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  • How can I override mod-php5's .php mapping to php4-cgi per VirtualHost or Directory?

    - by geocoo
    I am running Debian Linux with apache2 and libapache2-mod-php5 5.3.3-7. I have one VirtualHost which requires php4. So I researched and compiled php4-cgi. However, I cannot seem to: Override mod-php5's mapping of .php in that vhost (or even globally, without disabling php completley). Even find where that mapping is made, in hope of disabling it and enabling mod-php5 or php4-cgi per vhost. This is my php4-cgi mapping (Inside the one php4 vhost): ScriptAlias /php4 /usr/local/php4/bin <Directory /usr/local/php4/bin> Options +ExecCGI +FollowSymLinks </Directory> <Directory /www/test> AddHandler php4-cgi-script .php Action php4-cgi-script /php4/php Options +ExecCGI </Directory> This does not work, mod-php5 still runs all .php files in that vhost/directory. If I change the file extension in the AddHandler above from .php to .php4, then .php4 files do run php4-cgi as expected, but I can't change all the files in the app to .php4. I thought maybe I could disable the mod-php5's mapping in my vhost or directory, then do my cgi-config (as above) but many combinations of these in different contexts did not work: RemoveHandler .php RemoveType .php php_flag engine off (this seems to even disable my php4-cgi so that wont work) The only other place I can find any mapping is in /etc/mime.types, but commenting out the relevant lines and restarting apache2 does not affect mod-php5's .php mapping. I have searched as much as I can, it is now a mystery to me. Any help or direction would be greatly appreciated.

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  • Installing PHP4 on a Debian (lenny) 7 32bit box

    - by Asim
    I am trying to install PHP4 on a Debian 7 32bit box but I ran into the following root@php4:~# apt-get update Get:1 http://snapshot.debian.org lenny Release.gpg [189 B] Hit http://snapshot.debian.org lenny Release Ign http://snapshot.debian.org lenny Release Hit http://snapshot.debian.org lenny/main Sources/DiffIndex Hit http://snapshot.debian.org lenny/main i386 Packages/DiffIndex Hit http://ftp.us.debian.org wheezy Release.gpg Hit http://security.debian.org wheezy/updates Release.gpg Ign http://snapshot.debian.org lenny/main Translation-en_US Ign http://snapshot.debian.org lenny/main Translation-en Hit http://ftp.us.debian.org wheezy Release Hit http://security.debian.org wheezy/updates Release Hit http://ftp.us.debian.org wheezy/main i386 Packages Hit http://ftp.us.debian.org wheezy/main Translation-en Hit http://security.debian.org wheezy/updates/main i386 Packages Hit http://security.debian.org wheezy/updates/main Translation-en Fetched 189 B in 0s (229 B/s) Reading package lists... Done W: GPG error: http://snapshot.debian.org lenny Release: The following signatures couldn't be verified because the public key is not available: NO_PUBKEY A70DAF536070D3A1 I did the following to fix it gpg --keyserver hkp://subkeys.pgp.net --recv-keys A70DAF536070D3A1 gpg --export --armor A70DAF536070D3A1 | sudo apt-key add - Now I get the following KEYEXPIRED error and unsure how to fix. Even Google does not help root@php4:~# apt-get update Get:1 http://snapshot.debian.org lenny Release.gpg [189 B] Hit http://snapshot.debian.org lenny Release Ign http://snapshot.debian.org lenny Release Hit http://snapshot.debian.org lenny/main Sources/DiffIndex Hit http://security.debian.org wheezy/updates Release.gpg Hit http://snapshot.debian.org lenny/main i386 Packages/DiffIndex Hit http://ftp.us.debian.org wheezy Release.gpg Hit http://security.debian.org wheezy/updates Release Ign http://snapshot.debian.org lenny/main Translation-en_US Hit http://ftp.us.debian.org wheezy Release Hit http://security.debian.org wheezy/updates/main i386 Packages Ign http://snapshot.debian.org lenny/main Translation-en Hit http://ftp.us.debian.org wheezy/main i386 Packages Hit http://security.debian.org wheezy/updates/main Translation-en Hit http://ftp.us.debian.org wheezy/main Translation-en Fetched 189 B in 0s (275 B/s) Reading package lists... Done W: GPG error: http://snapshot.debian.org lenny Release: The following signatures were invalid: KEYEXPIRED 1246455239 Any help?

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  • PHP Mystery Theatre - HTML Element doesn't display when served with php5, restart server with php4 a

    - by togglemedia
    Get this...I start the server in php5 and a specific HTML element ('a' element with a background image) is nowhere to be seen, I reboot the server in php4 and the HTML element is displaying properly. I boot back and forth between php5 and php4 with absolute consistent results, not displaying in php5 and displaying in php4. The thing that blows my mind is that: the HTML is consistent between boots of php5/4, so both scenarios have the necessary HTML elements the CSS is consistent between boots of php5/4, so both scenarios have the necessary CSS definitions there is an identically styled sibling element, with different class name that displays properly the problem is reproducible between browsers, between platforms. I've tested it on every possible config, Mac/Windows, IE6,7,8, FireFox, Safari etc... When the server is booted in php5 there is one HTML element that just doesn't display/render. A stab in the dark, I turned on PHP error reporting in php5 (to see if there would be any clues there) and low and behold the HTML element is now rendering in php5. I turn off PHP error reporting, and restart php5 and the HTML element is still rendering in php5, and the problem has been fixed...and I can't get the problem to reproduce. This is why my curiosity has brought me here. I just spent about four hours scratching my head trying to figure out how this could be. And now, I ask any php web dev gurus out there...what the heck was this all about?

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  • Best way to implement try catch in php4

    - by rgz
    What is the closest you can get to a try-catch block in php4? I'm in the middle of a callback during an xmlrpc request and it's required to return a specifically structured array no matter what. I have to error check all accesses to external resources, resulting in a deep stack of nested if-else blocks, ugly.

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  • PHP4 No Error When Missing Method is Called

    - by INTPnerd
    I have come across an annoying problem while writing some PHP4 code. I renamed the method of a class but I forgot to rename it where it was being called from. The annoying part is it was hard to track down where the problem was because no error was triggered. The script simply aborted leaving the web page partially rendered. Is it normal for this not to trigger an error or is there something wacky going on here? If this is normal is there a way to force this kind of thing to cause an error?

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  • PHP4: Send XML over HTTPS/POST via cURL?

    - by starmonkey
    I wrote a class/function to send xml over https via PHP4/cURL, just wondering if this is the correct approach, or if there's a better one. Note that PHP5 is not an option at present. /** * Send XML via http(s) post * * curl --header "Content-Type: text/xml" --data "<?xml version="1.0"?>...." http://www.foo.com/ * */ function sendXmlOverPost($url, $xml) { $ch = curl_init(); curl_setopt($ch, CURLOPT_URL, $url); // For xml, change the content-type. curl_setopt ($ch, CURLOPT_HTTPHEADER, Array("Content-Type: text/xml")); curl_setopt($ch, CURLOPT_POST, 1); curl_setopt($ch, CURLOPT_POSTFIELDS, $xml); curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1); // ask for results to be returned if(CurlHelper::checkHttpsURL($url)) { curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, 0); curl_setopt($ch, CURLOPT_SSL_VERIFYHOST, 0); } // Send to remote and return data to caller. $result = curl_exec($ch); curl_close($ch); return $result; } cheers!

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  • PHP4 including file during session

    - by Matherw
    I am trying to put second language on my webpage. I decided to use different files for different languages told apart by path - language/pl/projects.ln contains Polish text, language/en/projects.ln - English. Those extensions are just to tell language files from other, the content is simple php: $lang["desc"]["fabrics"]["title"] = "MATERIALY"; $lang["desc"]["fabrics"]["short_text"] = "Jakis tam tekst na temat materialów"; $lang["desc"]["services"]["title"] = "USLUGI"; $lang["desc"]["services"]["short_text"] = "Jakis tam tekst na temat uslóg"; And then on the index page I use it like so: session_start(); if (isset($_SESSION["lang"])) { $language = $_SESSION["lang"]; } else { $language = "pl"; } include_once("language/$language/projects.ln"); print $lang["desc"]["fabrics"]["title"]; The problem is that if the session variable is not set everything works fine and array item content is displayed but once I change and set $_SESSION["lang"] nothing is displayed. I tested if the include itself works as it should by putting print "sth"; at the beginning of projects.ln file and that works all right both with $_SESSION["lang"] set and unset. Please help.

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  • PHP4 HTTP Post without cURL

    - by Luke
    I have the following code that works on PHP5 to send a HTTP POST without using cURL. I would like this to work on PHP 4.3.0 and above: $opts = array('http' => array( 'method' => 'POST', 'header' => "Content-type: application/x-www-form-urlencoded\r\n" . "Content-Type: application/json\r\n", 'content' => $query ) ); $context = stream_context_create($opts); $result = file_get_contents($url, false, $context); HTTP context is only supported on PHP5. Is there anyway to make this work with PHP 4.3.0 - I need a fallback method if PHP5 or cURL is not installed.

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  • HTML Element doesn't display when served with php5, restart server with php4 and Element displays.

    - by togglemedia
    Get this...I start the server in php5 and a specific HTML element ('a' element with a background image) is nowhere to be seen, I reboot the server in php4 and the HTML element is displaying properly. I boot back and forth between php5 and php4 with absolute consistent results, not displaying in php5 and displaying in php4. The thing that blows my mind is that: the HTML is consistent between boots of php5/4, so both scenarios have the necessary HTML elements the CSS is consistent between boots of php5/4, so both scenarios have the necessary CSS definitions there is an identically styled sibling element, with different class name that displays properly the problem is reproducible between browsers, between platforms. I've tested it on every possible config, Mac/Windows, IE6,7,8, FireFox, Safari etc... When the server is booted in php5 there is one HTML element that just doesn't display/render. A stab in the dark, I turned on PHP error reporting in php5 (to see if there would be any clues there) and low and behold the HTML element is now rendering in php5. I turn off PHP error reporting, and restart php5 and the HTML element is still rendering in php5, and the problem has been fixed...and I can't get the problem to reproduce. This is why my curiosity has brought me here. I just spent about four hours scratching my head trying to figure out how this could be. And now, I ask any php web dev gurus out there...what the heck was this all about?

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  • PHP5 without PHP4 compatibility

    - by Serty Oan
    This is the opposite question of this one. My purpose is that there is no need when you write only PHP5 code to have PHP4 compatibility. It is not helpful at all and due to differences between object models in those two version, it must be penalizing on interpreter performances (or am I wrong ?). So my questions are : is it possible to recompile PHP5 without retrocompatibility with PHP4 or to disable it in any other way to gain performances ? if it is not possible, is there a project somewhere which aim that goal ?

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  • preg_match not defined in php4 in openwrt?

    - by user2723949
    I have installed the packages PHP4 and PHP4-CGI in openwrt through command line (opkg install) when I tried this simple code by putting this in the openwrt www folder, <?php $subject = "abcdef"; $pattern = '/^def/'; preg_match($pattern, $subject, $matches, PREG_OFFSET_CAPTURE, 3); print_r($matches); ?> I got this error Fatal error: Call to undefined function: preg_match() in /www/phptest.php on line 4 but php manual says preg_match() is defined in php4 also.. What might be the reason for this error? Is there any other PHP module that is to be installed? Or is there any alternate method for preg_match?

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  • Moving from php4 to php5

    - by fastcodejava
    I tried moving to php5, I am getting into lot of issues. This class gives error : Class DatabaseConnection { // error here private $connection; private $result; // public function __construct() { $this->databaseName = $GLOBALS['configuration']['db']; } // other methods follow }

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  • Reliable strtotime() result for different languages

    - by Maksee
    There was always a strange bug in Joomla when adding new article with back-end displayed with a language other than English (for me it's Russian). The field "Finish Publishing" started to be current date instead of "Never" equivalent in Russian. For a site in php4 finally found that strtotime function returns different results for arbitrary words. For "Never" it always -1 and joomla relies on this result in the JDate implementation. But in other case it sometimes returns a valid date. For russian translation of Never (???????) it is the case, but also for single "N" it is the case, so if one decided to change the string to some other he or she would face the same issue. So the code below <?php echo "Res:".strtotime("N")."<br>"; echo "Res:".strtotime("Nev")."<br>"; echo "Res:".strtotime("Neve")."<br>"; echo "Res:".strtotime("Never")."<br>"; ?> Outputs: Res:1271120400 Res:-1 Res:-1 Res:-1 So what are the solutions would be in this case? I would like not to write language-specific date.php handler, but to modify date method of JDate class, but what are language-neutral changes would be in order to detect invalid string. Thank you

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  • php foreach question

    - by user295189
    I have the following code $oldID=-1; $column=0; foreach($pv->rawData as $data){ if ($oldID!= $data->relativeTypeID){ $oldID= $data->relativeTypeID; $column++; $row=1; } echo "Row: ".$row.": Column: ".$column.": ID".$data->relativeTypeID."<br>"; //if exists a description if($data->description){ //insert here in the array $pv->results[$data->relativeTypeID][$row][0]= $data->relation; $pv->results[$data->relativeTypeID][$row][1]= ''; $pv->results[$data->relativeTypeID][$row][2] =''; $pv->results[$data->relativeTypeID][$row][3] = ''; $row++; } } this generates this output Row: 1: Column: 1: ID1 Row: 2: Column: 1: ID1 Row: 1: Column: 2: ID2 Row: 2: Column: 2: ID2 Row: 3: Column: 2: ID2 Row: 4: Column: 2: ID2 Row: 5: Column: 2: ID2 Row: 6: Column: 2: ID2 Row: 7: Column: 2: ID2 Row: 8: Column: 2: ID2 Row: 9: Column: 2: ID2 Row: 10: Column: 2: ID2 Row: 11: Column: 2: ID2 Row: 1: Column: 3: ID3 Row: 1: Column: 4: ID4 Row: 1: Column: 5: ID8 Row: 2: Column: 5: ID8 Row: 3: Column: 5: ID8 Row: 1: Column: 6: ID10 Row: 2: Column: 6: ID10 Row: 3: Column: 6: ID10 Row: 4: Column: 6: ID10 ... .. what I want it to do is to stop at the top 4 columns so I want an output like this Row: 1: Column: 1: ID1 Row: 2: Column: 1: ID1 Row: 1: Column: 2: ID2 Row: 2: Column: 2: ID2 Row: 3: Column: 2: ID2 Row: 4: Column: 2: ID2 Row: 5: Column: 2: ID2 Row: 6: Column: 2: ID2 Row: 7: Column: 2: ID2 Row: 8: Column: 2: ID2 Row: 9: Column: 2: ID2 Row: 10: Column: 2: ID2 Row: 11: Column: 2: ID2 Row: 1: Column: 3: ID3 Row: 1: Column: 4: ID4 as you can see it stopped at column 4. Thanks

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  • GD DLL installation

    - by smokinguns
    I'm using the JpGraph library(PHP graphing Lib). I'm getting the foll error: "JpGraph Error Your PHP installation does not seem to have the required GD 2.x library enabled. Please see the PHP documentation, "Image" section. Make sure that "php_gd2.dll" statement is uncomment in the [modules] section in the php.ini file" I uncommented the php_gd2.dll statement in php.ini(in C:\WINDOWS) and set the 'extension_dir' as extension_dir = C:\PHP\extensions (I have a extensions directory and put the php_gd2.dll in this directory ). I restarted IIS and when I fire up my webapp, it doesnt comeup. I commented out the php_gd2.dll and restarted IIS again. The webapp came up. I dont understand what I'm doing wrong. Here is the server config: Windows Server 2003, with IIS Version 6. The PHP version is 4.3.8

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  • why this sql not working?

    - by user295189
    I have a query public static function TestQuery( $start=0, $limit=0){ $sql = " SELECT count(*) AS total FROM db.table1 JOIN db.table2 ON table1.fieldID = {$fieldID} AND table2.assigned = 'N'"; $qry = new SQLQuery; $qry-query($sql); if($row = $qry-fetchRow()){ $total = intval($row-total); } return $total; } which works fine but when I add the limit as below, then it doesnt work and gives me errors public static function TestQuery( $start=0, $limit=0){ $sql = " SELECT count(*) AS total FROM db.table1 JOIN db.table2 ON table1.fieldID = {$fieldID} AND table2.assigned = 'N'"; //this fails if($recordlimit 0) $sql .= "LIMIT {$startRecord}, {$recordLimit} "; // $qry = new SQLQuery; $qry-query($sql); if($row = $qry-fetchRow()){ $total = intval($row-total); } return $total; } Any help will be appreciated

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  • php random array

    - by user295189
    I am trying to get a random array like this srand((float) microtime() * 10000000); $input = array("Neo", "Morpheus", "Trinity", "Cypher", "Tank"); $rand_keys = array_rand($input, 2); echo $input[$rand_keys[0]] . "\n"; echo $input[$rand_keys[1]] . "\n"; it shows 2 numbers randomly if I have $rand_keys = array_rand($input, 2);$rand_keys = array_rand($input, 5); but since I want all 5 to show it doesnt work. whats causing that. I need to use array_rand. thanks

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  • Importing Variables from PHP 4 into Flash

    - by Glenn
    I'm trying to get variable importing from a PHP script before I implement it into a larger project. So far all I've gotten is headaches. //this is all thats in test.php other than the open and close brackets. Normally I'd have it doing a mysql_query and putting useful information into the print statement. print( "lamegame.net/test/test.php?val=foo&id=0000&name=Glenn"); All test.as has to do is access the three variables. The problem comes in that 'val' is undefined. The id and name variables however are just fine and return 0000 and Glenn respectively. package { import flash.display.MovieClip; import flash.text.TextField; import flash.events.; import flash.net.; public class test extends MovieClip { public function test() { super(); //prep request var request:URLRequest = new URLRequest("test.php"); request.method = URLRequestMethod.GET; var loader:URLLoader = new URLLoader(); //load request loader.dataFormat = URLLoaderDataFormat.VARIABLES; loader.addEventListener(Event.COMPLETE, dataLoaded); loader.load(request); } public function dataLoaded(e:Event):void { var name = e.target.data.name; var id = e.target.data.id; var val = e.target.data.val; trace("Val: " + val + "ID: " + id + " Name: " + name); } } }

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  • Intern working for Indian NGO - Help with PHP 4, advising staff

    - by Kevin Burke
    Hello, For the past three months I've been working for an Indian NGO (http://sevamandir.org), doing some volunteer work in the field but also trying to improve their website, which needs a ton of work. Recently I've been trying to fix the "subscribe to newsletter" button, which is broken. I used filter_var to filter the email input, but when I tried to test this out I got an error. Then I learned that the web host is still using php version 4.3.2 and register_globals is turned on. I've mentioned that they should upgrade their web host before (they are paying around $50 per year for Rediff Web Hosting, complete with 100MB storage and 1 MySQL database). That would add a lot of complexity for the IT staff of 3, who would have to update everyone's email information (I assume? this is a 250-person organization), and have me find a new web host and teach them about it. The staff isn't that sophisticated about web usage - the head guy still uses IE6, and the website's laid out in tables (they use Dreamweaver WYSIWYG to lay out pages). So I've got two options - use regular expressions to filter the email, which I'm not that skilled at doing (and would be more vulnerable to exploitation after I leave), turn off register globals and then try to teach the staff what I'm doing, or try to get them to upgrade their versions of PHP and MySQL and/or change web host. I'd appreciate some advice. Thanks for your help, Kevin

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  • working with arrays

    - by user295189
    I currently do a query which goes through the records and forms an array. the print_r on query gives me this print_r($query) yields the following: Array ( [0] = ( [field1] = COMPLETE [field2] = UNKNOWN [field3] = Test comment ) [1] = ( [field1] = COMPLETE [field2] = UNKNOWN [field3] = comment here ) [2] = ( [field1] = COMPLETE [field2] = UNKNOWN [field3] = checking ) [3] = ( [field1] = COMPLETE [field2] = UNKNOWN [field3] = testing ) [4] = ( [field1] = COMPLETE [field2] = UNKNOWN [field3] = working ) ) somehow I want to take this array and convert it back to php. So for example some thing like this $myArray = array( ...) then $myArray should yield the samething as the print_r($query) yeilds. Thanks

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  • CodeIgniter not working on PHP 4 server without .htaccess availability

    - by Rasiel
    Hi All, Basically i developed my app on a localhost wamp server with PHP 5. ON uploading to the actual host i notice that The server is running php 4.4.9 Everytime i upload my .htaccess file, the server removes it completely.. seems to not be allowed When i test out the set all i get is a 404 page not found Any help on how to make it work on this PHP 4 server?

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