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  • Newbie's problems with MySQL and php

    - by Mirage81
    I'm a real newbie with php and MySQL. Now I'm working on the following code which should search the database for eg. all the Lennons living in Liverpool. 1) How should I modify "get.php" to get the text "no results" to appear if there are no search results. 2) How should I modify "index.php" to get the option values (city and lastname) straight from the database instead of having to type them one by one? 3) Am I using mysql_real_escape_string the right way? 4) Any other mistakes in the code? index.php: <form action="get.php" method="post"> <p> <select name="city"> <option value="Birmingham">Birmingham</option> <option value="Liverpool">Liverpool</option> <option value="London">London</option> </select> </p> <p> <select name="lastname"> <option value="Lennon">Lennon</option> <option value="McCartney">McCartney</option> <option value="Osbourne">Osbourne</option> </select> </p> <p> <input value="Search" type="submit"> </p> </form> get.php: <?php $city = $_POST['city']; $lastname = $_POST['lastname']; $conn = mysql_connect('localhost', 'user', 'password'); mysql_select_db("database", $conn) or die("connection failed"); $query = "SELECT * FROM users WHERE city = '$city' AND lastname = '$lastname'"; $result = mysql_query($query, $conn); $city = mysql_real_escape_string($_POST['city']); $lastname = mysql_real_escape_string($_POST['lastname']); echo $rowcount; while ($row = mysql_fetch_row($result)) { if ($rowcount == '0') echo 'no results'; else { echo '<b>City: </b>'.htmlspecialchars($row[0]).'<br />'; echo '<b>Last name: </b>'.htmlspecialchars($row[1]).'<br />'; echo '<b>Information: </b>'.htmlspecialchars($row[2]); } } mysql_close($conn);

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  • php array code with regular expressions

    - by user551068
    there are few mistakes which it is showing as Warning: preg_match() [function.preg-match]: Delimiter must not be alphanumeric or backslash in array 4,9,10,11,12... can anyone resolve them <?PHP $hosts = array( array("ronmexico.kainalopallo.com/","beforename=$F_firstname%20$F_lastname&gender=$F_gender","Your Ron Mexico Name is ","/the ultimate disguise, is <u><b>([^<]+)<\/b><\/u>/s"), array("www.fjordstone.com/cgi-bin/png.pl","gender=$F_gender&submit=Name%20Me","Your Pagan name is ","/COLOR=#000000 SIZE=6> *([^<]*)<\/FONT>/"), array("rumandmonkey.com/widgets/toys/mormon/index.php","gender=$F_gender&firstname=$F_firstname&surname=$F_lastname","Your Mormon Name is ","/<p>My Mormon name is <b>([^<]+)<\/b>!<br \/>/s"), array("cyborg.namedecoder.com/index.php","acronym=$F_firstname&design=edox&design_click-edox.x=0&design_click-edox.y=0&design_click-edox=edo","","Your Cyborg Name is ","/<p>([^<]+)<\/p>/"), array("rumandmonkey.com/widgets/toys/namegen/10/","nametype=$brit&page=2&id=10&submit=God%20save%20the%20Queen!&name=$F_firstname%20$F_lastname","Your Very British Name is ","/My very British name is \&lt\;b\&gt;([^&]+)\&lt;\/b\&gt;\.\&lt;br/"), array("blazonry.com/name_generator/usname.php","realname=$F_firstname+$F_lastname&gender=$F_gender","Your U.S. Name is ","/also be known as <font size=\'\+1\'><b>([^<]+)<\/b>/s"), array("www.spacepirate.org/rogues.php","realname=$F_firstname%20$F_lastname&formentered=Yes&submit=Arrrgh","Your Space Pirate name is ","/Your pirate name is <font size=\'\+1\'><b>([^<]+)<\/b><\/font>/s"), array("rumandmonkey.com/widgets/toys/ghetto/","firstname=$F_firstname&lastname=$F_lastname","Your Ghetto Name is ","/<p align=\"center\" style=\"font-size: 36px\">\s*<br \/>\s*([^<]*)<br \/>/"), array("www.emmadavies.net/vampire/default.aspx","mf=$emgender&firstname=$F_firstname&lastname=$F_lastname&submit=Find+My+Vampire+Name","","Your Vampire Name is ","/<i class=\"vampirecontrol vampire name\">([^<]*)<\/i>/"), array("www.emmadavies.net/fairy/default.aspx","mf=$emgender&firstname=$F_firstname&lastname=$F_lastname&submit=Seek+Fairy","","Your Fairy Name is ","/<i class=\'ng fairy name\'>([^<]*)<\/i>/"), array("www.irielion.com/israel/reggaename.html","phase=3&oldname=$F_firstname%20$F_lastname&gndr=$reggender","","Your Rasta Name is ","/Yes I, your irie new name is ([^\n]*)\n/"), array("www.ninjaburger.com/fun/games/ninjaname/ninjaname.php","realname=$F_firstname+$F_lastname","Your Ninja Burger Name is ","/<BR>Ninja Burger ninja name will be<BR><BR><FONT SIZE=\'\+1\'>([^<]*)<\/FONT>/"), array("gangstaname.com/pirate_name.php","sex=$F_gender&name=$F_firstname+$F_lastname","Your Pirate Name is ","/<p><strong>We\'ll now call ye:<\/strong><\/p> *<h2 class=\"newName\">([^<]*)<\/h2>/"), array("www.xach.com/nerd-name/","name=$F_firstname+$F_lastname&gender=$F_gender","Your Nerd Name is ","/<p><div align=center class=\"nerdname\">([^<]*)<\/div>/"), array("rumandmonkey.com/widgets/toys/namegen/5941/","page=2&id=5941&nametype=$dj&name=$F_firstname+$F_lastname","Your DJ Name is ","/My disk spinnin nu name is &lt\;b&gt\;([^<]*)&lt\;\/b&gt\;\./"), array("pizza.sandwich.net/poke/pokecgi.cgi","name=$F_firstname%20$F_lastname&color=black&submit=%20send%20","Your Pokename is ","/Your Pok&eacute;name is: <h1>([^<]*)<\/h1>/") ); return $hosts; ?>

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  • Display all images from outside web root folder using PHP

    - by micmola
    Hello, I want to display all images that are stored outside my web root folder. Please help me. I am only able to display one image repeatedly. For example, if I have 5 images in my folder, only one image is displayed on my browser 5 times. Please help me on this. I've been working on this problem for over a month now. I'm a newbie. Help. Thank you. Here is the code I'm using. images.php <?php // Get our database connector require("includes/copta.php"); // Grab the data from our people table $sql = "select * from people"; $result = mysql_query($sql) or die ("Could not access DB: " . mysql_error()); $imgLocation = " /uploadfile/"; while ($row = mysql_fetch_array($result)) { $imgName = $row["filename"]; $imgPath = $imgLocation . $imgName; echo "<img src=\"call_images.php?imgPath=" . $imgName . "\" alt=\"\"><br/>"; echo $row['id'] . " " . $imgName. "<br />"; } ?> call_images.php <?php // Get our database connector require("includes/copta.php"); $imgLocation = '/ uploadz/'; $sql = "select * from people"; $result = mysql_query($sql) or die ("Could not access DB: " . mysql_error()); while ($row = mysql_fetch_array($result)) { $imgName = $row["filename"]; $imgPath = $imgLocation . $imgName; // Make sure the file exists if(!file_exists($imgPath) || !is_file($imgPath)) { header('HTTP/1.0 404 Not Found'); die('The file does not exist'); } // Make sure the file is an image $imgData = getimagesize($imgPath); if(!$imgData) { header('HTTP/1.0 403 Forbidden'); die('The file you requested is not an image.'); } // Set the appropriate content-type // and provide the content-length. header("Pragma: public"); header("Expires: 0"); header("Cache-Control: must-revalidate, post-check=0, pre-check=0"); header("Content-Type: image/jpg"); header("Content-length: " . filesize($imgPath)); // Print the image data readfile($imgPath); exit(); } ?>

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  • Included php file calling Javascript function

    - by Illes Peter
    Hi there! Here's the deal. I've got index.php which links to an internal JS file in it's header. index.php then includes another .php file, which outputs this: + add file. addFile() is a Javascript function defined in the external JS file. By doing this nothing happens, the included php does not "see" the JS function. Encapsulating the JS in the included PHP makes it all work. But I don't want to do it that way. Any ideas? EDIT: here's the source <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.1//EN" "http://www.w3.org/TR/xhtml11/DTD/xhtml11.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <title>Archie</title> <meta http-equiv="Content-Type" content="text/html; charset=utf-8"/> <link rel="stylesheet" href="/screen.css" type="text/css" media="screen"/> <script src="/lib/js/archie.js" type="text/javascript"></script> </head> <body> ... ... //included php starts here <form action="/lib/course.php" method="post"> <fieldset> <div id="addFileLocation"></div> <a href="#" onClick="addFile()">+ add file</a> <input type="hidden" id="addFileCount" value="0"/> </fieldset> </form> //ends here ... ... </body> </html> and the js: <script type="text/javascript"> //Dynamically add form fields //add file browser function addFile() { var location = document.getElementById('addFileLocation'); var num = document.getElementById('addFileCount'); var newnum = (document.getElementById('addFileCount').value -1)+ 2; num.value = newnum; var newname = 'addFile_'+newnum; var newelement = document.createElement('input'); newelement.setAttribute('name',newname); newelement.setAttribute('type','file'); location.appendChild(newelement); } </script>

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  • php and asp problem in uploading

    - by moustafa
    i have an ASP web services to change byte array that given from the client and change it to a file and save it in the web server the code is like this : [WebMethod] public string UploadFile(byte[] f, string fileName) { try { MemoryStream ms = new MemoryStream(f); String path="/myfile/"; String location=HttpContext.Current.Server.MapPath(path); FileStream fs = new FileStream(HttpContext.Current.Server.MapPath(path)+fileName, FileMode.Create); ms.WriteTo(fs); ms.Close(); fs.Close(); return "OK"; } catch (Exception ex) { return ex.Message.ToString(); } } the web services need byte array and file name.. i build the client in php upload.php the code is <html> <body> <form action="action1.php" method="post" enctype="multipart/form-data"> Pilih File Anda: <input type="file" name="myfile" /> <input type="submit" value="Upload" /> </form> </body> <html> and action1.php the code is: <?php require_once('nusoap.php'); $client = new nusoap_client('http://192.168.254.160/testuploadah/FileUploader.asmx?WSDL', 'wsdl','','', '', ''); $err = $client->getError(); if ($err) { echo '<h2>Constructor error</h2><pre>' . $err . '</pre>'; } if(is_uploaded_file($_FILES['myfile']['tmp_name'])){ $uploadFile = $_FILES['myfile']; ////how can read byte array of $uploadFile so i can send to web services??? ////are php only can send array or string ? $params[]->f=??????????????? $params[]->fileName=$_FILES['myfile']['name']; $result = $client->call('UploadFile', $params,'', '', false, true); if ($client->fault) { echo '<h2>Fault</h2><pre>'; print_r($result); echo '</pre>'; } else { //Check for errors $err = $client->getError(); if ($err) { //// Display the error echo '<h2>Error</h2><pre>' . $err . '</pre>'; } else { //// Display the result echo '<h2>Result</h2><pre>'; print_r($result); echo '</pre>'; } } } ?> how can i Send the byte array parameter to the web services,so the web services can started???? i still can resolve this problem,the web services always return an error because i can't send byte array

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  • Downloading large file with php

    - by Alessandro
    Hi, I have to write a php script to download potentially large files. The file I'm reporting here works fine most of the times. However, if the client's connection is slow the request ends (with status code 200) in the middle of the downloading, but not always at the very same point, and not at the very same time. I tried to overwrite some php.ini variables (see the first statements) but the problem remains. I don't know if it's relevant but my hosting server is SiteGround, and for simple static file requests, the download works fine also with slow connections. I've found Forced downloading large file with php but I didn't understand mario's answer. I'm new to web programming. So here's my code. <?php ini_set('memory_limit','16M'); ini_set('post_max_size', '30M'); set_time_limit(0); include ('../private/database_connection.php'); $downloadFolder = '../download/'; $fileName = $_POST['file']; $filePath = $downloadFolder . $fileName; if($fileName == NULL) { exit; } ob_start(); session_start(); if(!isset($_SESSION['Username'])) { // or redirect to login (remembering this download request) $_SESSION['previousPage'] = 'download.php?file=' . $fileName; header("Location: login.php"); exit; } if (file_exists($filePath)) { header('Content-Description: File Transfer'); header('Content-Type: application/octet-stream'); //header('Content-Disposition: attachment; filename='.$fileName); header("Content-Disposition: attachment; filename=\"$fileName\""); header('Content-Transfer-Encoding: binary'); header('Expires: 0'); header('Cache-Control: must-revalidate, post-check=0, pre-check=0'); //header('Pragma: public'); header('Content-Length: ' . filesize($filePath)); ob_clean(); flush(); // download // 1 // readfile($filePath); // 2 $file = @fopen($filePath,"rb"); if ($file) { while(!feof($file)) { print(fread($file, 1024*8)); flush(); if (connection_status()!=0) { @fclose($file); die(); } } @fclose($file); } exit; } else { header('HTTP/1.1 404 File not found'); exit; } ?>

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  • how to increment a javascript variable title that is within a php while loop

    - by steve
    I'm building multiple countdown clocks on one page. The number of countdown clocks varies from day to day so I need to call javascript several times from within "while" code in php to produce different clocks. The following code works but it's based on knowing how many clocks are needed before I start: <script language="javascript" src="countdown.js"></script> <script language="javascript"> var cd1 = new countdown('cd1'); cd1.Div = "clock1"; cd1.TargetDate = "<?php echo "$clocktime"; ?>"; cd1.DisplayFormat = "%%D%% days, %%H%% hours, %%M%% minutes, %%S%% seconds until event AAA happens"; </script> <div id="clockwrapper"><div id="clock1">[clock]</div></div> <script language="javascript" src="countdown.js"></script> <script language="javascript"> var cd2 = new countdown('cd2'); cd2.Div = "clock2"; cd2.TargetDate = "02/01/2011 5:30:30 PM"; cd2.DisplayFormat = "%%D%% days, %%H%% hours, %%M%% minutes, %%S%% seconds until event BBB happens..."; </script> <div id="clockwrapper"><div id="clock2">[clock]</div></div> So if I keep on calling the javascript above (the code with cd1 in it) all previous "cd1" clocks change to the latest clock because it is being overwritten. Somehow I need to call javascript from within my "while" loop in php and have cd1 become cd2, then cd3 so that the clocks work as they're supposed to. How do I go about doing this? I don't know how to call the javascript several times and increment the variable cd1 within the javascript. I tried something like this but couldn't get it to work. $id=mysql_result($result,$i,"id"); while($id){ $cd = ("$cd"."$id"); ?> <script language="javascript" src="countdown.js"></script> <script language="javascript"> var <?php echo "$cd"; ?> = new countdown('<?php echo "$cd"; ?>'); .... </script> <div id="clockwrapper"><div id="<?php echo "$cd"; ?>">[clock]</div></div> <?php $id=mysql_result($result,$i,"id"); } ?> Surely there is some easy way of getting around this that I don't know about. Thanks

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  • How to send variable to php with ajax?

    - by Dee1983
    <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> <html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en"> <head> <script type="text/javascript"> function load(thediv, thefile) { if (window.XMLHttpRequest) { xmlhttp = new XMLHttpRequest(); } else { xmlhttp = new ActiveXObject('Microsoft.XMLHTTP)'); } xmlhttp.onreadystatechange = function() { if (xmlhttp.readyState==4 && xmlhttp.status==200) { document.getElementById(thediv) .innerHTML = xmlhttp.responseText; } } xmlhttp.open('GET', thefile, true); xmlhttp.send(); } </script> </head> <body> <?php //connection to db and mysql query $result = mysql_query($query) or die(mysql_error()); $options=""; while ($row=mysql_fetch_array($result)) { $id=$row["idProducts"]; $thing=$row["country"]; $options.="<OPTION VALUE=\"$id\">".$thing.'</option>'; } mysql_close(); ?> <SELECT id="countrySearch" NAME=countrySearch onchange="load('divtest', 'step2.search.php')";> <OPTION VALUE=0>Choose <?=$options?> </SELECT> <div id="divtest"> test </div> </body> step2.search.php consists of: <?php echo "I want it to store the users selection as a variable for php to use"; ?> The problem I have is I want to store what the user selects from the drop down box and use it in php to do a mysql query using the variable from the user select form to form the WHERE part of the mysql statement. Then use ajax to put new data in "divtest". How can I store the user selection into a variable then send it to be used in step2.search.php?

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  • Parse error in PHP login form

    - by user225269
    I'm trying to have a login form in php. But my current code doesnt work. Here is the form: <form name="form1" method="post" action="loginverify.php"> <td><font size="3">Username:</td> <td></td> <td><input type="text" name="uname" value="" maxlength="15"/><br/></td> <td><font size="3">Password:</td> <td></td> <td><input type="text" name="pword" value="" maxlength="15"/><br/></td> <tr> <td>&nbsp;</td> <td>&nbsp;</td> <td><input type="submit" name="Submit" value="Login" /></td> </form> And the verify.php <?php session_start(); ?> <?php $host="localhost"; $username="root"; $password="nitoryolai123$%^"; $db_name="login"; $tbl="users"; $connection=mysql_connect($host, $username, $password) or die("cannot connect"); mysql_select_db($db_name, $connection) or die("cannot select db"); $user=$_POST['uname'] $pass=$_POST['pword'] $sql="SELECT Username, Password from users where Username='$user' and Password='$pass'"; $result=mysql_query[$sql]; $count=mysql_num_rows($result); if($count==1){ $SESSION['Username']=$user; echo"<a href='searchmain.php'> CONTINUE</a>"; } else{ echo"wrong username or password"; echo"<a href='loginform.php'>Back</a>"; } ?> Is there something wrong with my code. I get this parse error at line 15, which is this: $pass=$_POST['pword'] But when I try to remove it.It goes to line 16 or line 17 again. What do I do

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  • Page not redirecting properly(php)

    - by user225269
    I want to do the login page this way so that I won't be having trouble posting the username in the userpage. But everytime I try to access login.php. I get an error in firefox, that the page is not redirecting properly. What do I do? This works when I separate them into two. Into something like, login.php and verifylogin.php as the form action. But if I do it like this, I get redirection errors: <?php $host="localhost"; $username="root"; $password="nitoryolai123$%^"; $db_name="school"; $tbl_name="users"; mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); $uname = mysql_real_escape_string($_POST['username']); $pword = mysql_real_escape_string($_POST['password']); $SQL = "SELECT * FROM users WHERE username = '$uname' AND password = '$pword'"; $result = mysql_query($SQL); $num_rows = mysql_num_rows($result); if ($result) { if ($num_rows > 0) { session_start(); $_SESSION['login'] = "1"; header ("Location: userpage.php"); } else { session_start(); $_SESSION['login'] = ""; header ("Location: login.php"); } } else { $errorMessage = "Error logging on"; } ?> <tr> <form name="form1" method="post" action="login.php"> <td> <table> <tr> <td><strong><font size="2">Login User</strong></td> </tr> <tr> <td width="30" height="35"><font size="2">Username:</td> <td width="30"><input name="username" type="text" id="username" maxlength="17"></td> </tr> <tr> <td width="30" height="35" ><font size="2">Password:</td> <td width="30"><input name="password" type="password" id="password" maxlength="17"></td> </tr> <td><td align="right" width="30"><input type="submit" name="Submit" value="Submit" /></td> <td><input type="reset" name="Reset" value="Reset"></td></td> </tr> </form> please help, thanks.

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  • How to customize BsGridView to show links instead of BsButtonColumn?

    - by felipe.zkn
    Given the code below, I need to customize the third column to show two links instead of that BsButtonColumn. I didn't find any related documentation to get the answer. <?php $this->widget( 'bootstrap.widgets.BsGridView', array( 'id' => 'activity-translation-grid', 'dataProvider' => $model->search(), 'filter' => $model, 'columns' => array( 'id', 'name', array( 'class' => 'BsButtonColumn', ), ), ) ); ?>

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  • PHP Network Monitoring

    - by Vlad Patrascu
    Is there a way that I can monitor the traffic, Upload/Download (separately) using PHP? I`d like to echo out something like that: Upload: 523 GB | Download: 25 GB This should be based on the System Uptime, so if I restart the computer, the count should restart. Thanks in Advance.

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  • PHP bcompiler install

    - by dobs
    How to install bcompiler on Fedora, have this error: [root@server server]# pecl install channel://pecl.php.net/bcompiler-0.9.1 downloading bcompiler-0.9.1.tgz ... Starting to download bcompiler-0.9.1.tgz (47,335 bytes) .............done: 47,335 bytes 10 source files, building running: phpize Configuring for: PHP Api Version: 20090626 Zend Module Api No: 20090626 Zend Extension Api No: 220090626 building in /var/tmp/pear-build-server/bcompiler-0.9.1 running: /var/tmp/bcompiler/configure checking for grep that handles long lines and -e... /bin/grep checking for egrep... /bin/grep -E checking for a sed that does not truncate output... /bin/sed checking for cc... cc checking whether the C compiler works... yes checking for C compiler default output file name... a.out checking for suffix of executables... checking whether we are cross compiling... no checking for suffix of object files... o checking whether we are using the GNU C compiler... yes .... /var/tmp/bcompiler/bcompiler.c:2174: ?????????: expected ‘struct zend_arg_info *’ but argument is of type ‘const struct _zend_arg_info *’ /var/tmp/bcompiler/bcompiler.c: ? ??????? ‘apc_serialize_zend_class_entry’: /var/tmp/bcompiler/bcompiler.c:3100: ??????????????: ???????????? ???????? ????????????? ?????????? ???? /var/tmp/bcompiler/bcompiler.c:3107: ??????????????: ???????? ????????? 1 ‘apc_serialize_zend_function_entry’ ???????? ????????????? ?????????? ???? /var/tmp/bcompiler/bcompiler.c:2874: ?????????: expected ‘struct zend_function_entry *’ but argument is of type ‘const struct _zend_function_entry *’ /var/tmp/bcompiler/bcompiler.c: ? ??????? ‘apc_deserialize_zend_class_entry’: /var/tmp/bcompiler/bcompiler.c:3324: ??????????????: ???????? ????????? 1 ‘apc_deserialize_zend_function_entry’ ???????? ????????????? ?????????? ???? /var/tmp/bcompiler/bcompiler.c:2900: ?????????: expected ‘struct zend_function_entry *’ but argument is of type ‘const struct _zend_function_entry *’ make: *** [bcompiler.lo] ?????? 1 ERROR: 'make' failed yum install bzip2-libs bzip2-devel - Fix it...

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  • Mac Os X 10.7 + PHP OCI8 + MAMP PRO

    - by Mike
    i followed this article: http://www.enavigo.com/2012/01/04/enabling-oracle-oci8-php-extension-on-os-x-snow-leopard/ and im getting this error: Error 324 (net::ERR_EMPTY_RESPONSE): The server closed the connection without sending any data. Anytime i try to connect to an oracle database using OCI8. I'm not sure that oci8 is install properly. Can someone provide steps for installation of OCI8 using Mamp Pro?

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  • Mac Os X 10.7 + PHP OCI8 + MAMP PRO + ERR_EMPTY_RESPONSE

    - by gorelative
    i followed this article: http://www.enavigo.com/2012/01/04/enabling-oracle-oci8-php-extension-on-os-x-snow-leopard/ and im getting this error: Error 324 (net::ERR_EMPTY_RESPONSE): The server closed the connection without sending any data. Anytime i try to connect to an oracle database using OCI8. I'm not sure that oci8 is install properly. Can someone provide steps for installation of OCI8 using Mamp Pro?

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  • downgrade PHP 5.3 to 5.2 on lenny

    - by Ron
    Unfortunately I did upgrade PHP to version 5.3, but it end up breaking up some web apps, now I'm trying to go back to 5.2. I removed both sources php53.dotdeb.org from /etc/apt/sources.list and I did apt-get update && apt-get upgrade, but it didn't downgrade anything. Any ideas on how to go back will be appreciated Thanks

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  • What files to backup on Lighttpd+MySQL+PHP server

    - by Tomaszs
    I have a VPS with CentOS 5. I would like to create backup of: all my config files tweaks of database, php, server a databases cron settings website files installed applications and their settings (?) What files should i take into account? I don't want to miss any file that will be necessary to restore fast my webserver in case of any failure. And I don't want to create whole backup because entire VPS has like 30 GB of data.

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  • Installing libcurl and libxml for PHP on windows

    - by mauro.dec
    Hello! Ive done a fair ammount of googling but didnt find any answers yet so I decided to ask here. I need to do some programming with PHP and Google Checkout. I need to install libcurl and libxml on a server running WAMP, but being a Linux user myself I have no idea how to install these libs. Has anyone here done this before? Thanks in advance.

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  • PHP ignores upload_tmp_dir?

    - by Matthias Vance
    LS, I am using IIS7 with PHP (FastCGI). I set up the upload_tmp_dir to "X:\Temp" instead of leaving it empty, but it's still using "C:\Windows\Temp" for some reason. I did give the following users full rights : NETWORK SERVICE, , IIS_IUSRS. I also restarted IIS after I made the change. Kind regards, Matthias Vance

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  • PHP & MySQL on Mac OS X: Access denied for GUI user

    - by Eirik Lillebo
    Hey! This question was first posted to Stack Overflow, but as it is perhaps just as much a server issue I though it might be just as well to post it here also. I have just installed and configured Apache, MySQL, PHP and phpMyAdmin on my Macbook in order to have a local development environment. But after I moved one of my projects over to the local server I get a weird MySQL error from one of my calls to mysql_query(): Access denied for user '_securityagent'@'localhost' (using password: NO) First of all, the query I'm sending to MySQL is all valid, and I've even testet it through phpMyAdmin with perfect result. Secondly, the error message only happens here while I have at least 4 other mysql connections and queries per page. This call to mysql_query() happens at the end of a really long function that handles data for newly created or modified articles. This basically what it does: Collect all the data from article form (title, content, dates, etc..) Validate collected data Connect to database Dynamically build SQL query based on validated article data Send query to database before closing the connection Pretty basic, I know. I did not recognize the username "_securityagent" so after a quick search I came across this from and article at Apple's Developer Connection talking about some random bug: Mac OS X's security infrastructure gets around this problem by running its GUI code as a special user, "_securityagent". Then I tried put a var_dump() on all variables used in the mysql_connect() call, and every time it returns the correct values (where username is not "_securityagent" of course). Thus I'm wondering if anyone has any idea why 'securityagent' is trying to connect to my database - and how I can keep this error from occurring when I call mysql_query(). Update: Here is the exact code I'm using to connect to the database. But a little explanation must follow: The connection error happens at a call to mysql_query() in function X in class_1 class_1 uses class_2 to connect to database class_2 reads a config file with the database connection variables (host, user, pass, db) class_2 connect to the database through the following function: var $SYSTEM_DB_HOST = ""; function connect_db() { // Reads the config file include('system_config.php'); if (!($SYSTEM_DB_HOST == "")) { mysql_connect($SYSTEM_DB_HOST, $SYSTEM_DB_USER, $SYSTEM_DB_PASS); @mysql_select_db($SYSTEM_DB); return true; } else { return false; } }

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  • How to troubleshoot performance issues of PHP, MySQL and generic I/O

    - by jbx
    I have a WordPress based website running on a shared hosting. Its response time is very decent (around 2s to retrieve the HTML page and 5s to load all the resources). I was planning to move it to a dedicated virtual server (Ubuntu 12.04 LTS), which should theoretically improve things and make them more consistent given its not shared. However I observed severe performance degredation, with the page taking 10seconds to be generated. I ruled out network issues by editing /etc/hosts on the server and mapping the domain to 127.0.0.1. I used the Apache load tester ab to get the HTML, so JS, CSS and images are all excluded. It still took 10 seconds. I have Zpanel installed on the server which also uses MySQL, and its pages come up quite fast (1.5s) and also phpMyAdmin. Performing some queries on the wordpress database directly through phpMyAdmin returns them quite fast too, with query times in the 10 to 30 millisecond region. Memory is also sufficient, with only 800Mb being used of the 1Gb physical memory available, so it doesn't seem to be a swap issue either. I have also installed APC to try to improve the PHP performance, but it didn't have any effect. What else should I look for? What could be causing this degradation in performance? Could it be some kind of I/O issue since I am running on a cloud based virtual server? I wish to be able to raise the issue with my provider but without showing actual data from some diagnosis I am afraid he will just blame my application. UPDATE with sar output (every second) when I did an HTTP request: 02:31:29 CPU %user %nice %system %iowait %steal %idle 02:31:30 all 0.00 0.00 0.00 0.00 0.00 100.00 02:31:31 all 2.22 0.00 2.22 0.00 0.00 95.56 02:31:32 all 41.67 0.00 6.25 0.00 2.08 50.00 02:31:33 all 86.36 0.00 13.64 0.00 0.00 0.00 02:31:34 all 75.00 0.00 25.00 0.00 0.00 0.00 02:31:35 all 93.18 0.00 6.82 0.00 0.00 0.00 02:31:36 all 90.70 0.00 9.30 0.00 0.00 0.00 02:31:37 all 71.05 0.00 0.00 0.00 0.00 28.95 02:31:38 all 14.89 0.00 10.64 0.00 2.13 72.34 02:31:39 all 2.56 0.00 0.00 0.00 0.00 97.44 02:31:40 all 0.00 0.00 0.00 0.00 0.00 100.00 02:31:41 all 0.00 0.00 0.00 0.00 0.00 100.00 My suspicion that this comes from I/O related issue is also because a caching plugin I use to reduce the amount of queries to the database, by precompiling PHP pages is actually making things worse instead of better. It seems that file access is making things worse instead.

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  • Compiling PHP with LDAP support on Ubuntu 12.10

    - by Andrew Ellis
    I am trying to compile PHP on Ubuntu 12.10 with LDAP support. I have run: apt-get install libldap2-dev That installs the header files to /usr/include. However, when attempting to compile it is unable to locate the header files. I have tried to with --with-ldap=/usr/include as well and it still fails with: configure: error: Cannot find ldap.h I also tried symlinking with the following and I still get the same error: ln -s /usr/lib/ldap* /usr/lib/ Thanks in advance for your help.

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  • PHP FastCGI/XML/DOM Configure

    - by James
    Guys, any ideas why when I configure PHP 5.3.1, these options fail? Notice: Following unknown configure options were used: --with-xml --with-dom --enable-fastcgi --enable-discard-path --enable-force-cgi-redirect

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