|
PHP - Php Die() Doesn't End Script??
I'm a complete newb at programming so please bear with me
i'm trying to check to make sure my sql query $result=mysql_query("SELECT screenname FROM users WHERE privacy='0'"); shows the correct results so i did a print_r($result); die(); but when i reload the page, it shows a syntax error after the die(); isn't the script suppose to completely stop so syntax errors after die() shouldn't matter? also i'm expecting the $result to be an array of names and i wanted to do a query to another table to grab comments from each of those screennames, is the only way to do a for loop of mysql queries? Thanks Similar Tutorials Why doesn't this code update two mysql databases (on two servers)? What am I doing wrong? //SERVER 1 $link = mysql_connect("localhost","usern1","pw1"); mysql_select_db("db_one1"); //SERVER 2 $link = mysql_connect("xxx.xxx.xx.xxx","usern2","pw2"); mysql_select_db("db_one2"); $query = "INSERT INTO db1(subject, search, News, img) VALUES('$hsubject','$key','$news','$img')"; $result = mysql_query($query); $query = "INSERT INTO db2(subject, search, News, img) VALUES('$hsubject','$key','$news','$img')"; $result = mysql_query($query); $sentOk = "The data has been added to the database."; echo "sentOk=" . $sentOk; This topic has been moved to Other Web Server Software. http://www.phpfreaks.com/forums/index.php?topic=331552.0 I have a page that shows entries in a guestbook I'm making, and below the entries there is supposed to be a form to write an entry. Except, none of the HTML after the script to show the entries shows up on the page. I have no clue what's wrong. Here is the script to show the entries. $n is name, $d is date, $s is site (optional), and $m is message. $file = fopen("posts.txt", 'rb'); flock($file, LOCK_SH); while(!feof($file)){ $entry = fgetcsv($file, 0, '|'); if(empty($entry)){exit;} $d = $entry[1]; $n = $entry[2]; $s = $entry[3]; $m = $entry[4]; echo ' <table style="border: #3399AA 1px solid;"><tr style="background: #3399AA; font: bold 10px verdana,sans-serif; color: #FFFFFF;"> <td width="170">'.$n.'</td> <td align="right" width="170">'.$d.'</td> </tr> '; if($s != 'none'){ echo ' <tr><td colspan="2" style="font: 10px verdana,sans-serif; color: #3399AA;"> <b>Site: </b><a href="'.$s.'">'.$s.'</a> <center><div style="width: 250; height: 1px; background: #3399AA; margin: 10px;"></div></center></td></tr> '; } echo ' <tr> <td colspan="2" style="font: 10px verdana,sans-serif; color: #3399AA;">'.$m.'</td> </tr></table><br> '; } flock($file, LOCK_UN); fclose($file); The file it is reading looks like this: Code: [Select] |11:51 am, 3rd Nov 2010|Memoria|none|Hello. :) |11:51 am, 3rd Nov 2010|Memoria|http://sitehere|Hello again. |11:51 am, 3rd Nov 2010|Memoria|none|How are you doing? Thanks! Hi all, I have two issue with script. 1. It works in PhpEd and with apache but doesn't work at remote server with apache. Error is well known - "Warning: Cannot modify header information - headers already sent " 2. when I added more than 20 records like $var = $_POST['var']; it stops work local. Error is same. adminka_wrapp.php Code: [Select] <?php $addFormatName = $_POST['addFormatName']; $addFormatDes = $_POST['addFormatDes']; $page = $_GET['page']; if (isset($addFormatName) && isset($addFormatDes)) { $page = 'add_formats'; } else if ($page == NULL) { $page = "user_access_log"; } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <link rel="stylesheet" type="text/css" href="../style/adminkaview.css"> </head> <body> <div class="col-wrap1"> <div class="col-wrap2"> <div class="col1"> <div class="content" id="c1"> <p><a href="?page=view_add_format">Format and extensions</a></p> </div> </div> <div class="col2"> <div class="content" id="c2"> <?php require("$page.php");?> </div> </div> <div class="clear"></div> </div> </div> </body> </html> view_add_format.php Code: [Select] <?php require_once "../function.php"; $q = connect("SELECT `formats`.* FROM `xxx`.`formats` "); echo "<form method='post' action='adminka_wrapp.php'> <table> <tr> <td>id</td> <td>Formats Name</td> <td>Description</td> </tr>"; while($rowResult = $q->fetch_assoc()) { echo "<tr>"; $id = $rowResult["id"]; $formatName = $rowResult["f_name"]; $description = $rowResult["description"]; echo "<td>".$id."</td>" ."<td>".$formatName."</td>" ."<td>".$description."</td>"; echo "</tr>"; } echo " <tr> <td> <input type='submit' value='AddFormat' name='submitAddFormat'> </td> <td> <input type='text' name='addFormatName' maxlength='20' size='5'> </td> <td> <input type='text' name='addFormatDes' maxlength='20' size='5'> </td> "; echo "</table> </form>"; ?> add_formats.php Code: [Select] <?php require "../function.php"; if (isset($addFormatName) && isset($addFormatDes) ) { $q = connect("INSERT INTO `xxx`.`formats` (`id` ,`f_name` ,`description` ) VALUE (NULL ,'$addFormatName' ,'$addFormatDes' ) "); header("Location: adminka_wrapp.php?page=view_add_format"); exit(); } ?> function.php Code: [Select] <?php function connect($query) { $db = new mysqli('127.0.0.1', 'xxx', 'xxx', 'xxx'); if (mysqli_connect_errno()) { printf("Connect failed: %s\n", mysqli_connect_error()); exit(); } $result = $db->query($query); $db->close(); return $result; } ?> Hello, I am trying to pick up php again and just exercising my skills. So I have it so that it fills my form with the values of what I want to edit, and when I click the edit button, it doesn't edit any of the information. When I echo out $result, I get a MYSQL query string that has the same values as the table, so its not getting the new values that are edited. <?php @mysql_connect('localhost', 'root', '') or die("Could not connect to Mysql Server. " . mysql_error()); @mysql_select_db('tutorials') or die("Could not connect to Database. " . mysql_error()); if(isset($_GET['edit'])) { $id = $_GET['edit']; $query = "SELECT `username`, `password` FROM `users` WHERE `id` = '$id'"; $result = mysql_query($query); $row = mysql_fetch_array($result); $name = $row['username']; $password = $row['password']; } if(isset($_POST['edit'])) { $id = $_GET['edit']; $query = "UPDATE `users` SET `username` = '$name', `password` = '$password' WHERE `id` = '$id'"; $result = mysql_query($query); echo $query; if(!$result) { echo mysql_error(); }else{ echo 'updated post'; } } ?> <form method="POST" action="" > <input type="text" name="name" value="<?php echo $name; ?>" /> First name <br /> <input type="text" name="password" value="<?php echo $password; ?>" /> Last name <br /> <input type="submit" name="edit" value="edit" /> </form> I believe it has something to do with the values of $name and $password in the form conflicting with the first if isset and the second if isset. Thanks for any help possible Hello, The following is my situation where I seem to get a 500 error code from the linux server: i have an 'index' file like this: Code: [Select] <?php require("includes/config.php"); $a = $_REQUEST['a']; switch ($a) { case "home": include("frontpage/main.php"); case "user-process": include("user-process.php"); } ?> config.php is something like this: Code: [Select] <?php require(includes/classes/session.class.php); require(includes/classes/user.class.php); require(includes/classes/db.class.php); ... ?> Now if we fall into the case "home" it works fine. Instead, if we fall into user-process it writes to the logs file Fatal Error: Class User does not exist bla bla bla. Why doesn't it exist ? every class is included in the config.php file then index.php includes first config.php ( which has all the classes) and then includes the requested page. I also have a .htaccess file which is as follows: Code: [Select] RewriteEngine On RewriteRule ^([^/\.]+)?/?([^/\.]+)?/?([^/\.]+)?/?([^/\.]+)?/?([^/\.]+)?/?([^/\.]+)?/?([^/\.]+)?$ index.php?a=$1&b=$2&c=$3&d=$4&e=$5&f=$6&g=$7 [NC,L] which is used to access in a SEO friendly way the pages that users request. Goal: To have a gallery that downloads images from the folder I previously uploaded to in a previous script. Bug: When I load the page the thumbnail comes up as broken and when I click on the thumbnail to get the bigger picture it comes up with the following error message: "Firefox doesn't know how to open this address, because the protocol (c) isn't associated with any program." <?php include 'db.inc.php'; //connect to MySQL $db = mysql_connect(MYSQL_HOST, MYSQL_USER, MYSQL_PASSWORD) or die ('Unable to connect. Check your connection parameters.'); mysql_select_db(MYSQL_DB, $db) or die(mysql_error($db)); //change this path to match your images directory $dir ='C:/x/xampp/htdocs/images'; //change this path to match your thumbnail directory $thumbdir = $dir . '/thumbs'; ?> <html> <head> <title>Welcome to our Photo Gallery</title> <style type="text/css"> th { background-color: #999;} .odd_row { background-color: #EEE; } .even_row { background-color: #FFF; } </style> </head> <body> <p>Click on any image to see it full sized.</p> <table style="width:100%;"> <tr> <th>Image</th> <th>Caption</th> <th>Uploaded By</th> <th>Date Uploaded</th> </tr> <?php //get the thumbs $result = mysql_query('SELECT * FROM images') or die(mysql_error()); $odd = true; while ($rows = mysql_fetch_array($result)) { echo ($odd == true) ? '<tr class="odd_row">' : '<tr class="even_row">'; $odd = !$odd; extract($rows); echo '<td><a href="' . $dir . '/' . $image_id . '.jpg">'; echo '<img src="' . $thumbdir . '/' . $image_id . '.jpg">'; echo '</a></td>'; echo '<td>' . $image_caption . '</td>'; echo '<td>' . $image_username . '</td>'; echo '<td>' . $image_date . '</td>'; echo '</tr>'; } ?> </table> </body> </html> Any help appreciated. This topic has been moved to Other Programming Languages. http://www.phpfreaks.com/forums/index.php?topic=312147.0
if ($count==1){
header("Location:store.php");
}
very simple I have issolated it and it doesn't redirect maybe u can see where my mistake is
Hi everyone! I've been working on a php script to replace links that contain a query with direct links to the files they would redirect to. Hi i have this upload script which works fine it uploads image to a specified folder and sends the the details to the database. but now i am trying to instead make a modify script which is Update set so i tried to change insert to update but didnt work can someone help me out please this my insert image script which works fine but want to change to modify instead Code: [Select] <?php mysql_connect("localhost", "root", "") or die(mysql_error()) ; mysql_select_db("upload") or die(mysql_error()) ; // my file the name of the input area on the form type is the extension of the file //echo $_FILES["myfile"]["type"]; //myfile is the name of the input area on the form $name = $_FILES["image"] ["name"]; // name of the file $type = $_FILES["image"]["type"]; //type of the file $size = $_FILES["image"]["size"]; //the size of the file $temp = $_FILES["image"]["tmp_name"];//temporary file location when click upload it temporary stores on the computer and gives it a temporary name $error =array(); // this an empty array where you can then call on all of the error messages $allowed_exts = array('jpg', 'jpeg', 'png', 'gif'); // array with the following extension name values $image_type = array('image/jpg', 'image/jpeg', 'image/png', 'image/gif'); // array with the following image type values $location = 'images/'; //location of the file or directory where the file will be stored $appendic_name = "news".$name;//this append the word [news] before the name so the image would be news[nameofimage].gif // substr counts the number of carachters and then you the specify how how many you letters you want to cut off from the beginning of the word example drivers.jpg it would cut off dri, and would display vers.jpg //echo $extension = substr($name, 3); //using both substr and strpos, strpos it will delete anything before the dot in this case it finds the dot on the $name file deletes and + 1 says read after the last letter you delete because you want to display the letters after the dot. if remove the +1 it will display .gif which what we want is just gif $extension = strtolower(substr($name, strpos ($name, '.') +1));//strlower turn the extension non capital in case extension is capital example JPG will strtolower will make jpg // another way of doing is with explode // $image_ext strtolower(end(explode('.',$name))); will explode from where you want in this case from the dot adn end will display from the end after the explode $myfile = $_POST["myfile"]; if (isset($image)) // if you choose a file name do the if bellow { // if extension is not equal to any of the variables in the array $allowed_exts error appears if(in_array($extension, $allowed_exts) === false ) { $error[] = 'Extension not allowed! gif, jpg, jpeg, png only<br />'; // if no errror read next if line } // if file type is not equal to any of the variables in array $image_type error appears if(in_array($type, $image_type) === false) { $error[] = 'Type of file not allowed! only images allowed<br />'; } // if file bigger than the number bellow error message if($size > 2097152) { $error[] = 'File size must be under 2MB!'; } // check if folder exist in the server if(!file_exists ($location)) { $error[] = 'No directory ' . $location. ' on the server Please create a folder ' .$location; } } // if no error found do the move upload function if (empty($error)){ if (move_uploaded_file($temp, $location .$appendic_name)) { // insert data into database first are the field name teh values are the variables you want to insert into those fields appendic is the new name of the image mysql_query("INSERT INTO image (myfile ,image) VALUES ('$myfile', '$appendic_name')") ; exit(); } } else { foreach ($error as $error) { echo $error; } } //echo $type; ?> I'm having trouble echoing $year in my script. Listed below is the script, just below ,$result = mysql_query("SELECT * FROM $dbname WHERE class LIKE '%$search%'") or die(mysql_error());, in the script I try to echo $year. It doesn't show up in the table on the webpage. Everything else works fine. Any help wold be appreciated greatly. Thanks in advance. <?php include 'config2.php'; $search=$_GET["search"]; // Connect to server and select database. mysql_connect($dbhost, $dbuser, $dbpass)or die("cannot connect"); mysql_select_db("vetman")or die("cannot select DB"); $result = mysql_query("SELECT * FROM $dbname WHERE class LIKE '%$search%'") or die(mysql_error()); // store the record of the "" table into $row //$current = ''; echo "<table align=center border=1>"; echo "<br>"; echo "<tr>"; echo "<td align=center>"; ?> <div style="float: center;"><a><h1><?php echo $year; ?></h1></a></div> <?php echo "</td>"; echo "</tr>"; echo "</table>"; // keeps getting the next row until there are no more to get if($result && mysql_num_rows($result) > 0) { $i = 0; $max_columns = 2; echo "<table align=center>"; echo "<br>"; while($row = mysql_fetch_array($result)) { // make the variables easy to deal with extract($row); // open row if counter is zero if($i == 0) echo "<tr>"; echo "<td align=center>"; ?> <div style="float: left;"> <div><img src="<?php echo $image1; ?>"></div> </div> <?php echo "</td>"; // increment counter - if counter = max columns, reset counter and close row if(++$i == $max_columns) { echo "</tr>"; $i=0; } // end if } // end while } // end if results // clean up table - makes your code valid! if($i > 0) { for($j=$i; $j<$max_columns;$j++) echo "<td> </td>"; echo '</tr>'; } mysql_close(); ?> </table> I'm trying to use this script known as SimpleImage.php that can be found here <a href="http://www.white-hat-web-design.co.uk/articles/php-image-resizing.php">link</a> I'm trying to include what is on the bottom of the page to my existing script can anyone help me I've tried several ways but its not working. Code: [Select] <?php session_start(); error_reporting(E_ALL); ini_set('display_errors','On'); //error_reporting(E_ALL); // image upload folder $image_folder = 'images/classified/'; // fieldnames in form $all_file_fields = array('image1', 'image2' ,'image3', 'image4'); // allowed filetypes $file_types = array('jpg','gif','png'); // max filesize 5mb $max_size = 5000000; //echo'<pre>';print_r($_FILES);exit; $time = time(); $count = 1; foreach($all_file_fields as $fieldname){ if($_FILES[$fieldname]['name'] != ''){ $type = substr($_FILES[$fieldname]['name'], -3, 3); // check filetype if(in_array(strtolower($type), $file_types)){ //check filesize if($_FILES[$fieldname]['size']>$max_size){ $error = "File too big. Max filesize is ".$max_size." MB"; }else{ // new filename $filename = str_replace(' ','',$myusername).'_'.$time.'_'.$count.'.'.$type; // move/upload file $target_path = $image_folder.basename($filename); move_uploaded_file($_FILES[$fieldname]['tmp_name'], $target_path); //save array with filenames $images[$count] = $image_folder.$filename; $count = $count+1; }//end if }else{ $error = "Please use jpg, gif, png files"; }//end if }//end if }//end foreach if($error != ''){ echo $error; }else{ /* -------------------------------------------------------------------------------------------------- SAVE TO DATABASE ------------------------------------------------------------------------------------ -------------------------------------------------------------------------------------------------- */ ?> hey guys im really just after a bit of help/information on 2 things (hope its in the right forum).
1. basically I'm wanting to make payments from one account to another online...like paypal does...im wondering what I would need to do to be able to do this if anyone can shine some light please?
2.as seen on google you type in a query in the search bar and it generates sentences/keywords from a database
example:
so if product "chair" was in the database
whilst typing "ch" it would show "chair" for a possible match
I know it would in tale sql & json but im after a good tutorial/script of some sort.
if anyone can help with some information/sites it would be much appreciated.
Thank you
Hello, I stored a fsockopen function in a separate "called.php" file, in order to run it as another thread when it needs. The called script should return results to the "master.php" script. I'm able to run the script to get the socket working, and I'm able to get results from the called script. I tried for hours but I can't do the twice both My master.php script (with socket working): Code: [Select] <?php $command = "(/mnt/opt/www/called.php $_SERVER[REMOTE_ADDR] &) > /dev/null"; $result = exec($command); echo ("result = $result\r\n"); ?> and my called.php script Code: [Select] #!/mnt/opt/usr/bin/php-cli -q <?php $device = $_SERVER['argv'][1]; $port = "8080"; $fp = fsockopen($device, $port, $errno, $errstr, 5); fwrite($fp, "test"); fclose($fp); echo ("normal end of the called.php script"); ?> In the master script, if I use Code: [Select] $command = "(/mnt/opt/www/called.php $_SERVER[REMOTE_ADDR] &) > /dev/null"; the socket works, but I have nothing in $result (note also that I don't anderstand why the ( ... &) are needed!?) and if I use Code: [Select] $command = "/mnt/opt/www/called.php $_SERVER[REMOTE_ADDR]"; I have the correct text "normal end of the called.php script" in $result but the socket connection is not performed (no errors in php logs) Could you help me to find a way to let's work the two features correctly together? Thank you. Please delete. I've joined querys because I'm making my own forums. However, when it gets to the echoin out part, it shows the category and only ONE of the sub categorys. Code: [Select] <div id="forumContainer"> <?php $lastboard = ''; $forumQ = mysql_query("SELECT f1.cat_name as catName, f1.cat_id as catID, f2.cat_id as subCatID, f2.sub_id as subID, f2.sub_name as subName, f2.sub_desc as subDesc FROM `forum_cats` as f1 LEFT JOIN `forum_sub` as f2 ON f1.cat_id = f2.cat_id GROUP BY f2.cat_id "); while($forumF = mysql_fetch_assoc($forumQ)) { if($forumF['catName'] != $lastboard) { echo '<div class="forumCat">'.$forumF['catName'].'</div>'; $lastboard = $forumF['catName']; } echo '<div class="forumSub">'.$forumF['subName'].'</div>'; } ?> </div> Is there something wrong with this query? elseif($_POST["titlee"] && $_POST["contente"]) { $titlee = $_POST['titlee']; $contente = $_POST['contente']; $to = $_POST['edit']; mysql_query("UPDATE custom_pages SET title='$titlee' AND content='$contente' WHERE id='$to'"); echo '<div class="post"> <div class="postheader"><h1>Updated</h1></div> <div class="postcontent"> <p>Your custom page has been updated.</p> </div> <div class="postfooter"></div> </div> '; } Hi, can anybody check on this code? I don't have any errors in it, except that the supposedly alert after I deleted an information is not working. Thank you very much. **This is a modified code provided by a friend. It worked originally (I just lost the original file since I sold my old computer xD Here's the code: <?php include("dbconnection.php"); if(mysql_num_rows($result) > 0) { $query = "SELECT * FROM records WHERE id = '".$_GET["id"]."'"; $result = mysql_query($query, $connection); $status = mysql_result($result, 0, "status"); $territory = mysql_result($result, 0, "territory"); $job_title = mysql_result($result, 0, "job_title"); $area_of_work = mysql_result($result, 0, "area_of_work"); $employer = mysql_result($result, 0, "employer"); $department = mysql_result($result, 0, "department"); $location = mysql_result($result, 0, "location"); $date_posted = mysql_result($result, 0, "date_posted"); $closing_date = mysql_result($result, 0, "closing_date"); $gender = mysql_result($result, 0, "gender"); $first_name = mysql_result($result, 0, "first_name"); $last_name = mysql_result($result, 0, "last_name"); $telephone_number = mysql_result($result, 0, "telephone_number"); $title = mysql_result($result, 0, "title"); $address_1 = mysql_result($result, 0, "address_1"); $address_2 = mysql_result($result, 0, "address_2"); $address_3 = mysql_result($result, 0, "address_3"); $city = mysql_result($result, 0, "city"); $country = mysql_result($result, 0, "country"); $postal_code = mysql_result($result, 0, "postal_code"); $website = mysql_result($result, 0, "website"); $email_address = mysql_result($result, 0, "email_address"); $cg_comment = mysql_result($result, 0, "cg_comment"); $date_emailed = mysql_result($result, 0, "date_emailed"); $mailing_comments= mysql_result($result, 0, "mailing_comments"); $telesales_comments= mysql_result($result, 0, "telesales_comments"); } if(isset($_POST["btnSubmit"])) { $id = $_POST["id"]; $query = "DELETE FROM records WHERE id = '".$id."'"; mysql_query($query) or die(mysql_error()); echo "<script> alert('You have successfully deleted a record'); window.location = 'view_client.php'; </script>"; } ?> Never had this one before. Here's a string from a URL: $urltext = Product_Name_'with_single_quotes'_"_B Code: [Select] $name = str_replace( "_", " ", $urltext ); echo 'raw: ' . $name . "<br>"; $name = mysql_real_escape_string( $name ); echo 'mysql_real_escaped: ' . $name . '<br>'; Doesn't seem possible, but both of the "echos" return the same string. My query fails because there are no backslashes in the SQL statement. What's going on here? raw: Product Name 'with single quotes' " B mysql_real_escaped: Product Name 'with single quotes' " B |
|||||||