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PHP - Unable To Align Properly
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! Similar Tutorialshi there, the code below is suppose to display something like the attachement "code2" but instead it displays something like attachement "code1" please assist in find what is wrong with my echo lines.
<?php
$counter = 2;
$sqlq="select * from state WHERE status = 0 ";
$categorysqlq = mysql_query($sqlq);
$varq = mysql_num_rows($categorysqlq);
while($catfetchq = mysql_fetch_array($categorysqlq))
{
$cnty = $catfetchq[0];
$sqllq="select * from vehicle WHERE country = '$cnty' ";
$categorysqllq = mysql_query($sqllq);
$numsql = "select * from branchaddr WHERE state = '$cnty' ";
$numquery = mysql_query($numsql);
$varqa = mysql_num_rows($numquery);
$cntyfetchq= mysql_fetch_array($numquery);
if($varq != 0){
if($counter == 2){
echo "<tr><td><a href=\"transport2.php?id=".$cntyfetchq['state'].">".$catfetchq[1]."(<span style=\"color:red\">".$varqa."</span>)</a></td>";
$counter--;
}
else{
echo "<td><a href=\"transport2.php?id=".$cntyfetchq['state'].">".$catfetchq[1]."(<span style=\"color:red\">".$varqa."</span>)</a></td></tr>";
$counter = 2;
}
}
}
?>
Attached Files
code1.png 12.65KB
0 downloads
code2.png 25.34KB
0 downloadsCode: [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 ? 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($user, 0) == '1') ? true : false; } Hi all,
I am not sure why my header is not displaying the header image after using the CSS
I have a png file that repeats horizotally.
Please help
<!DOCTYPE html> <html> <head> <meta charset="UTF-8"> <title>Wikigets The online store</title> <style type="text/css"> body { margin:0 px;} #pageTop { background: url(style/headerline1.png); height:110 px; } </style> </head> <body> <div id="pageTop"> </div> <div id="pageMiddle"></div> <div id="pageBottom"></div> </body> </html> headerline1.png 2.82KB 0 downloads headerline1.png 2.82KB 0 downloads Hello, I've never really used the update command before for mysql and I'm attempting to use it and struggling a little bit. I'm trying to use mysqli prepared statements.. here's the code that I have thus far: if($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? 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
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>
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 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.** ==> MySQL version 5.0.91-community So once upon a time I was working on a CMS. I worked on it for 3 months, had to wipe my computer - but didn't worry about it because I backed up the files on a flash drive... which got smushed by some boxes when I moved <.< awesome -.- Anyway, a year later (ie: today), I've decided to start again from scratch. For me - for whatever reason - the simplest things always seem to give me the most trouble. I decided to start with a "user" table - naturally - and began working on the first page: user_view.php and added an include file that connects to the database (connectDB.php). Since I started today, everything is simple - no design elements yet - just good 'ol code. The View Users Page is simply a page that grabs all the rows from Table: users and prints the information to the screen, as if you were viewing a Members page on a forum. This Page works fine. No complaints, does everything I need it to, prints all the rows and cols from Table Users. Then I created the Edit User Page. When I did this a year ago, and I wanted to print out all the data from the table, I used the while loop. When I wanted to only grab certain variables from a table, I just created a new set of variables, as show below. Using the same method from a year ago, I started working on printing various data to the screen (such as "Welcome $username"). 5 minutes of typing... refresh page... nada. So I stripped away just about every tag but the essentials and it still wouldn't work Code: [Select] <?php ... $userID = 3; $sql = "SELECT * FROM users WHERE userID='$userID'"; $result = mysql_query($sql); $rows = mysql_fetch_row($result); echo $rows['username']; ... ?> Normally, this would just grab the username from a row in Table users where the userID = 3 But instead - it does nothing. I just see a blank page. It's a simple 5 lines of code and it's like "im just gonna do w.e the fk i wanna do mmmk thxforplaying" PS: Keep in mind there is nothing wrong with the connection to the database, and as already stated I'm able to print ALL values from the Users Table using this same method: Code: [Select] <?php ... $user_sql = "SELECT * FROM users ORDER BY username ASC"; $user_result = mysql_query($user_sql); while ( $user_rows = mysql_fetch_array($user_result) ) ... ?> 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> my firefox won't open up .tpl format im running apache...how can i view that format on my apache please? 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";
}
?>
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 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 In the given code below I am getting this error Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\samples\getcustomers.php on line 20 WHY? Code: [Select] <?php $value = $_GET["q"]; $con = mysql_connect('localhost', 'root', ''); //db_connect(); if (!$con) { die('Could not connect: ' . mysql_error()); } $db_selected = mysql_select_db("ajaxDb",$con); $sql = "SELECT * FROM 'cutomers' WHERE Name = '".$value."'"; // echo $sql; $result = mysql_query($sql); echo "<table border='1'> <tr> <th>Name</th> <th>Address</th> <th>Country</th> </tr>"; while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['Name'] . "</td>"; echo "<td>" . $row['Address'] . "</td>"; echo "<td>" . $row['Country'] . "</td>"; echo "</tr>"; } echo "</table>"; mysql_close($con); ?> 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 Hi all, Newbie here, i am having a problem to get my images to show which are stored in mysql database as a mediumblob. I get id number to print in table ut am just getting empty square with red cross in where my image should be. Is my code incorrect or is it something else? Appreciate your help with this. I have included both of the pages codes i am using. Thanks Tony image2.php <?php include("common.php"); error_reporting(E_ALL); $link = mysql_connect(host,username,password) or die("Could not connect: " . mysql_error()); mysql_select_db(db) or die(mysql_error()); $sql = "SELECT id FROM photos"; $result = mysql_query("$sql") or die("Invalid query: " . mysql_error()); ?> <table border="1"><tr><td>id</td><td>image</td></tr> <?php while($row=mysql_fetch_assoc($result)){ print '<tr><td>'.$row['id'].'</td><td>'; print '<img src="image1.php?id='.$row['id'].'height="75" width="100"">'; } echo '</td></tr></table>' ?> image1.php <?php ob_start(); include("common.php"); mysql_connect(host,username,password) or die(mysql_error()); mysql_select_db(db) or die(mysql_error()); $query = mysql_query("SELECT imgage FROM photos WHERE id={$_GET['image_id']}"; $row = mysql_fetch_array($query); $content = $row['image']; header('Content-type: image/jpg'); echo $content; } ob_end_flush(); ?> |
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