Tracking Forums, Newsgroups, Maling Lists
Home Scripts Tutorials Tracker Forums
  Advanced Search
  HOME    TRACKER    PHP


SuperbHosting.net have generously sponsored dedicated servers to ensure a reliable and scalable dedicated hosting solution for BigResource.com.





Warning: Mysql_result(): Unable To Jump To Row 0 On MySQL Result Index


i got this message:

Warning: mysql_result(): Unable to jump to row 0 on MySQL result index 6 in c:inetpubwwwrootDreamWeaverSiteswebadminwa.php on line 2296

for these lines of code:

$sql11="delete from Booking where Booking_Id='$Booking_Id'";
$result11=mysql_query($sql11);.




View Complete Forum Thread with Replies

Related Forum Messages:
Warning: Mysql_result() [function.mysql-result]: Unable To Jump To Row 0
i want to grab the value of a sql entry. but the thing is, sometimes this value is <NULL>. i thought, no biggy, i'll just have an if statement:

if (is_numeric(mysql_result($query, 0))) {
//do something
} else {
//do something else using mysql_result($category_id_query, 0);
}

however, this doe not work. i get this error: Warning: mysql_result() [function.mysql-result]: Unable to jump to row 0 on MySQL result index 3 is there a way to check to see if mysql_result() is <NULL> without throwing an error.

View Replies !
Unable To Jump To Row 0 On MySQL Result
im making a login program, and i get this error.

"Warning: Unable to jump to row 0 on MySQL result index 2 in /var/www/html/login.php on line 16"

Code:

View Replies !
Warning: MySQL: Unable To Save Result Set In
I'm running PHP/MySQL/Apache on my home system on Windows for testing purposes. As a run a specific query, I get this error:

Warning: MySQL: Unable to save result set in C:apachehtdocsfinal.php on line 129

Warning: Supplied argument is not a valid MySQL result resource in C:apachehtdocsfinal.php on line 132

I know the second result means the specified query cannot be found, and I'm guessing for some reason MySQL cannot save the query. Here is my code. PHP Code:

View Replies !
Warning: MySQL: Unable To Save Result Set
I'm getting this error:

Warning: MySQL: Unable to save result set in /www/hosts/wwwroot/mainPage.php on line 11

Here is what line 11 looks like:

$articles = mysql_query("select entryID from articles where siteID like 'senior'");

View Replies !
Mysql_fetch_array Error - Warning: 4 Is Not A MySQL Result Index
I have written a function to bulid a table of url links stored in MySQL. In my database I have two tables: tblLinkTypes and tblLinks. My Output should look like this:

Link Type1
Link Title2 URL2

Link Type2
Link Title1 URL1

However I get this error when the code gets to the bottom (see //errors here).
Warning: 4 is not a MySQL result index in /home/httpd/inc/dbWhiteWater.inc on line 275
Code:-

View Replies !
Warning: Smarty Error: Unable To Read Resource: "index.tpl" In
I am using "Smarty 2.1 Version" in php. It is giving following error

Warning: Smarty error: unable to read resource: "index.tpl" in
/home/httpd/tsils/parag/tradesite/Smarty.class.php on line 1095

the directory which is containing templates i.e. templates and templates_c is
having 777 permission.

Also tried with full absolute path in the Smarty.class.php. like
var $compile_dir = '/home/httpd/tsils/parag/tradesite/templates_c'

View Replies !
Unable To Jump To Row
I have a tariff which is atm 44 rows with 15 columns of numbers. Now i am trying to pull as individual price from the tariff based on user input: Code:

View Replies !
Unable To Jump Row 0
I am getting the following error:

Warning: mysql_result(): Unable to jump to row 0 on MySQL result index 2 in /home/content/i/n/n/**/html/tracker/connect.php on line 35

Here is my coding from line 34 on down. Code:

View Replies !
Pagination :: Unable To Jump To Row 0
I am running a script to find the "Previous" and "Next" records, the problem is when I get to record 1 I get the error below, same with when I reach the last record in the db.
PHP Code:

Warning: mysql_result(): Unable to jump to row 0 on MySQL result index 18

Here is the script running it, do I need to add to the query? or how can I use IF/ELSE?PHP Code:

<?php
$id = $_GET['id'];
// Retrieve your record according to id
// Display Record
// Links to Previous Records
$query = "SELECT id FROM jokes WHERE category = '$category' AND id < $id ORDER BY id DESC LIMIT 1" ; // For Previous
$result = mysql_query($query) ;
..........

View Replies !
Mysql Unable To Save Result Set In Phpdev Folder
i have this problem. i keep getting this error. Code:

mysql unable to save result set in phpdev folder search2.php on line 25 got error 127 from table handler

what this means? or how to solve it?

View Replies !
Error Message - 0 Is Not A MySQL Result Index
Could anyone explain what this might mean?

Warning: 0 is not a MySQL result index

I have a simple form set up where a user types in a keyword and then a page is displayed based on the keyword. See below:

HTML page:
<html>
<head>
<title></title>
</head>
<body>
<form action="submitform.php3" method="GET">
<p><strong><font face="verdana, arial, ms sans serif" size="1">id#</font>:</strong> <input
type="text" name="keyword" size="15" maxlength="25"> <input type="submit" value="Go!"> </p>
</form>
</body>
</html>

submitform.php3 page:
<?php
mysql_connect ("localhost", "username", "password");
mysql_select_db (dbname);
$result=mysql_query("SELECT $url FROM table
WHERE keyword LIKE '%$keyword%'");
list($url)=mysql_fetch_row($result);
?>

The concept is really simple. A user types in an keyword and that page is displayed. There is a database set up with a table that has two columns. One is "keyword" and the other is "url" For example row 1 the keyword is abc, and the url is http://www.abc.com.

View Replies !
Nested MySQL Queries Create Issue With Validity Of Result Index?
I am iterating through a result set to generate a second set of queries but
no matter what I do I get the error

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result
resource

even though if I echo the query to the browser and cut and paste it into the
command line I get valid results.

Do I have to store the first result in an array before doing the second set?

while ($i < mysql_num_rows($result)) {
$PIDs[$id]['ClientName'] = $client;
$PIDs[$id]['ProjectName'] = mysql_result($result, $i, 'Name');
$PIDs[$id]['PID'] = mysql_result($result, $i, 'PID');
$thisPID = $PIDs[$id]['PID'];
$PIDresult = ("SELECT * from hours WHERE PID = '$thisPID'") or
die(mysql_error());
$j=0;
while ($j < mysql_num_rows($PIDresult)) {
$PIDs[$id]['PID']['Comments'] = mysql_result($PIDresult, $j,
'Comments');
$j++;
}
$i++;
}
}

View Replies !
Warning: Mysql_fetch_array(): 9 Is Not A Valid MySQL Result Resource
I have just got a strange error when trying to construct my sql query using pagination:
Quote: Warning: mysql_fetch_array(): 9 is not a valid MySQL result resource in manage-buyers.php on line 54 Line 54 = PHP Code:

View Replies !
Warning: Supplied Argument Is Not A Valid MySQL Result Resource
<?
Warning: Supplied argument is not a valid MySQL result resource
if (mysql_num_rows($result) >0 )
!Didnt have this problem before


session_start();

if ($username && $password)
{
// if the user has just tried to log in

$db_conn = mysql_connect("localhost", "*****", "*******");
mysql_select_db("micro_db.adf", $db_conn);
$query = "select * from users "
."where name='$username' "
." and pass=password('$password')";
$result = mysql_query($query, $db_conn);
if (mysql_num_rows($result) >0 )
{
// if they are in the database register the username
$valid_user = $username;
session_register("valid_user");
}
}
?>

View Replies !
Warning: Mysql_fetch_row(): Supplied Argument Is Not A Valid MySQL Result
I removed the setcookie and ran the code and it shows this Warning.


Warning: mysql_fetch_row(): supplied argument is not a valid MySQL
result resource in /home/public_html/post1.php on line 77

This is the code again.

View Replies !
Warning: Mysql_fetch_array(): Supplied Argument Is Not A Valid MySQL Result
Here is my script:

<?php

$dbh=mysql_connect("localhost", "my_username", "my_password") or die ('No can do!');
mysql_select_db ("my_databasename");
$query = "SELECT * FROM contacts";
while($i = mysql_fetch_array($query)){
echo $i[first];
echo $i[last];
echo $i[email];
}
?>

Get this error message:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home2/my_username/my_domain/my_databasename/test.php on line 6.

View Replies !
Warning: Mysql_num_rows(): Supplied Argument Is Not A Valid MySQL Result
I am building an aunction site but have an error on this line. The error is:

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in C:Program FilesxampphtdocsAUNCTIOINindex.php on line 304 And like 304 is:

$num_auction = mysql_num_rows($result);

View Replies !
Warning: Mysql_data_seek(): Supplied Argument Is Not A Valid MySQL Result Resource
Warning: mysql_data_seek(): supplied argument is not a valid MySQL result resource

View Replies !
Warning: Mysql_fetch_array(): Supplied Argument Is Not A Valid MySQL Result Resource
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/xswapco/public_html/techsupport/account.php on line 41

This is the error I get when I try to run this code:

View Replies !
Warning: Mysql_fetch_assoc(): Supplied Argument Is Not A Valid MySQL Result Resource
Im trying to create a login form that ask for login name and password. once submited, the form submits to the same page again but assigns the value 1 to the URL varible. An If statement gets the URL varible and begins to check if the login name and password is correct. See script below

Below starts a session. if the user session is not defined the display the login form....

View Replies !
What Does This Error Mean? Warning: Mysql_fetch_array(): Supplied Argument Is Not A Valid MySQL Result
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/public_html/index.php on line 32. What does this error mean? What's causing it?

View Replies !
Warning Mysql_fetch_array(); Supplied Argument Is Not A Valid MySQL Result Resource?
i am getting the following error with this bit of code? Warning mysql_fetch_array(); supplied argument is not a valid MySQL result resource? Code:

<?php
/* Connect to database */

$extra = 'blah'
$size= &#391;'
$paper= 'matte'

$db= mysql_connect("localhost", "", "");
mysql_select_db("", $db);


$sql="select type_id FROM type WHERE extra=".$extra." AND size=".$size. " AND paper=".$paper ;
$type_id=mysql_query($sql);
............

View Replies !
Warning: Mysql_fetch_row(): Supplied Argument Is Not A Valid MySQL Result Resource
I up grade a module on my site and when i clicked on icon in admin panel it gave me this error-Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in /home/vash1/public_html/includes/sql_layer.php on line 299 im dyslexic and i did search thru forums before i posted at least till my eyes got worse -any ideas.

View Replies !
How To Solve The Error: Warning: Mysql_fetch_array(): Supplied Argument Is Not A Valid MySQL Result
I keep getting this:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in ....... 

while($row = mysql_fetch_array($result, MYSQL_ASSOC))

what can i do? i still new in php.

View Replies !
Warning: Mysql_num_rows(): Supplied Argument Is Not A Valid MySQL Result Resource In *** On Line 9
I keep getting an error result saying:

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in *** on line 9

Warning: Cannot modify header information - headers already sent by (output started at ***:9) in **** on line 12

I'm pretty new to PHP so I'm thinking its a problem with syntax. Any help is appreciated. Code is below.

<?PHP

$ni = trim($_POST['ni']);

require 'db_connect.php'

$sql = "SELECT * FROM $db_table WHERE ni = '$ni'";
$query = "mysql_query($sql) or die (mysql_error())";
if (mysql_num_rows($query) < 1) {

require 'db_close.php'
header("location: DESIRED LOCATION");
}
else{
require 'db_close.php'
header("Location: EXIT ");
}

?>

View Replies !
Warning: Mysql_fetch_array(): Supplied Argument Is Not A Valid MySQL Result Resource Code Error
im trying to access the 'update' time from my database by using the below code.

$result = mysql_query("SHOW TABLE STATUS FROM logtable;");
while($array = mysql_fetch_array($result)) {
print_r($array[Update_time]); // Will print information about each table stored in database

The only problem is I get this error. What am I doing wrong??

'Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in...'

View Replies !
Warning: Mysql_num_rows(): Supplied Argument Is Not A Valid MySQL Result Resource In /mounted-storage/
my code:

$username = mysql_real_escape_string($username);
$result= mysql_query("SELECT * from `admin_login WHERE username=".$username);
if (mysql_num_rows($result) != 0)
{
echo ("That Username is already being used. Please try a different username.");
}

I know that there is a row that has the same username as $username, but i keep getting an error like: Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /mounted-storage/home23a/sub001/sc21473-GRUR/www/inphp/members/register.php on line 56 Line 56 being:

if (mysql_num_rows($result) != 0)

View Replies !
Service Provider Problem: Warning: Mysql_fetch_array(): Supplied Argument Is Not A Valid MySQL Result
Strange one this, I have checked the forum and tried everything I can that is related and it still won't work.

I keep getting the following error on all my code:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource

I have just moved service provider and the code worked perfectly on my old site but no longer works with my new provider the code is as below. Code:

View Replies !
Warning: Mysql_fetch_array(): Supplied Argument Is Not A Valid MySQL Result Resource In /home/scaspco/public_html/testing/reviews.php On Line 41
I'm trying to outtput records from a mysql table and I get an error here

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/scaspco/public_html/testing/reviews.php on line 41
in

PHP Code:

<?php
$i = 0;
while ($result = mysql_fetch_array($query)) {
    if($i++ % 2) {
        $bgcolor = "category";
    } else {
        $bgcolor = "#DFE6EF";
    }
    print "<tr bgcolor="$bgcolor">";
    print "<TD>".$result['ASP_Name']."</TD>";
    print "<TD>".$result['Username']."</TD>";
...............

View Replies !
Warning: Mysql_fetch_row(): Supplied Argument Is Not A Valid MySQL Result Resource In /home/dyingtwi/public_html/functions/matchdb.php On Line 61
1: Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in /home/dyingtwi/public_html/functions/matchdb.php on line 61

2: Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in /home/dyingtwi/public_html/functions/playerdb.php on line 69

3: Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in /home/dyingtwi/public_html/functions/teamdb.php on line 71

View Replies !
Warning: Mysql_fetch_array(): Supplied Argument Is Not A Valid MySQL Result Resource In /home/szauerne/public_html/sites/yourcapecoral/wizard/listings.php On Line 364.
PHP Code:

$number = mysql_query("SELECT * $housesTable WHERE house_id='".$select."'");
$row = mysql_fetch_array($number);

returns

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/szauerne/public_html/sites/yourcapecoral/wizard/listings.php on line 364.

View Replies !
Warning: Ftp_put(): Unable To Access
I'm using the following code on my website, but something seems to be wrong, cause everytime I try to upload something, I get this error:

Warning: ftp_put(): Unable to access in /home/cjbcc/public_html/members/koerel/party/uploads.php on line 16

<html>
<head></head>
<body>
<?
$ftp = ftp_connect("koerel.cjb.cc");
$source_file = $_POST['bestand'];
$logged = ftp_login($ftp,"***","***");
if ((!$ftp) || (!$logged)){
echo "der is iets mis!";
} else {
echo "je bent ingelogd";
}
if(isset($_FILES['bestand'])) {
    ftp_put($ftp,"x.gif",$source_file,FTP_BINARY);
    echo "Het bestand is opgeslagen";
} else {
    echo "Selecteer een bestand";
}
?>
<form action="" method="post" enctype="multipart/form-data">
<input type="file" name="bestand"><br>
<input type="submit" name="submit" value="Upload">
</form>
</body>
</html>

View Replies !
Warning: Unable To Open 'php2' For Reading:
I using PHP 3.0.16, Apache Web Server. and MySQL 3.22.34 in my Win 98 machine. The problem that i had is when i want to upload a picture in php the error Warning: Unable to open 'php2' for reading:Unknown error in crogram filesapache groupapachehtdocsmansteradd.php3 on line 23.

The sample of my coding is paste below. Can somebody help me to figure out what is the problem. Code:

View Replies !
Warning: Stream_select(): Unable To Select [0]: No Error In ...
<?php
while (true) {
$read = array(STDIN);
$write = $except = NULL;
if (stream_select($read, $write, $except, 0)) {
}
}
?>

When I run this in PHP4, I get a "warning: stream select(): unable to
select [0]: No error" error. In PHP5, I don't.

View Replies !
Warning: (null)(): Unable To Obtain IDispatch Interface
I'm working on a site using COM to open word and excel documents, but i have an error with the CLSID, it says :

"Warning: (null)(): Unable to obtain IDispatch interface for CLSID {xxxxxxxx-xxxx-xxxx-xxxx-xxxxxxxxxxxx}: Server execution failed in monsite.php on line 21 Word could not be started"

and line 21 is : $word = new COM("word.Application",$server) or die("Word could not be started"); i work on a win2k server with IIS.

View Replies !
Warning: Unable To Open 'none' For Reading: No Such File Or Directory
I'm using copy($userfile,$dir.$userfile_name); to upload files and this works fine for small files. But if I have to upload large files (5-10mb) then I get the error message:

Warning: Unable to open 'none' for reading: No such file or directory

Are you not able to upload large files? Is there a fix to this?

View Replies !
PDFlib :: Warning: Dl(): Unable To Load Dynamic Library
I am a user of a Linux machine (not root) and I have installed PDFlib somewhere in my home
directory. So I can't change the '/etc/php.ini' file.

If I use the next statement "dl('!home/lib/PDFlib.so')" in my script I get the next error :

Warning: dl(): Unable to load dynamic library
'/usr/lib/php/extensions/local/usr/people/emami/lib/PDFlib.so' -
/usr/lib/php/extensions/local/usr/people/emami/lib/PDFlib.so: cannot
open shared object file

And it is understandable, because he are looking for a directory which is not exist.

View Replies !
Warning : Unknown(): Unable To Load Dynamic Library
I install php 4.3.1 on Windows XP (Use Apache 2.0.44) I open c:windowsphp.ini and uncomment line extension=php_gd2.dll When I restart Apache It has error pop up every time I open php file in browser like this:

Warning : Unknown(): Unable to load dynamic library './php_gd2.dll' - The specified module could not be found.

View Replies !
PHP Warning: Unknown(): Unable To Load Dynamic Library
I'm running a dedicated server with windows 2003 server with IIS, anyway recently a strange message started to pop out in my face each time I tried to access a page have query to the database.

---------------
PHP Warning: Unknown(): Unable to load dynamic library 'E:Program FilesSWsoftPleskAdditionalPhpextensionsphp_curl.dll' - Access is denied. in Unknown on line 0
-------------------

View Replies !
PHP Warning: Unable To Load Dynamic Library 'Chp4php_nsmail.dll'
I have my extention_dir = Chp4, and my extention=dllname all right, but when I try and load the page it says.

(for example)

PHP Warning: Unable to load dynamic library 'Chp4php_nsmail.dll' - One of the library files needed to run this application cannot be found.

All of the modules are located in that directory and for some reason it's still giving me this error, can anyone help?

View Replies !
PHP Warning: Unknown(): Unable To Load Dynamic Library Php_mhash.dll
I just downloaded PHP-4.3.3 for win32 and got the extensions folder and
dropped it intot he root of the PHP directory and set it's path in the
PHP.ini:

extension_dir = "c:/php/extensions/"

But I am still getting the error that it can't find the php_mhash.dll though
I can see that it is indeed in the folder....

View Replies !
Using Form To Post Info And Pic's (Warning: Unable To Fork)
I have created an admin form which allows the admin to update info and pic's to his site but I am having trouble with the update_pic script. When I pass along the form to the php script it gives me the following error message:

"Warning: Unable to fork [C:WINDOWSTEMPphp30.tmp /full/path/to/joesauto/images/anime_dvd_icongif.gif] in c:inetpubwwwrootadd_data.php on line 39"

This is line 39: exec("cp $picture /full/path/to/joesauto/images/$picture_name");
This is the form I used which was taken from this site. Code:

View Replies !
SquirrelMail Error: Warning: Fsockopen(): Unable To Connect To [ip Address]:143
If anybody has experienced the following SquirrelMail error:

Warning: fsockopen(): unable to connect to xx.xx.xx.xx:143 in
/home/virtual/siteX/fst/var/www/squirrelmail/functions/imap_general.php
on line 172

or something similar, then listen up, here's the solution. PHP 4.3.x
has known bugs with file handle and socket leak issues. This is true
up to PHP 4.3.3. The only solution for now is to revert to PHP 4.2.x.
I'm told that this issue only affects systems with 250+ SquirrelMail
sites.

View Replies !
Warning: Unable To Include 'cgi-bin/hello.cgi' - Request Execution Failed
I run PHP pages on my webserver. What I am trying to do is run a Perl script within a PHP page (and return the output to the webpage). The test script I am trying to run is this... Code:

#!/usr/bin/perl
print "Content-type: text/html";
print "HELLO WORLD!!<br>";

When I run this using <?php virtual("cgi-bin/hello.cgi") ?>, all I get is this error message....

"Warning: Unable to include 'cgi-bin/hello.cgi' - request execution failed"

Any ideas how I can run this script in my PHP page?

View Replies !
Warning: Unable To Create '/home/jj/picparty.com/users/ss/tmp/phproJvoQ': No
I am attempting to setup a file upload and I am getting an error message which is stumping me Here is the error message we get when we attempt to upload an image

Warning: Unable to create '/home/jj/picparty.com/users/ss/tmp/phproJvoQ': No
such file or directory in /home/jj/picparty.com/add.php3 on line 6 Code:

View Replies !
Fput Function (Warning: Unable To Find File Identifier 0 In)
I made a script for logging the emails that they send on my site. It works fine on windows 98 version, but when I upload it to a Linux computer it fails. Here are the first lines of the script:

$file = fopen("log/index.html", "a+");
$today = date("G:i, l jS Y");
if ($argv[0] == "contact"){
if ($name == ""){
fputs($file, "Text comes here";
fclose$file);}

Then the output is:

Warning: fopen("log/index.html","a+") - Permission denied in /home/jonathan/public_html/beftubbies/email.php3 on line 3

Warning: Unable to find file identifier 0 in /home/jonathan/public_html/beftubbies/email.php3 on line 30

View Replies !
Warning: Ldap_bind(): Unable To Bind To Server: Invalid Credentials In
I have some problems with ldap and php fot my authentification. I don't understand why. My DN is correct in my ldap server Can you help me please?

My php script :

$login = $_POST['pseudo'];
$passe = $_POST['passe'];
$ldapServer = "localhost";

$loginDN = "uid=".$login.", o=people, dc=toto, dc=org";
$connexion = ldap_connect($ldapServer)
or die ("Impossible de se connecter au seveur LDAP");

if ($connexion) {
$result= ldap_bind($connexion, $loginDN, $passe);
// Identification
if ($result) {
echo "Connexion LDAP réussie";
} else {
echo "Connexion LDAP échouée";
}
}

My error message :

Warning: ldap_bind(): Unable to bind to server: Invalid credentials in
d:www-rootphpauth_ldap.php on line 15 ...

View Replies !
Warning: Undefined Index: 2 In
wat does this error means?

Warning: Undefined index: 2 in C:Inetpubwwwroot emplateadmin.php on line 453

if i manually install php,(setting manually), this error wil not appear. but if i use the .exe to install, (means i dont' have to do the settings), this error will appear.

View Replies !
PHP Warning: PHP Startup: Unable To Load Dynamic Library 'c:PHPExtphp_mysql.dll'
Why do I keep getting this error mesgage on my phpinfo() page trying to
get PHP and MySQL to communicate? I am running this in IIS.

Here's what I have -

My php.ini resides in my C:WINNT directory. In it, I have removed the
'' from the extension=php_mysql.dll.

Also in the php.ini file I set extension_dir = c:PHPExt

In c:PHPExt I have the following files:

libmysqli.dll
libmySQL.dll
php5apache2.dll (Do I need this since I am using IIS)?
php5ts.dll
php_mysql.dll

View Replies !
Exec() Function Warning: Unable To Fork (Access Violation At 77F81B4D)
I am trying to execute a DOS command with the exec() function. All I get is fork errors or access errors, regardless of what file I try to execute. Example:

<?php
exec("sync.bat");
`sync.bat`;?>

Warning: Unable to fork [sync.bat] in C:testexec.php on line 3
Warning: Unable to execute 'sync.bat' in C:testexec.php on line 4
PHP has encountered an Access Violation at 77F81B4D


Id doesn't seem to matter what kind of file I try to execute, .exe, .bat or .com. Any ideas? Has ANYONE executed an external program under php/Windows?

View Replies !
PHP/MySql Jump Script
I have also read in a affiliate forum (quote below)that putting &afsrc=1 at the end of a link will help click through, how might this also be implemented? "Recently, I put the &afsrc=1 on all my links across my largest sites. Damned if sales have not increased by 10-20% since then. (it has been about 2 weeks at about 13000 clicks a day)"

<?php
$link = mysql_connect ("localhost", "user", "pass");
mysql_select_db ("datebase");
$result = mysql_query ("SELECT link FROM id WHERE number=$locn");
$row = mysql_fetch_array($result);
header("Location: " . $row["link"]);
mysql_close ($link);
?>

View Replies !
Checkbox Value - Result In An Undefined Index Error Message
I have some checkboxes which if you leave unchecked, result in an Undefined index error message. How can I get round this so there is some kind of default value that is entered apart from the checked or unchecked value: PHP Code:

<td>VDE (Volunteer Development England) Quality Mark</td>
        <td>&nbsp;</td>
        <td><input type="checkbox" name="list_quality_sys_29" value="Heard of this"></td>
        <td><input type="checkbox" name="list_quality_sys_30" value="Relates to organisation"></td>

View Replies !

Copyright © 2005-08 www.BigResource.com, All rights reserved