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

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Join Date: Nov 2007

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nav33n

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

#11

Dec 7th, 2007

 Help with Code Tags 
 php Syntax (Toggle Plain Text) 
$user_id=$_POST['UserID'];
$password=$_POST['Password'];
$query="select * from Customer where
UserID = '$user_id' and Password = '$password'";
echo $query;
$result=mysql_query($query) or die(mysql_error());
$user_id=$_POST['UserID'];
$password=$_POST['Password'];
$query="select * from Customer where
UserID = '$user_id' and Password = '$password'";
echo $query;
$result=mysql_query($query) or die(mysql_error());

Check what $query prints. Execute the same in phpmyadmin.

oh, and please, next time you post your code, put it in [code] tags.

Last edited by nav33n; Dec 7th, 2007 at 8:54 am.

Ignorance is definitely not bliss!

*PM asking for help will be ignored*

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Join Date: May 2009

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luqyan

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

#12

May 13th, 2009

y need some help pls
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in c:\xxxxxxxxx\index.php on line 7
No Template Selected

 Help with Code Tags 
 php Syntax (Toggle Plain Text) 
require("dbconfig.php");
$con =mysql_connect ($db_host,$db_user,$db_pass) or die("Error connecting");
mysql_select_db($db_name,$con);
$tem="select dir from template where active='1'";
$temr=mysql_query($tem,$con);
$temn=mysql_num_rows($temr);   line 7 
if($temn==0)
require("dbconfig.php");
$con =mysql_connect ($db_host,$db_user,$db_pass) or die("Error connecting");
mysql_select_db($db_name,$con);
$tem="select dir from template where active='1'";
$temr=mysql_query($tem,$con);
$temn=mysql_num_rows($temr);   line 7 
if($temn==0)

Last edited by peter_budo; May 15th, 2009 at 7:14 am. Reason: Keep It Organized - For easy readability, always wrap programming code within posts in [code] (code blocks) and [icode] (inline code) tags.

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

#13

May 13th, 2009

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Originally Posted by luqyan

y need some help pls
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in c:\xxxxxxxxx\index.php on line 7
No Template Selected

require("dbconfig.php");
$con =mysql_connect ($db_host,$db_user,$db_pass) or die("Error connecting");
mysql_select_db($db_name,$con);
$tem="select dir from template where active='1'";
$temr=mysql_query($tem,$con);
$temn=mysql_num_rows($temr); line 7
if($temn==0)

Excuse me. Did you not see the GIANT GLARING STICKIED POST that said "FAQ: Supplied argument not valid resource"? How about you go back and take a look at that.

GCS d- s+ a-->? C++(++++) UL+++ P+>+++ L+++ E--- W+++
N+ o K w++(---) O? !M- V PS+>++ PE+ Y+ PGP !t- 5? X- R tv+
b+>++ DI+ D G++>+++ e+ h+>++ r y+
PMs asking for help will not be answered, post on the forums. That's what they're there for.

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chirag87

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#14

32 Days Ago

$rsEmpresa = mysql_query($SQLEmpresa,$conexion) or die(mysql_error());
//add the text shown in BOLD, this will atleast display what error you are having

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Originally Posted by janzaldo

when im test in my local machine i dont have any error but when im trying to test in web i have this error, the same DB, the same tables, the same data in both structures any idea????

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/virtual/site244/fst/var/www/html/categoria.php on line 107
Help with Code Tags

PHP Syntax (Toggle Plain Text)
<?php
    require('conexion2.php');
    $SQLEmpresa= "SELECT empresa.cve_Empresa, Nombre, Direccion, pagina, telefono 
	             FROM empresa, empresa_Categoria  
		WHERE ".$_GET['cve_Categoria']." = empresa_Categoria.cve_Categoria 
		 AND empresa_Categoria.cve_Empresa = empresa.cve_Empresa";
    $rsEmpresa = mysql_query($SQLEmpresa,$conexion);
	$reg = array();
-->line 107	while ($reg = mysql_fetch_array($rsEmpresa))
	   {		
			$nombre =  $reg["Nombre"]; 
			$direccion =  $reg["Direccion"]; 
			$telefono =  $reg["telefono"]; 						
			$pagina = $reg["pagina"];
			$cve_Empresa = $reg["cve_Empresa"];	   					
?>
<?php require('conexion2.php'); $SQLEmpresa= "SELECT empresa.cve_Empresa, Nombre, Direccion, pagina, telefono FROM empresa, empresa_Categoria WHERE ".$_GET['cve_Categoria']." = empresa_Categoria.cve_Categoria AND empresa_Categoria.cve_Empresa = empresa.cve_Empresa"; $rsEmpresa = mysql_query($SQLEmpresa,$conexion); $reg = array(); -->line 107 while ($reg = mysql_fetch_array($rsEmpresa)) { $nombre = $reg["Nombre"]; $direccion = $reg["Direccion"]; $telefono = $reg["telefono"]; $pagina = $reg["pagina"]; $cve_Empresa = $reg["cve_Empresa"]; ?>

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network18

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#15

32 Days Ago

your query is failing for sure, echo the query and post it here.
or else execute the query directly inside the mysql and post ,if any error you got there

"The discipline of writing something down is the first step towards making it happen."

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srinivasjinde srinivasjinde is offline

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#16

21 Days Ago

please help me...even i m getting this error:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\xampp\htdocs\admin_loggedin_search_admin.php on line 35

[code=php]
<?php
$con = mysql_connect("localhost","root","");
if(!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("polling",$con);
$result = mysql_query("SELECT * FROM administrator WHERE $_POST[searchby] like '%$_POST[query]%'");
$found=0;
[LINE 35]while($sri=mysql_fetch_array($result))
{
if($sri['name']!="")
{
$found++;
echo "     ".$sri['name']."->".$sri['username']."<br />";
}
}
if($found!=0)
{echo "Found ".$found." results!";}
mysql_close($con)
?>
[code/]

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venkat0904 venkat0904 is offline

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#17

21 Days Ago

your query doesnt seems right to me... try this

 Help with Code Tags 
 PHP Syntax (Toggle Plain Text) 
$query = "SELECT * FROM administrator WHERE".$_POST[searchby]."like  '%".$_POST[query]."%'";
$result = mysql_query($query);
$query = "SELECT * FROM administrator WHERE".$_POST[searchby]."like  '%".$_POST[query]."%'";
$result = mysql_query($query);

always try to create n store ur query in a separate variable.. helps in debugging.
Hope this helps... cheers!!

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Originally Posted by srinivasjinde

please help me...even i m getting this error:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\xampp\htdocs\admin_loggedin_search_admin.php on line 35
Help with Code Tags

PHP Syntax (Toggle Plain Text)
<?php
$con = mysql_connect("localhost","root","");
if(!$con)
  { 
  die('Could not connect: ' . mysql_error());
  }
mysql_select_db("polling",$con);
$result = mysql_query("SELECT * FROM administrator WHERE $_POST[searchby] like '%$_POST[query]%'");
$found=0;
[LINE 35]while($sri=mysql_fetch_array($result))
{
 if($sri['name']!="")
 {
 $found++;
 echo "&nbsp;&nbsp;&nbsp;&nbsp; ".$sri['name']."->".$sri['username']."<br />";
 }
}
if($found!=0)
{echo "Found ".$found." results!";}
mysql_close($con)
?>
<?php $con = mysql_connect("localhost","root",""); if(!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("polling",$con); $result = mysql_query("SELECT * FROM administrator WHERE $_POST[searchby] like '%$_POST[query]%'"); $found=0; [LINE 35]while($sri=mysql_fetch_array($result)) { if($sri['name']!="") { $found++; echo "     ".$sri['name']."->".$sri['username']."<br />"; } } if($found!=0) {echo "Found ".$found." results!";} mysql_close($con) ?>

Last edited by venkat0904; 21 Days Ago at 1:48 am.

Gimme reputation points if u find my post helpful.
use [code] tags wherever applicable
dont start a new thread unless u cant find the topic already on forum.
mark a thread "solved" as soon as u get a solution

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srinivasjinde srinivasjinde is offline

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#18

21 Days Ago

@venkat
doesnt help...now i am getting an parse error in the line select * .......

are you sure of the syntax in the sql query??
one more thing....$_post[query] ... returns the string entered in a search box named query.....

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venkat0904 venkat0904 is offline

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#19

21 Days Ago

ohhh m so sorry... didn't include the spaces.. my bad.
just add a space after "where" clause and before "like".
should be--

 Help with Code Tags 
 PHP Syntax (Toggle Plain Text) 
$query = "SELECT * FROM administrator WHERE ".$_POST[searchby]." like  '%".$_POST[query]."%'";
$query = "SELECT * FROM administrator WHERE ".$_POST[searchby]." like  '%".$_POST[query]."%'";

n do make sure that ur "searchby" parameter feeds the column name exactly as in ur table..(upper n lower case are considered different)

try this n lemme know..
cheers!!

•

Originally Posted by srinivasjinde

@venkat
doesnt help...now i am getting an parse error in the line select * .......

are you sure of the syntax in the sql query??
one more thing....$_post[query] ... returns the string entered in a search box named query.....

Last edited by venkat0904; 21 Days Ago at 3:29 am.

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srinivasjinde srinivasjinde is offline

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#20

20 Days Ago

The prob. is still there....anyway the search functionality is working well....the only thing is that the warning msg doesnt look good....and when i do the first search by entering a query....i get the search results and the error msg disappears...