public/captcha', 'font' => APPLICATION_PATH . '/../public/fonts/LiberationSansRegular.ttf', ) )); I've checked permissions, and all directories mentioned above are accessible to root. Has anyone had a similar problem or have an idea how I can fix this? Kind Regards, Mike Hello....I am using ajax,jquery with php to insert data in db.....but the script is not working. form.html: Code: [Select] <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Untitled Document</title> <style type="text/css"> label{ display:block; } </style> <script type="text/javascript" src="jquery-1.5.min.js"> </script> <script type="text/javascript"> $(function() { $('#submit').click(function(){ $('#container').append('<img src="ajax-loader.gif" id="loading" alt="image" />'); var name=$('#name').val(); var email=$('#email').val(); var d=$('#d').val(); var m=$('#m').val(); var y=$('#y').val(); var add=$('#add').val(); var phone=$('#phone').val(); $.ajax({ url: 'process.php', type: 'POST', data: 'name=' + name + '&email=' + email + '&d=' + d + '&m=' + m + '&y=' + y + '&add=' + add + '&phone=' + phone, success: function(result){ $('#response').remove(); $('#container').append('<p id="response">' + result + '</p>'); $('#loading').fadeOut(500,function(){ $(this).remove(); }); } }); return false; }); }); </script> </head> <body> <h2>User Registeration</h2> <form action="process.php" method="POST"> <div id="container"> Name:<br> <input type="text" name="name" id="name" /><br> Email:<br> <input type="text" name="email" id="email" /><br> Date Of Birth:<br> <input type="text" id="d" name="d" size="2" /> <input type="text" id="m" name="m" size="2" /> <input type="text" id="y" name="y" size="4" /><br> Address:<br> <input type="text" id="add" name="add" /><br> Phone:<br> <input type="text" id="phone" name="phone" /><br> <input type="submit" name="submit" id="submit" value="GO!" /> </div> </form> </body> </html> process.php My db id named "jquery" and table is "tab" Code: [Select] <?php $conns=mysql_connect("localhost","root",""); if(!$conns) echo "error in connection"; mysql_select_db("jquery", $conns); $name=$_POST['name']; $email=$_POST['email']; $m=$_POST['m']; $d=$_POST['d']; $y=$_POST['y']; $add=$_POST['add']; $phone=$_POST['phone']; echo $name,"<br>"; echo $email,"<br>"; echo $m, "<br>"; echo $d,"<br>"; echo $y,"<br>"; echo $add, "<br>"; echo $phone,"<br>"; $query="INSERT INTO tab (name,email,d,m,y,add,phone) VALUES ('$name',$email','$d','$m','$y','$add','$phone')"; if(!mysql_query($query,$conns)) echo "Error"; else echo "DATA inserted"; ?> The output shown is: pulkit pulkit@gmail.com 1 26 1991 lucknow 987576787 Error For some values i entered. I wrote a very simple web form that allows my user to view text files from within their Internet browser. Occasionally, the search criteria entered returns more than one file. So I want to implement a feature whereby the text files returned by the search are compressed into a ZIP. I got a prototype working but it only compresses the first file. The second or third files are ignored. Here's my code Code: [Select] <HTML><body><form name="myform" method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>"> <fieldset><label for="DBRIDs">RIDs</label><input type="text" id="DBRIDs" name="DBRIDs" > </fieldset></form></body></HTML> <?php function check_search() { if (isset($_POST['submit'])) {if (!empty($_POST['DBRIDs'])) { $results = getFiles(); } } else $errors = "Please enter something before you hit SUBMIT."; return Array($results, $errors); } function getFiles() { $result = null; $ZIPresult = null; if (empty($_POST['DBRIDs'])) { return null; } $mydir = MYDIR; $dir = opendir($mydir); $DBRIDs = $_POST['DBRIDs']; $getfilename = mysql_query("select filename from search_table where rid in (" . $DBRIDs . ")") or die(mysql_error()); while ($row = mysql_fetch_array($getfilename)) { $filename = $row['filename']; $result .= '<tr><td><a href="' . basename($mydir) . '/' . $filename . '" target="_blank">' . $filename . '</a></td></tr>'; $ZIPresult .= basename($mydir) . '/' . $filename; } if ($result) { $result = "<table><tbody><tr><td>Search Results.</td></tr> $result</table>"; shell_exec("zip -9 SearchResult.zip ". $ZIPresult ." > /dev/null "); } return $result; } The hyperlinks pointing to the file(s) are generated just fine. The ZIP file however only contains the first file listed in the results. How can I get the ZIP file to capture ALL the files returned?? Thanks for your input. **PS: The new ZipArchive() library/class is not available on our production environment so I chose to use the Unix utility ZIP instead.** Hello. I am coding a remember me feature. Everything is working, except i am being logged in using cookies even when i want to use a session. To login using a cookie i must select the checkbox, for sessions i must leave it blank. Here is my code, if someone could spot a mistake i would be really grateful. Login page Code: [Select] <?php ob_start(); // starting session... session_start(); // requiring connection... require("functions.php"); // assigning variables... $username = mysql_real_escape_string($_POST['username']); $password = mysql_real_escape_string($_POST['password']); $submit = mysql_real_escape_string($_POST['submit']); $rememberme = $_POST['rememberme']; // querying database... $query = mysql_query("SELECT * FROM users WHERE username = '$username'"); $numrows = mysql_num_rows($query); if ($numrows != 0) { while ($row = mysql_fetch_assoc($query)) { $db_username = $row['username']; $db_password = $row['password']; } } // verifying login details... if ($submit) { if (!empty($username) && !empty($password)) { if ($username == $db_username && $password == $db_password) { if ($rememberme = "on") { setcookie("username", $username, time() + 7200); header('Location: tablets.php'); } else { $_SESSION['username'] = $db_username; $url = $_SESSION['origin'] ? $_SESSION['origin'] : "main.php"; unset($_SESSION['origin']); header('Location: ' . $url); exit; } } else { echo "Incorrect login details"; } } else { echo "You must type in username and password"; } } ob_end_flush(); ?> Login form Code: [Select] <?php session_start(); require("connect.php"); if (isset($_SESSION['username']) || isset($_COOKIE['username'])) { header('Location: main.php'); } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en"> <head> <title>Login</title> <link rel="stylesheet" type="text/css" href="form.css" /> </head> <body> <form method="post" action="login.php"> <div class="box"> <h1>Login</h1> <label> <span>Username</span> <input type="text" class="input_text" name="username" id="name" /> </label> <label> <span>Password</span> <input type="password" class="input_text" name="password" id="password" /> </label> <input type="checkbox" name="rememberme" /> <label> <input type="submit" class="button" value="Login" name="submit" /> </label> </div> </form> </body> </html> Hey guys, After my php script a HTML code follows. But i can't edit the HTML code because it isn't recognised. It does show up, however i can't edit it. Problem: Unable to edit HTML code below php scripts Code: <?php function createNewFile($name,$mail,$subject,$comments,$count,$date,$other="",$up="0") { global $settings; $header=implode('',file('header.txt')); $footer=implode('',file('footer.txt')); $content=' ?> <!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> <html> <head> <title>'.$subject.'</title> <meta content="text/html; charset=windows-1250"> Thanks in advance HI, When I trying to covnert 3gp to FLV I got this error: FFmpeg version SVN-r12665, Copyright (c) 2000-2008 Fabrice Bellard, et al. configuration: --enable-gpl --enable-postproc --enable-swscale --enable-avfilt er-lavf --enable-pthreads --enable-liba52 --enable-avisynth --enable-libfaac --e nable-libfaad --enable-libgsm --enable-memalign-hack --enable-libmp3lame --enabl e-libnut --enable-libtheora --enable-libvorbis --enable-libx264 --enable-libxvid --cpu=i686 --extra-ldflags=-static libavutil version: 49.6.0 libavcodec version: 51.54.0 libavformat version: 52.13.0 libavdevice version: 52.0.0 built on Apr 2 2008 22:35:11, gcc: 4.2.3 Seems stream 0 codec frame rate differs from container frame rate: 29.97 (30000/ 1001) -> 15.00 (15/1) Input #0, mov,mp4,m4a,3gp,3g2,mj2, from 'uploads/v_1.3gp': Duration: 00:01:03.6, start: 0.000000, bitrate: 189 kb/s Stream #0.0(und): Video: h263, yuv420p, 176x144 [PAR 12:11 DAR 4:3], 15.00 t b(r) Stream #0.1(und): Audio: samr / 0x726D6173, 8000 Hz, mono WARNING: The bitrate parameter is set too low. It takes bits/s as argument, not kbits/s Output #0, flv, to 'uploads/a.flv': Stream #0.0(und): Video: flv, yuv420p, 320x240 [PAR 1:1 DAR 4:3], q=2-31, 20 0 kb/s, 15.00 tb(c) Stream #0.1(und): Audio: libmp3lame, 22050 Hz, mono, 0 kb/s Stream mapping: Stream #0.0 -> #0.0 Stream #0.1 -> #0.1 Unsupported codec (id=73728) for input stream #0.1 What should I do? Thanks First of all, here's what i want it to look:' I just want that the related image must align towards left, and its information is just in front of it. But, what my code does is place the image on above line, and the information on the link below. Here's how it looks with my code. I will be able to make the changes with the File Information. The PHP which is being used is: 1. To fetch if preview is available: Code: [Select] $pre = ''; if ($prew==0) { if ($ext == 'bmp') $pre = 'Impossible Preview <br>'; if ($ext == 'gif' or $ext == 'jpeg' or $ext == 'jpg' or $ext == 'png' or $ext == 'JPG' or $ext == 'GIF' or $ext == 'PNG'or $ext == 'JPEG') $pre = '<img style="align:left;margin: 1px;" src="im.php?bab=1&id='.$file_info['id'].'" alt=""/>'; }2. To insert the preview image: Code: [Select] if($pre!=NULL) echo '<div class="block">'.$pre.'</div>'; 3. And finally the code to fetch file information: Code: [Select] echo '<div class="fileName"><a href="load.php?id='.$file_info[id].'"><font color="red">'.$file_info['name'].''.$extension.'</font></a>|'; if($ext =='txt') { echo '<a href="read.php?id='.$file_info['id'].'&id2='.$id.'"><font color="red">Read</font></a>';} echo $new_info.''; if(!empty($file_info['fastabout'])) echo str_replace("\n", '<br>',$file_info['fastabout']); echo '</div>'; echo '<tr><div class="t_block">'.$ico.'<a href="view.php?id='.$file_info[id].'"><strong>File Info</strong></a></div></tr></td>'; I just can't figure out what to insert with the code. I think it must be some table formatting (TR/TD), but because I'm a noob with PHP, I failed in all my attempts. Please, if anyone could help me out! Hi All,
I have this issue when upload a file using an uploader i made. For the life of me I cant figure out why it won't write the file.
Error:
[Sat Jun 21 20:45:40 2014] [error] [client xxxxxxx] AH01215: PHP Warning: move_uploaded_file() [<a href='function.move-uploaded-file'>function.move-uploaded-file</a>]: Unable to move '/tmp/phpUsCyXG' to '/home/sites/xxxxxx.co.uk/public_html/yourphotos/uploads/' in /home/sites/xxxxxxx.co.uk/public_html/yourphotos/index.php on line 31My PHP
<?php
//if they DID upload a file...
$message = '';
if($_FILES['photo']['name'])
{
$valid_file = true;
//if no errors...
if(!$_FILES['photo']['error'])
{
//now is the time to modify the future file name and validate the file
$new_file_name = strtolower($_FILES['photo']['tmp_name']); //rename file
if($_FILES['photo']['size'] > (26214400)) //can't be larger than 25MB
{
$valid_file = false;
$message = 'Oops! Your file\'s size is to large.';
echo $_FILES['photo']['size'];
}
if($_FILES['photo']['size'] < (1572864)) //can't be smaller than 1.5MB
{
$valid_file = false;
$message = 'Oops! Your file\'s size is to small.';
echo $_FILES['photo']['size'];
}
//if the file has passed the test
if($valid_file)
{
//move it to where we want it to be
move_uploaded_file($_FILES['photo']['tmp_name'], '/home/sites/xxxxxxxx.co.uk/public_html/yourphotos/uploads/');
$message = 'Congratulations! Your file was accepted.';
}
}
//if there is an error...
else
{
//set that to be the returned message
$message = 'Ooops! Your upload triggered the following error: '.$_FILES['photo']['error'];
}
}
?>
<html>
<body>
<form action="index.php" method="post" enctype="multipart/form-data">
Your Photo: <input type="file" name="photo" size="25" />
<input type="submit" name="submit" value="Submit" />
<?PHP echo $message; ?>
</form>
</body>
</html>
I have set the uploads file to have write permissions as well.
Sam
Edited by samtwilliams, 21 June 2014 - 02:51 PM. Hi, I've a simple code which uses a class from here. When I execute it, I get this error QuoteFatal error: Uncaught Error: Class 'phpMQTT' not found in C:\Apache24\htdocs\mqtt\test.php:10 Stack trace: #0 {main} thrown in C:\Apache24\htdocs\mqtt\test.php on line 10 and I cannot understand why my code doesn't see the class. Both files, the class.php and test.php, are in the same folder. Any help is appreciated.
TIA
<?php
require("phpMQTT.php");
$server = "192.168.0.250"; // change if necessary
$port = 1883; // change if necessary
$username = ""; // set your username
$password = ""; // set your password
$client_id = "phpMQTT-publisher"; // make sure this is unique for connecting to sever - you could use uniqid()
$mqtt = new phpMQTT($server, $port, $client_id);
if ($mqtt->connect(true, NULL, $username, $password)) {
echo 'hereeeee';
$mqtt->publish("room/lamp", "on" . date("r"), 0);
$mqtt->close();
} else {
echo "Time out!\n";
}
?>
Hallo everyone... here is my following code. i am fetching the picture from the file but unable to fetch it.
kindly help me. error in bold underline...
Thanks
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Untitled Document</title>
</head>
<body>
<?php
session_start();
$uid='';
include("connection.php");
$query=mysql_query("SELECT * FROM users WHERE uid='".$_GET['uid']."'");
$row=mysql_fetch_array($query);
$image=$row['images_path'];
?>
<form action="edit.php" method="post">
<table border="1px" align="center" width="85%">
<tr>
<td colspan="3"><?php include'header.php' ?></td>
</tr>
<input type="hidden" name="uid" value="<?=$_GET['uid'];?>" />
<tr>
<td colspan="3">Welcome <?php echo $_GET['uid']; ?> </td>
</tr>
<tr height="30px;">
<td width="37%" style="padding-left:250PX;">User Name</td>
<td width="63%"> <input type="text" name="uname" value="<?php echo $row['uname']?>" /></td>
</tr>
<tr height="30px;">
<td width="37%" style="padding-left:250PX;">User Email</td>
<td width="63%"> <input type="text" name="email" value="<?php echo $row['email'];?>" /></td>
</tr>
<tr height="30px;">
<td width="37%" style="padding-left:250PX;">Mobile</td>
<td width="63%"> <input type="text" name="mob" value="<?php echo $row['mob'];?>" /></td>
</tr>
<tr height="30px;">
<td width="37%" style="padding-left:250PX;">Type</td>
<td width="63%"> <input type="text" name="type" value="<?php echo $row['type'];?>" /></td>
</tr>
<tr height="30px;">
<td width="37%" style="padding-left:250PX;">Status</td>
<td width="63%"> <input type="text"name="status" value="<?php echo $row['status'];?>" /></td>
</tr>
<tr height="30px;">
<td width="37%" style="padding-left:250PX;">Picture</td>
<td><?php echo "<img border=\"0\" src=\"".$row['images_path']."\" width=\"102\" alt=\"your name\" height=\"91\">"; ?></td>
</tr>
</tr>
<tr height="50px;">
<td colspan="3" style="padding-left:350px;"><input type="submit" name="submit" /> <input type="reset" value="Clear" /></td>
</tr>
<tr>
<td colspan="3"><?php include'footer.php' ?></td>
</tr>
</table>
<div align="left"><a href="home_users.php">Home</a></div>
</form>
</body>
</html>
Hi All. New at this, form is working perfect, then added recapthca, now it only handles recaptcha and not the form. not sure where the error is
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