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  • Upload image from URL to FTP server using PHP

    - by user1807556
    I want to upload a picture from another site to my FTP server using PHP. Example: File to upload("http://page.mi.fu-berlin.de/krudolph/stuff/stackoverflow.png") FTP-path("pictures/") This is what I've already tried: 1 $image = file_get_contents("http://img.youtube.com/vi/Rz8KW4Tveps/1.jpg"); file_put_contents("imgfolder/imgID.jpg", $image); 2 copy('http://img.youtube.com/vi/Rz8KW4Tveps/1.jpg', 'imgfolder/imgID.jpg'); 3 <?php set_time_limit (24 * 60 * 60); if (!isset($_POST['submit'])) die(); $file = fopen ($url, "rb"); if ($file) { $newf = fopen ($newfname, "wb"); if ($newf) while(!feof($file)) { fwrite($newf, fread($file, 1024 * 2000 ), 1024 * 2000 ); } } if ($file) { fclose($file); } if ($newf) { fclose($newf); } ?> 4 http://www.teckdevil.com/php-server-to-server-transfer-script-to-remotely-transfer-files/ 5 (kinda the same as first linked) Download files directly to my server # I don't get any errors when I'm running the scripts and I have chmod the directory to 777.

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  • PHP Dropdown menu [closed]

    - by rShetty
    <br><h2>Select a Tag</h2></br> <?php $con = mysql_connect("localhost","root",""); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("portal", $con); $query = "SELECT tag_name FROM tags"; $result = mysql_query($query); ?> <select name="tag_name" id="abc"> <option size=30 selected>Select</option> <?php while($array = mysql_fetch_assoc($result)){ ?> <option value ="<?php echo $array['tag_name'];?>"><?php echo $array['tag_name'];?> </option> <?php } ?> </select> <br><br> This is a snippet of code for getting the dropdown menu in the page. I have a database named portal and table named tags with tag_name as the attribute. Do help me to find the error in the program. I am not getting the tag_names in the dropdown menu

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  • How would i output a message that says "values inserted to table" ?

    - by ranlo
    <?php error_reporting(0); $con=mysql_connect("localhost","root",""); if (!$con) { die('could not connect:'.mysql_error()); } mysql_select_db("final?orgdocs",$con); $org_name = $_POST["org_name"]; $org_type = $_POST["org_type"]; $org_code = $_POST["org_code"]; $description = $_POST["description"]; $stmt = "INSERT INTO organization VALUES('".$org_name."','".$org_type."','".$org_code."','".$description."')"; $result = mysql_query("SELECT * FROM organization WHERE org_name = '$org_name' "); echo '<TABLE BORDER = "1">'; $result1 = $result; while ($row = mysql_fetch_array($result1)){ echo '<TR>'.'<TD>'.'Organization Name'.'</TD>'.'<TD>'.'Organization Type'.'</TD>'.'<TD>'.'Organization Code'.'</TD>'.'<TD>'.'Description'.'</TD>'.'<TD>'.'Constitution'.'</TD>'; echo '</TR>'; echo '<TR>'.'<TD>'.$row['org_name'].'</TD>'.'<TD>'.$row['org_type'].'</TD>'; echo '<TD>'.$row['org_code'].'</TD>'.'<TD>'.$row['description'].'</TD>'.'<TD>'; echo '</TR>'; } echo '</TABLE>'; ?>

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  • problem during data fetch

    - by nectar
    here is my code $sql="SELECT * FROM $tbl_name WHERE ownerId='$UserId'"; $result=mysql_query($sql,$link)or die(mysql_error()); $row = mysql_fetch_array($result, MYSQL_ASSOC); <?php while($row = mysql_fetch_array($result, MYSQL_ASSOC)) { echo "<tr>"; echo "<td>".$row['pinId']."</td>"; echo "<td>".$row['usedby']."</td>"; echo "<td>".$row['status']."</td>"; echo "</tr>"; } ?> it is ignoring the first record means if 4 rows are in $row its ignoring the 1st one rest three are coming on page. ownerId is not primary key.

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  • PHP delete script, return to 'viewsubjects.php?classroom_id=NO VALUE'

    - by Derek
    Hi, As the title states... I am deleting a 'subject' from a 'classroom' I view classrooms, then can click on a classroom to view the subject for that classroom. So the link where I am viewing subjects looks like: viewsubjects.php?classroom=23 When the user selects the delete button (in a row) to remove a subject from a class, I simply want the user to be redirected back to the list of subjects for the classroom (exactly where they were before!!) So I though this is simply a case of calling up the classroom ID within my delete script. Here is what I have: EDIT: corrected spelling mistake in code (this was not the problem) $subject_id = $_GET['subject_id']; $classroom_id = $_GET['classroom_id']; $sql = "DELETE FROM subjects WHERE subject_id=".$subject_id; $result = mysql_query($sql, $connection) or die("MySQL Error: ".mysql_error()); header("Location: viewsubjects.php?classroom_id=".$classroom_id); exit(); The subject is being removed from the DB, but when I am redirected back the URI is displaying with an empty classroom ID like: viewsubjects.php?classroom_id= Is there a way to carry the classroom ID through successfully through the delete script so it can be displayed after, allowing the user to be redirected back to the page? Thanks for any help!

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  • Help Optimizing MySQL Table (~ 500,000 records).

    - by Pyrite
    I have a MySQL table that collects player data from various game servers (Urban Terror). The bot that collects the data runs 24/7, and currently the table is up to about 475,000+ records. Because of this, querying this table from PHP has become quite slow. I wonder what I can do on the database side of things to make it as optomized as possible, then I can focus on the application to query the database. The table is as follows: CREATE TABLE IF NOT EXISTS `people` ( `id` bigint(20) unsigned NOT NULL AUTO_INCREMENT, `name` varchar(40) NOT NULL, `ip` int(4) unsigned NOT NULL, `guid` varchar(32) NOT NULL, `server` int(4) unsigned NOT NULL, `date` int(11) NOT NULL, PRIMARY KEY (`id`), UNIQUE KEY `Person` (`name`,`ip`,`guid`), KEY `server` (`server`), KEY `date` (`date`), KEY `PlayerName` (`name`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 COMMENT='People that Play on Servers' AUTO_INCREMENT=475843 ; I'm storying the IPv4 (ip and server) as 4 byte integers, and using the MySQL functions NTOA(), etc to encode and decode, I heard that this way is faster, rather than varchar(15). The guid is a md5sum, 32 char hex. Date is stored as unix timestamp. I have a unique key on name, ip and guid, as to avoid duplicates of the same player. Do I have my keys setup right? Is the way I'm storing data efficient? Here is the code to query this table. You search for a name, ip, or guid, and it grabs the results of the query and cross references other records that match the name, ip, or guid from the results of the first query, and does it for each field. This is kind of hard to explain. But basically, if I search for one player by name, I'll see every other name he has used, every IP he has used and every GUID he has used. <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post"> Search: <input type="text" name="query" id="query" /><input type="submit" name="btnSubmit" value="Submit" /> </form> <?php if (!empty($_POST['query'])) { ?> <table cellspacing="1" id="1up_people" class="tablesorter" width="300"> <thead> <tr> <th>ID</th> <th>Player Name</th> <th>Player IP</th> <th>Player GUID</th> <th>Server</th> <th>Date</th> </tr> </thead> <tbody> <?php function super_unique($array) { $result = array_map("unserialize", array_unique(array_map("serialize", $array))); foreach ($result as $key => $value) { if ( is_array($value) ) { $result[$key] = super_unique($value); } } return $result; } if (!empty($_POST['query'])) { $query = trim($_POST['query']); $count = 0; $people = array(); $link = mysql_connect('localhost', 'mysqluser', 'yea right!'); if (!$link) { die('Could not connect: ' . mysql_error()); } mysql_select_db("1up"); $sql = "SELECT id, name, INET_NTOA(ip) AS ip, guid, INET_NTOA(server) AS server, date FROM 1up_people WHERE (name LIKE \"%$query%\" OR INET_NTOA(ip) LIKE \"%$query%\" OR guid LIKE \"%$query%\")"; $result = mysql_query($sql, $link); if (!$result) { die(mysql_error()); } // Now take the initial results and parse each column into its own array while ($row = mysql_fetch_array($result, MYSQL_NUM)) { $name = htmlspecialchars($row[1]); $people[] = array( 'id' => $row[0], 'name' => $name, 'ip' => $row[2], 'guid' => $row[3], 'server' => $row[4], 'date' => $row[5] ); } // now for each name, ip, guid in results, find additonal records $people2 = array(); foreach ($people AS $person) { $ip = $person['ip']; $sql = "SELECT id, name, INET_NTOA(ip) AS ip, guid, INET_NTOA(server) AS server, date FROM 1up_people WHERE (ip = \"$ip\")"; $result = mysql_query($sql, $link); while ($row = mysql_fetch_array($result, MYSQL_NUM)) { $name = htmlspecialchars($row[1]); $people2[] = array( 'id' => $row[0], 'name' => $name, 'ip' => $row[2], 'guid' => $row[3], 'server' => $row[4], 'date' => $row[5] ); } } $people3 = array(); foreach ($people AS $person) { $guid = $person['guid']; $sql = "SELECT id, name, INET_NTOA(ip) AS ip, guid, INET_NTOA(server) AS server, date FROM 1up_people WHERE (guid = \"$guid\")"; $result = mysql_query($sql, $link); while ($row = mysql_fetch_array($result, MYSQL_NUM)) { $name = htmlspecialchars($row[1]); $people3[] = array( 'id' => $row[0], 'name' => $name, 'ip' => $row[2], 'guid' => $row[3], 'server' => $row[4], 'date' => $row[5] ); } } $people4 = array(); foreach ($people AS $person) { $name = $person['name']; $sql = "SELECT id, name, INET_NTOA(ip) AS ip, guid, INET_NTOA(server) AS server, date FROM 1up_people WHERE (name = \"$name\")"; $result = mysql_query($sql, $link); while ($row = mysql_fetch_array($result, MYSQL_NUM)) { $name = htmlspecialchars($row[1]); $people4[] = array( 'id' => $row[0], 'name' => $name, 'ip' => $row[2], 'guid' => $row[3], 'server' => $row[4], 'date' => $row[5] ); } } // Combine people and people2 into just people $people = array_merge($people, $people2); $people = array_merge($people, $people3); $people = array_merge($people, $people4); $people = super_unique($people); foreach ($people AS $person) { $date = ($person['date']) ? date("M d, Y", $person['date']) : 'Before 8/1/10'; echo "<tr>\n"; echo "<td>".$person['id']."</td>"; echo "<td>".$person['name']."</td>"; echo "<td>".$person['ip']."</td>"; echo "<td>".$person['guid']."</td>"; echo "<td>".$person['server']."</td>"; echo "<td>".$date."</td>"; echo "</tr>\n"; $count++; } // Find Total Records //$result = mysql_query("SELECT id FROM 1up_people", $link); //$total = mysql_num_rows($result); mysql_close($link); } ?> </tbody> </table> <p> <?php echo $count." Records Found for \"".$_POST['query']."\" out of $total"; ?> </p> <?php } $time_stop = microtime(true); print("Done (ran for ".round($time_stop-$time_start)." seconds)."); ?> Any help at all is appreciated! Thank you.

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  • After mysql_query, no result output

    - by Jerry
    I have a simple mysql_query() update command to update mysql. When a user submits my form, it will jump to an update page to update the data. The problem is that there's supposed to be some data shown after the update, but it comes out blank. My form <form id="form1" method="POST" action="scheduleUpdate.php" > <select name=std1> <option>AA</option> <option>BB</option> <option>CC</option> </select> <select name=std2> <option>DD</option> <option>EE</option> <option>FF</option> </select> .......//more drop down menu but the name is std3..std4..etc... ....... </form> scheduleUpdate.php //$i is the value posted from my main app to tell me how many std we have for($k=0;$k<$i;$k++){ $std=$_POST['std'.$k]; //if i remove the updateQuery, the html will output.I know the query is the problem but i //couldn't fix it.. $updateQuery=mysql_query("UPDATE board SET student='$std' WHERE badStudent='$std' or goodStudent='$std'",$connection); //no output below this line at all if($updateQuery){ DIE('mysql Error:'+mysql_error()); } } // I have bunch of HTML here....but no output at all!!!! MySQL will be updated after I hit submit, but it doesn't shown any HTML.

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  • Ordering by number of rows?

    - by Rob
    Alright, so I have a table outputting data from a MySQL table in a while loop. Well one of the columns it outputs isn't stored statically in the table, instead it's the sum of how many times it appears in a different MySQL table. Sorry I'm not sure this is easy to understand. Here's my code: $query="SELECT * FROM list WHERE added='$addedby' ORDER BY time DESC"; $result=mysql_query($query); while($row=mysql_fetch_array($result, MYSQL_ASSOC)){ $loghwid = $row['hwid']; $sql="SELECT * FROM logs WHERE hwid='$loghwid' AND time < now() + interval 1 hour"; $query = mysql_query($sql) OR DIE(mysql_error()); $boots = mysql_num_rows($query); //Display the table } The above is the code displaying the table. As you can see it's grabbing data from two different MySQL tables. However I want to be able to ORDER BY $boots DESC. But as its a counting of a completely different table, I have no idea of how to go about doing that. I would appreciate any help, thank you.

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  • Problem with php GD image not creating

    - by ThinkingInBits
    With the following code, my browser is returning 'image not created or saved'. I assure you the location of the image exists. So the line within the if statement is returning false for some reason, and yes GD is installed. Any ideas? if ($img = @imagecreatefromjpeg("/var/www/images/upload/1/1.jpg")) { $image_width = imagesx($img); $image_height = imagesy($img); $target_width = 150; $target_height = 150; if ($image_width > $image_height) { $percentage = ($target_width / $image_width); } else { $percentage = ($target_height / $image_height); } $new_image_width = round($image_width * $percentage); $new_image_height = round($image_height * $percentage); $imgResized = imagecreatetruecolor($new_image_width, $new_image_height); imagecopyresampled($imgResized, $img, 0, 0, 0, 0, $new_image_width, $new_image_height, $image_width, $image_height); imagejpeg($imgResized, $this->path, 100); imagedestroy($img); imagedestroy($imgResized); $this->storeThumbnailLocation(); } else { die ("image was not created or saved"); }

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  • PHP & form validation

    - by user887515
    I have a form that I'm submitting via ajax, and I want to return a message of the list of fields that were empty. I've got that all done and dusted, but it just seems really long winded on the PHP side of things. How can I do the below in a less convoluted way? <?php if(empty($_POST["emailaddress"])){ $error = 'true'; $validation_msg = 'Country missing.'; if(empty($error_msg)){ $error_msg .= $validation_msg; } else{ $error_msg .= '\n' . $validation_msg; } } if(empty($_POST["password"])){ $error = 'true'; $validation_msg = 'Country missing.'; if(empty($error_msg)){ $error_msg .= $validation_msg; } else{ $error_msg .= '\n' . $validation_msg; } } if(empty($_POST["firstname"])){ $error = 'true'; $validation_msg = 'First name missing.'; if(empty($error_msg)){ $error_msg .= $validation_msg; } else{ $error_msg .= '\n' . $validation_msg; } } if(empty($_POST["lastname"])){ $error = 'true'; $validation_msg = 'Last name missing.'; if(empty($error_msg)){ $error_msg .= $validation_msg; } else{ $error_msg .= '\n' . $validation_msg; } } if($error){ header('HTTP/1.1 500 Internal Server Error'); header('Content-Type: application/json'); die($error_msg); } ?>

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  • Zip multiple database PDF blob files

    - by Michael
    I have a database table that contains numerous PDF blob files. I am attempting to combine all of the files into a single ZIP file that I can download and then print. Please help! <?php include '../config.php'; include '../connect.php'; $session = $_GET['session']; $query = "SELECT $tbl_uploads.username, $tbl_uploads.description, $tbl_uploads.type, $tbl_uploads.size, $tbl_uploads.content, $tbl_members.session FROM $tbl_uploads LEFT JOIN $tbl_members ON $tbl_uploads.username = $tbl_members.username WHERE $tbl_members.session = '$session'"; $result = mysql_query($query) or die('Error, query failed'); $files = array(); while(list($username, $description, $type, $size, $content) = mysql_fetch_array($result)) { $files[] = "$username-$description.pdf"; } $zip = new ZipArchive; $zip->open('file.zip', ZipArchive::CREATE); foreach ($files as $file) { $zip->addFile($file); } $zip->close(); header('Content-Type: application/zip'); header('Content-disposition: attachment; filename=filename.zip'); header('Content-Length: ' . filesize($zipfilename)); readfile($zipname); exit(); ?>

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  • How to check Isavailablity in ajax?

    - by udaya
    Hi This is what i have in my view page <td width=""><input type="text" name="txtUserName" id="txtUserName" /></td> <td><input type="button" name="CheckUsername" id="CheckUsername" value="Check Availablity" onclick="Check_User_Name();"/></td> Onclick of the button the Check_User_Name function in my ajax.js loads This is the Check_User_Name function function Check_User_Name(source) { var UserName = document.getElementById('txtUserName').value; if(window.ActiveXObject) User_Name = new ActiveXObject("Microsoft.XMLHTTP"); else if(window.XMLHttpRequest) User_Name = new XMLHttpRequest(); var URL = newURL+"ssit/system/application/views/ssitAjax.php"; URL = URL +"?CheckUsername="+UserName; User_Name.onreadystatechange = User_Name_Fun; User_Name.open("GET",URL,true); User_Name.send(null); } function User_Name_Fun() { document.getElementById('User_div').innerHTML=User_Name.responseText; } Then I can have the value in echo username then the $result has all the user name Hows can i checkIsavailablity of username from here if(($_GET['CheckUsername']!="") || (isset($_GET['CheckUsername']))) { echo $UserName = $_GET['CheckUsername'];//echo username $_SESSION['state'] = $State; $queryres = "SELECT dUser_name FROM tbl_login WHERE dIsDelete='0'"; $result = mysql_query($queryres,$cn) or die("Selection Query Failed !!!");

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  • PHP login, getting wrong count value from query / fetch array

    - by Chris
    Hello, *EDIT*Thanks to the comments below it has been figured out that the problem lies with the md5, without everything works as it should. But how do i implent the md5 then? I am having some troubles with the following code below to login. The database and register system are already working. The problem lies that it does not find any result at all in the query. IF the count is 0 it should redirect the user to a secured page. But this only works if i write count = 0, but this should be 0 , only if the user name and password is found he should be directed to the secure (startpage) of the site after login. For example root (username) root (password) already exists but i cannot seem to properly login with it. <?php session_start(); if (!empty($_POST["send"])) { $username = ($_POST["username"]); $password = (md5($_POST["password"])); $count = 0; $con = mysql_connect("localhost" , "root", ""); mysql_select_db("testdb", $con); $result = mysql_query("SELECT name, password FROM user WHERE name = '".$username."' AND password = '".$password."' ") or die("Error select statement"); $count = mysql_num_rows($result); if($count > 0) // always goes the to else, only works with >=0 but then the data is not found in the database, hence incorrect { $row = mysql_fetch_array($result); $_SESSION["username"] = $row["name"]; header("Location: StartPage.php"); } else { echo "Wrong login data, please try again"; } mysql_close($con); } ?>

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  • php memory how much is too much

    - by Rob
    I'm currently re-writing my site using my own framework (it's very simple and does exactly what I need, i've no need for something like Zend or Cake PHP). I've done alot of work in making sure everything is cached properly, caching pages in files so avoid sql queries and generally limiting the number of sql queries. Overall it looks like it's very speedy. The average time taken for the front page (taken over 100 times) is 0.046152 microseconds. But one thing i'm not sure about is whether i've done enough to reduce php memory usage. The only time i've ever encountered problems with it is when uploading large files. Using memory_get_peak_usage(TRUE), which I THINK returns the highest amount of memory used whilst the script has been running, the average (taken over 100 times) is 1572864 bytes. Is that good? I realise you don't know what it is i'm doing (it's rather simple, get the 10 latest articles, the comment count for each, get the user controls, popular tags in the sidebar etc). But would you be at all worried with a script using that sort of memory getting hit 50,000 times a day? Or once every second at peak times? I realise that this is a very open ended question. Hopefully you can understand that it's a bit of a stab in the dark and i'm really just looking for some re-assurance that it's not going to die horribly come re-launch day.

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  • PHP: How to know if a user is currently downloading a file?

    - by metrobalderas
    I'm developing a quick rapidshare-like site where the user can download files. First, I created a quick test setting headers and using readfile() but then I found in the comments section there's a way to limit the speed of the download, which is great, here's the code: $local_file = 'file.zip'; $download_file = 'name.zip'; // set the download rate limit (=> 20,5 kb/s) $download_rate = 20.5; if(file_exists($local_file) && is_file($local_file)) { header('Cache-control: private'); header('Content-Type: application/octet-stream'); header('Content-Length: '.filesize($local_file)); header('Content-Disposition: filename='.$download_file); flush(); $file = fopen($local_file, "r"); while(!feof($file)) { // send the current file part to the browser print fread($file, round($download_rate * 1024)); // flush the content to the browser flush(); // sleep one second sleep(1); } fclose($file);} else { die('Error: The file '.$local_file.' does not exist!'); } But now my question is, how to limit the number of downloads at the same time? How can I check there's still a connection with some user's IP? Thanks.

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  • How can I add one line into all php files' beginning?

    - by Tom
    So, ok. I have many php files and one index.php file. All files can't work without index.php file, because I include them in index.php. For example. if somebody click Contact us the URL will become smth like index.php?id=contact and I use $_GET['id'] to include contacts.php file. But, if somebody find the file's path, for example /system/files/contacts.php I don't want that that file would be executed. So, I figured out that I can add before including any files in index.php line like this $check_hacker = 1 and use if in every files beginning like this if($check_hacker <> 1) die();. So, how can I do it without opening all files and adding this line to each of them? Is it possible? Because I actually have many .php files. And maybe there is other way to do disable watching separate file? Any ideas? Thank you.

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  • How do you protect yourself from runaway memory consumption bringing down the PC?

    - by romkyns
    Every now and again I find myself doing something moderately dumb that results in my program allocating all the memory it can get and then some. This kind of thing used to cause the program to die fairly quickly with an "out of memory" error, but these days Windows will go out of its way to give this non-existent memory to the application, and in fact is apparently prepared to commit suicide doing so. Not literally of course, but it will starve itself of usable physical RAM so badly that even running the task manager will require half an hour of swapping (after all the runaway application is still allocating more and more memory all the time). This doesn't happen too often, but when it does it's disastrous. I usually have to reset my machine, causing data loss from time to time and generally a lot of inconvenience. Do you have any practical advice on making the consequences of such a mistake less dire? Perhaps some registry tweak to limit the max amount of virtual memory an app is allowed to allocate? Or some CLR flag that will limit this only for the current application? (It's usually in .NET that I do this to myself.) ("Don't run out of RAM" and "Buy more RAM" are no use - the former I have no control over, and the latter I've already done.)

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  • why this condition not work

    - by migo
    *strong text*I noticed that the first condition does not work if (empty($ss)) { echo "please write your search words"; } but the second work else if ($num < 1 ) { echo "not found any like "; full code if (empty($ss)) { echo "please write your search words"; } else if ($num < 1 ) { echo "not found any like "; }else { $sql=("SELECT * FROM student WHERE snum = $ss "); $rs = mysql_query($sql) or die(mysql_error()); while($data=mysql_fetch_array($rs)){ ? ??????? ????? ????? ???? ???? ?????? 100 100 100 100 100 ?????? ???????? ????? ?????? ????? ??????? ?????? ??????? } }; ?

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  • nothrow or exception ?

    - by Muggen
    I am a student and I have small knowledge on C++, which I try to expand. This is more of a philosophical question.. I am not trying to implement something. Since #include <new> //... T * t = new (std::nothrow) T(); if(t) { //... } //... Will hide the Exception, and since dealing with Exceptions is heavier compared to a simple if(t), why isn't the normal new T() not considered less good practice, considering we will have to use try-catch() to check if a simple allocation succeeded (and if we don't, just watch the program die)?? What are the benefits (if any) of the normal new allocation compared to using a nothrow new? Exception's overhead in that case is insignificant ? Also, Assume that an allocation fails (eg. no memory exists in the system). Is there anything the program can do in that situation, or just fail gracefully. There is no way to find free memory on the heap, when all is reserved, is there? Incase an allocation fails, and an std::bad_alloc is thrown, how can we assume that since there is not enough memory to allocate an object (Eg. a new int), there will be enough memory to store an exception ?? Thanks for your time. I hope the question is in line with the rules.

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  • Incorrect usage of UPDATE and ORDER BY

    - by nico55555
    I have written some code to update certain rows of a table with a decreasing sequence of numbers. To select the correct rows I have to JOIN two tables. The last row in the table needs to have a value of 0, the second last -1 and so on. To achieve this I use ORDER BY DESC. Unfortunately my code brings up the following error: Incorrect usage of UPDATE and ORDER BY My reading suggests that I can't use UPDATE, JOIN and ORDER BY together. I've read that maybe subqueries might help? I don't really have any idea how to change my code to do this. Perhaps someone could post a modified version that will work? while($row = mysql_fetch_array( $result )) { $products_id = $row['products_id']; $products_stock_attributes = $row['products_stock_attributes']; mysql_query("SET @i = 0"); $result2 = mysql_query("UPDATE orders_products op, orders ord SET op.stock_when_purchased = (@i:=(@i - op.products_quantity)) WHERE op.orders_id = ord.orders_id AND op.products_id = '$products_id' AND op.products_stock_attributes = '$products_stock_attributes' AND op.stock_when_purchased < 0 AND ord.orders_status = 2 ORDER BY orders_products_id DESC") or die(mysql_error()); }

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  • Warning: Trim expects

    - by user1257518
    I'm getting this warning Warning: trim() expects parameter 1 to be string, array given in .. which is my trim line. The full code is functioned to send an error when fields are empty. However, this error appears saying every field is empty, but only the 'native' field is meant to be required so thats my 2nd problem. Thanks for any help! session_start(); $err = array(); $user_id = intval($_SESSION['user_id']); // otherwise if (isset($_POST['doLanguage'])) { $link = mysql_connect(DB_HOST, DB_USER, DB_PASS) or die("Couldn't make connection."); // check if current user is banned $the_query = sprintf("SELECT COUNT(*) FROM users WHERE `banned` = '0' AND `id` = '%d'", $user_id); $result = mysql_query($the_query, $link); $user_check = mysql_num_rows($result); // user is ok if ($user_check > 0) { // check for empty fields foreach ($_POST as $key => $val) { $value = trim($val); if (empty($value)) { $err[] = "ERROR - $key is required"; } } // no errors if(empty($err)) { for($i = 0; $i < count($_POST["other"]); $i++) $native = mysql_real_escape_string($_POST['native'][$i]); $other = mysql_real_escape_string($_POST['other'][$i]); $other_list = mysql_real_escape_string($_POST['other_list'][$i]); $other_read = mysql_real_escape_string($_POST['other_read'][$i]); $other_spokint = mysql_real_escape_string($_POST['other_spokint'][$i]); $other_spokprod = mysql_real_escape_string($_POST['other_spokprod'][$i]); $other_writ = mysql_real_escape_string($_POST['other_writ'][$i]); // insert into the database $the_query = sprintf("INSERT INTO `language` (`user_id`,`native`,`other`,`other_list`,`other_read`, `other_spokint` ,`other_spokprod`,`other_writ` ) VALUES ('%d','%s','%s','%s','%s','%s','%s','%s')", $user_id,$native,$other,$other_list,$other_read, $other_spokint,$other_spokprod,$other_writ); // query is ok? if (mysql_query($the_query, $link) ){ // redirect to user profile header('Location: myaccount.php?id=' . $user_id); } } } }

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  • Image download from mysql results

    - by rozatrra
    i need to give my users the opportunity to download all the images I display in my project. images are displayed from a mysql query like this: $query = mysql_query("SELECT tl.customername, tl.visitdate, tl.employeename, pz.webpath from table tl inner join pictures pz on pz.visitid = tl.visitid and pz.groupid = tl.groupid inner join agenti ag on ag.idh = tl.employeeid WHERE tl.visitdate >= '$from' AND tl.visitdate <= '$to' AND tl.employeename like '$r_employee' AND tl.customerowner like '$r_customer' AND tl.customername like '$r_customername' AND tl.visitdate like '$r_date' group by pz.webpath order by tl.customername") or die(mysql_error()); while( $associate = mysql_fetch_assoc($query)) { echo '<li> <figure> <img src="../core/includes/timthumb.php?src='.$associate['webpath'].'&w=200&h=200" /> <figcaption> <h3>'.$associate['customername'].'</h3> <h6>'.$associate['employeename'].'</h6> <h6>'.$associate['visitdate'].' </h6> '; echo '<a class="fancybox" rel="gallery" href="'.$associate['webpath'].'" title=" '.$associate['visitdate'].' / '.$associate['customername'].'">Big picture</i></a>'; echo '</figcaption> </figure> </li>'; $zip->addFromString(pathinfo ( urldecode($associate['webpath']), PATHINFO_BASENAME), urldecode($associate['webpath'])); } How can i add a download button which will save all the images as zip on user computer?

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  • How do I select value from DropDown list in PHP??? Problem

    - by sandy
    Hello .. I want to know the error in this code The following code retrieves the names of the members of the database query in the dropdownlist But how do I know who you selected.... I want to send messages only to the members that selected form dropdown list <?php include ("connect.php"); $name = $_POST['sector_list']; echo $name ; ?> <form method="POST" action="" > <input type="hidden" name="sector" value="sector_list"> <select name="sector_list" class="inputstandard"> <option size ="40" value="default">send to </option> <?php $result = mysql_query('select * from members ') or die (mysql_error()); while ($row = mysql_fetch_assoc($result)) { echo '<option size ="40" value=" '. $row['MemberID'] . '" name="' . $row['MemberName']. '">' . $row['MemberName']. '</option>'; } ?> </select> </form> I hope somebody can help me

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  • PHP PDO changes remote host to local hostname

    - by Wade Urry
    I'm trying to connect to a remote mysql server using PDO. However, regardless of the hostname or ip address i supply in the dsn, when the script is run it always reverts the address to the hostname of the local server where the webserver is running. Google suggests this could be something to do with SELinux and apaches ability to connect to remote databases, however i have SELinux disabled. Distro: Ubuntu 11.04 x64 Apache version: 2.2.17 PHP Version: PHP 5.3.5-1ubuntu7.11 with Suhosin-Patch (cli) Edit: Added code as requested. Though i dont believe this is an issue with my coding as it works fine on the local server, but doesnt allow remote connection. public function db_connect($driver, $dbhost, $dbname, $user, $pass) { $dsn = $driver . ':host=' . $dbhost . ';dbname=' . $dbname; try { $this->DB = new PDO($dsn, $user, $pass); } catch (PDOException $err) { print 'Database Connection Failed: ' . $err->getMessage(); die(); } } $remote_db = new DB('mysql', 'remote_server.domain.tld', 'database_name', 'user_name', 'password'); This is the error message i am receiving. Database Connection Failed: SQLSTATE[28000] [1045] Access denied for user 'user_name'@'local_server.domain.tld' (using password: YES)

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  • Backup VM : Copy virtual disk xxx.vmdk The virtual disk is either corrupted or not a supported forma

    - by boiteavinc
    Hi I'm french. I'm student and I use ESXI4 for my studies. When I try backup my VMs, with ghettoVCBg2.pl and vSphere Management Assistant, I get the following error message on Vsphere Client : "Copy virtual disk xxx.vmdk The virtual disk is either corrupted or not a supported format". I have no error in the log VCB ; 03-23-2010 08:31:56 -- debug: Main: Login by vi-fastpass to: esxi 03-23-2010 08:31:56 -- debug: copyTask: Task START 03-23-2010 08:31:56 -- debug: copyTask: waiting for next job and sleep ... 03-23-2010 08:32:00 -- info: Initiate backup for AD_DNS_DHCP found on esxi 03-23-2010 08:32:09 -- debug: AD_DNS_DHCP original powerState: poweredOn 03-23-2010 08:32:09 -- debug: Creating Snapshot "ghettoVCBg2-snapshot-2010-03-23" for AD_DNS_DHCP 03-23-2010 08:33:19 -- info: AD_DNS_DHCP has 1 VMDK(s) 03-23-2010 08:33:19 -- debug: backupVMDK: Backing up "Raptor1 AD_DNS_DHCP/AD_DNS_DHCP.vmdk" to "Backup_VM VM/AD_DNS_DHCP/AD_DNS_DHCP-2010-03-23/$ 03-23-2010 08:33:19 -- debug: backupVMDK: Signal copyThread to start 03-23-2010 08:33:19 -- debug: backupVMDK: Backup progress: Elapsed time 0 min 03-23-2010 08:33:19 -- debug: copyTask: Wake up and follow the white rabbit, with status: doCopy 03-23-2010 08:33:19 -- debug: CopyThread: Start backing up VMDK(s) ... 03-23-2010 08:33:25 -- debug: copyTask: send copySuccess message ... 03-23-2010 08:33:25 -- debug: copyTask: waiting for next job and sleep ... 03-23-2010 08:34:20 -- debug: backupVMDK: Successfully completed backup for Raptor1 AD_DNS_DHCP/AD_DNS_DHCP.vmdk Elapsed time: 1 min 03-23-2010 08:34:22 -- debug: Removing Snapshot "ghettoVCBg2-snapshot-2010-03-23" for AD_DNS_DHCP 03-23-2010 08:34:24 -- debug: checkVMBackupRotation: Starting ... 03-23-2010 08:34:26 -- debug: Purging Backup_VM VM/AD_DNS_DHCP/AD_DNS_DHCP-2010-03-23--1 due to rotation max 03-23-2010 08:34:28 -- info: Backup completed for AD_DNS_DHCP! 03-23-2010 08:34:28 -- debug: Main: Disconnect from: esxi 03-23-2010 08:34:28 -- debug: Main: Calling final clean up 03-23-2010 08:34:28 -- debug: cleanUP: Thread clean up starting ... 03-23-2010 08:34:28 -- debug: cleanUp: Send exit to copyThread 03-23-2010 08:34:28 -- debug: copyTask: Wake up and follow the white rabbit, with status: exit 03-23-2010 08:34:28 -- debug: copyTask: die ... 03-23-2010 08:34:28 -- debug: cleanUp: Join passed 03-23-2010 08:34:28 -- info: ============================== ghettoVCBg2 LOG END ============================== My ghetto conf file is : VM_BACKUP_DATASTORE = "Backup_VM" VM_BACKUP_DIRECTORY = "VM" VM_BACKUP_ROTATION_COUNT = "3" DISK_BACKUP_FORMAT = "thin" ADAPTER_FORMAT = "lsilogic" POWER_VM_DOWN_BEFORE_BACKUP = "0" VM_SNAPSHOT_MEMORY = "1" VM_SNAPSHOT_QUIESCE = "1" LOG_LEVEL = "info" VM_VMDK_FILES = "all" I tried several I tried several DISK_BACKUP_FORMAT, I have same error. Despite the error, even when I get files on the NFS share. But when I try to open the vmx file with vmware workstation. I get this error: Can not open the disk 'G: \ Backup ESX \ VM \ AD_DNS_DHCP \ AD_DNS_DHCP-2010-03-23 - 1 \ AD_DNS_DHCP.vmdk' or one of the snapshot disks it depends on. Reason: The called function can not be performed on partial chains. Please open the parent virtual disk. I have no snapshot on my VM on ESXi. Can you help me ?

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