mysql_fetch_array(): supplied argument is not a valid MySQL result resource in c:\php

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April 11th, 2003, 08:14 AM

sarahnavas2003

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mysql_fetch_array(): supplied argument is not a valid MySQL result resource in c:\php

Need help this keeps running an error. Anyone know why?
Thanks

<?php
while($row = mysql_fetch_array($result))
{
?>

April 13th, 2003, 05:48 AM

jaguar_wolf

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try this

PHP Code:


		
			
<?

$sql = "SELECT * FROM `mytable` WHERE `id` = '1'";



db_connect();//connecting mysql db function i wrote



$result = mysql_query($sql) or die(mysql_error());



while($row = mysql_fetch_array($result)){



 // do some thing here



} 



mysql_close(db_connect());

?>

April 14th, 2003, 10:54 AM

torrent

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Re: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in c:\php

Quote:

Originally posted by sarahnavas2003
Need help this keeps running an error. Anyone know why?
Thanks

<?php
while($row = mysql_fetch_array($result))
{
?>

Your SQL query is failing. Do what jaguar_wolf says and use the or die mysql_error();. This will display the problem.

July 29th, 2004, 06:00 PM

tim_ver

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question

I am new to this and have the same error. When I goto put this code in

<?
$sql = "SELECT * FROM `mytable` WHERE `id` = '1'";

db_connect();//connecting mysql db function i wrote

$result = mysql_query($sql) or die(mysql_error());

while($row = mysql_fetch_array($result)){

// do some thing here

}

mysql_close(db_connect());
?>

Do I need to remove other code or edit anything? I want to get this right
thanks for the help.

January 21st, 2005, 08:15 PM

denisecamargo

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Hi all ! I got this code for showing random buttons on my personal weblog. but I keep getting that error. Any help ?

Code:

<? include("config.php");
 $linkcount = 6;
 ?>
 <?
 $result=mysql_query ("SELECT * FROM $table_link ORDER BY RAND() LIMIT $linkcount");
 if ($row=mysql_fetch_array($result)) {
 do {
 $width=$row["width"];
 $height=$row["height"];
 if ($width < 10){
 $insert_width = "";
 } else { $insert_width=" width=\"$width\""; }
 if ($height < 10){
 $insert_height = "";
 } else { $insert_height=" height=\"$height\""; }
 ?>
 <a href="<?=$row["url"]?>" target="new"><img border="0" src="<?=$base_url?><?=$row["image"]?>" alt="<?=$row["name"]?>"<?=$insert_width?><?=$insert_height?>></a>
 <?
 } while($row = mysql_fetch_array($result));
 } else {print "There are no links in this category.";}
 ?>

And this is the include

Code:

<? include ("/home/canela/public_html/fan/random.php") ?>

the buttons are within a subdomain on my personal domain.

Thanks.

January 24th, 2005, 12:04 PM

Madpawn

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Here's what's going on with 'not a valid result resource' errors. Take the following example:

PHP Code:


		
			
 $query = 'SELECT * from table';
   
   $result = mysql_query($query);
   
   while($row = mysql_fetch_array($result))
   { echo $row['foo']; }

In the above, the mysql_query() function is sending the specific query to the database and returning a 'resource identifier', which is being set as the value of $result. These are not the results themselves -- if you echo $result at this point, you'll get 'Resource id #x'. The mysql_fetch_array() function in your while loop takes this resource identifier and creates an array of the actual db contents that you can use.

If your mysql_query() function fails to get a resource id, it returns a boolean FALSE instead. FALSE, of course, isn't a resource identifier, so when you try to pass it to mysql_fetch_array(), it will cough up the 'not a valid result resource' error. So all of you are getting failures in your mysql_query() function.

Using 'or die(mysql_error())' at the end of your mysql_query() call, like jaguar_wolf posted, will, instead of setting $result to FALSE, kill the script and, more importantly, return a more descriptive error message.

Probably 80%+ of the time, a failed mysql_query() is caused by a syntax error in the query itself. Be especially mindful of any variables you're using in your query -- if they're not being set correctly, they can cause a syntax error. A failure to connect to the database can also cause mysql_query() to fail, so keep that in mind. Either way, the mysql_error() function will give you more information.

Comments on this post

rootcat agrees: A very detailed post which let me solve the problem.Thanks

January 26th, 2005, 06:47 AM

angryopium

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Hello all,
I got a strange case...
I use a getElements method to which I pass the table and the name of the field I want to sort results with...

here's the code:

Code:

function getElements($table, $sort){
 		$query = "SELECT * FROM $table ORDER BY $sort";
 		$connection = mysql_connect($this->HOST, $this->USERNAME, $this->PASSWORD);
 		$SelectedDB = mysql_select_db($this->DBNAME);
 		$result = mysql_query($query) or die(mysql_error()); 
 		$i=0;
 		while ( $element = mysql_fetch_array($result) ) {
 				$elements[$i] = $element;
 			  $i++;
 		}
 		return($elements);
 	}

OK... I use this method in a main page:

Code:

require_once("eMgr.php"); // the php class
 $mgr=new eMgr(); // instantiation
 // Management code:
 $groups=$mgr->getElements("gruppi", "descrizione");
 	if (count($groups > 0)){
 	  foreach($groups as $group){
 ?>
   <tr>
 	<td>
 	<td align="center">[mod] - [del]
   </tr>
 <?		
 	  }
 	}

and everything is ok...

now I open a new window (javascript) to insert a new group and I use again the same method to fill the options of html select. The code in the new window is:

Code:

require_once("eMgr.php");
 $m = new eMgr();
 $groups = $m->getElements("gruppi","descrizione");

the error returned is:
No Database Selected

I really don't know what's wrong! I tryed closing connection and setting $result to NULL .... But the error is always the same!

Any suggestions???

Thx

January 26th, 2005, 09:21 AM

MadCowDzz

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this sounds like a new question.
You might get better results making a new thread for it.

March 11th, 2008, 08:20 PM

toxinhead

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GUYS mac

#10

February 15th, 2009, 01:39 AM

Sergiu(^_^)

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Quote:

Originally Posted by sarahnavas2003

Need help this keeps running an error. Anyone know why?
Thanks

<?php
while($row = mysql_fetch_array($result))
{
?>

you might have forgoten a " " space in you query...
I're run into this problem shortly after I watched some vid. tutorials and that was the problem( took me 1 h to figure it out )

.

example:
...
$select = "SELECT X,Y";

$from = "FROM TABLE";

$query = $select.$from;

$result = mysql_query($query);

while($row = mysql_fetch_array($result))
...

so try to put a space before the closing qoute in each query..select,from, etc.

hope this helps

#11

June 27th, 2009, 03:11 PM

Kodiak3000

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Thanks!

Quote:

Originally Posted by Madpawn

Using 'or die(mysql_error())' at the end of your mysql_query() call, like jaguar_wolf posted, will, instead of setting $result to FALSE, kill the script and, more importantly, return a more descriptive error message.

Man, that so saved my butt. All it was was a capital letter ('Products' instead of 'products'), and since I didn't write the database, I didn't see it, but without the much clearer error message, I'd have been looking for it all day.

Thanks a ton!

#12

August 30th, 2009, 04:24 PM

BrandNew

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Similar issue, I have been trying...

to correct it myself using your advice, but since I am relatively new, perhaps I just do not have enough knowledge to see where the problems lie.

The following code -

php Code:

Original - php Code
function symbol_currency(){ $symbol_row=mysql_fetch_array(mysql_query("select symbol from site_var where status='1'")); if($symbol_row['symbol']!='') return $symbol_row['symbol']; else return '$'; }

function symbol_currency(){
 
$symbol_row=mysql_fetch_array(mysql_query("select symbol from site_var where status='1'"));
if($symbol_row['symbol']!='')
return $symbol_row['symbol'];
else
return '$';
}

is returning the error -

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

Any help would be greatly appreciated.

Last edited by Itsacon : August 31st, 2009 at 02:45 AM. Reason: added code-tags

#13

August 31st, 2009, 02:47 AM

Itsacon

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Most likely, the mysql_query() call is failing for some reason. Split the row into two seperate commands and add error checking in between the query and the result-handling (Never assume a function will go right. Lots of stuff can go wrong).

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#14

November 7th, 2009, 03:37 PM

beazleybub

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Thank you

I know this is an old thread but I am new to php and personally wanted to say thank you to jaguar_wolf.

or die mysql_error(); taught me a new thing!

#15

February 22nd, 2010, 04:24 PM

vanillashake

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I had the same error but my problem was in the query:

$sql = "SELECT * FROM table WHERE column1= ' " . $_GET['name'] . " ' ";

I used only column1 = ' . $_GET['name'] . ' when I was suposed to open another set of " ".

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