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  #1  
Old April 11th, 2003, 08:14 AM
sarahnavas2003 sarahnavas2003 is offline
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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))
{
?>

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  #2  
Old April 13th, 2003, 05:48 AM
jaguar_wolf jaguar_wolf is offline
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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());
?>

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  #3  
Old April 14th, 2003, 10:54 AM
torrent torrent is offline
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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.

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  #4  
Old July 29th, 2004, 06:00 PM
tim_ver tim_ver is offline
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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.


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  #5  
Old January 21st, 2005, 08:15 PM
denisecamargo denisecamargo is offline
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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.

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  #6  
Old January 24th, 2005, 12:04 PM
Madpawn Madpawn is offline
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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

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  #7  
Old January 26th, 2005, 06:47 AM
angryopium angryopium is offline
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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

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  #8  
Old January 26th, 2005, 09:21 AM
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MadCowDzz MadCowDzz is offline
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this sounds like a new question.
You might get better results making a new thread for it.

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  #9  
Old March 11th, 2008, 08:20 PM
toxinhead toxinhead is offline
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GUYS mac

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  #10  
Old February 15th, 2009, 01:39 AM
Sergiu(^_^) Sergiu(^_^) is offline
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Cool

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

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  #11  
Old June 27th, 2009, 03:11 PM
Kodiak3000 Kodiak3000 is offline
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Thumbs up 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!

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  #12  
Old August 30th, 2009, 04:24 PM
BrandNew BrandNew is offline
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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
  1. function symbol_currency(){
  2.  
  3. $symbol_row=mysql_fetch_array(mysql_query("select symbol from site_var where status='1'"));
  4. if($symbol_row['symbol']!='')
  5. return $symbol_row['symbol'];
  6. else
  7. return '$';
  8. }


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

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  #13  
Old August 31st, 2009, 02:47 AM
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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  
Old November 7th, 2009, 03:37 PM
beazleybub beazleybub is offline
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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!


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  #15  
Old February 22nd, 2010, 04:24 PM
vanillashake vanillashake is offline
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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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  #16  
Old May 13th, 2010, 10:27 AM
zedo4 zedo4 is offline
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Need help !

Please help me I have an error executing the script
mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/xxxxx/xxxxx/xxxxxxxx/include/bit_functions.inc.php on line 4

php code:

<?php
function symbol_currency(){

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

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  #17  
Old May 22nd, 2010, 12:57 PM
carlightexpress carlightexpress is offline
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I have the same error, however I know why it is failing, but I do not know what is the cause for the failing,

$subcat=$_POST['subcat'];
$results=array();
$result=mysql_query("SELECT * FROM `subcategory` where `subcategory`=$subcat") or die(mysql_error());
while($a_row = mysql_fetch_array($result, MYSQL_ASSOC)) array_push($results, $a_row);

Say for example, the returned value for $subcat is "145,1.9JTD,105 BHP"

The SQL Query decides to look for "1.9JTD,105 BHP" instead which obviously returns no results, I do not know why this is happening, in the previous page for the drop down list it displays perfectly, any ideas anyone?

I did an echo on $subcat and got the correct value "Value of $subcat = 145,1.9JTD,105 BHP "

So why is the first part of the value being truncated when performing the query?

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  #18  
Old August 24th, 2011, 02:06 PM
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Hi I'm, new here and don't have much coding experience. For my job I need to put a blog feed at the bottom of our homepage. I can do that, but then it causes me to get this error at the bottom where links used to be. The mysql works without the feed, but not with it and I don't know why.

I read most of the posts in here and tried or die mysql_error(); and whatnot but it still gives me the same general error.

Here is the code:

PHP Code:
<?
$dbnavrat 
mysql_query("SELECT * FROM t_pages_link WHERE position='footer' ORDER BY vaha");
while(
$h mysql_fetch_array($dbnavrat)){
  
$name getPageLink($h[id]);
  
$link $h["link"];
  if (
substr($link,0,4) != "http"){$link $server.$link;}
  echo 
"<li><a href=\"$link\" title=\"$name\">$name</a></li>\n";
}

?>


Any help would be greatly appreciated. Thanks.

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  #19  
Old August 27th, 2011, 12:44 AM
pegasus pegasus is offline
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database not selected

i have a returned msg on my php showing "database not selected" after executing the php. before i inserted "or die(mysql_error());", it had a return error as below. tks !

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/username/public_html/testing/phpfile.php on line 39

"<?php

$con = mysql_connect(localhost, 'dbuser', 'dbpswd');
if (!$con)
{
die('Could not connect: ' . mysql_error());
}

mysql_select_db('dbname', $con);
$sql = "SELECT * FROM `dbtablename` LIMIT 0, 30 ";
$result = mysql_query($sql) or die(mysql_error());
echo "<table border='1'>

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  #20  
Old November 4th, 2011, 03:20 AM
IdusOrtus IdusOrtus is offline
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Talking

Quote:
Originally Posted by Sergiu(^_^)
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



Thank you, Sergiu(^_^)!

I was encountering the thread topic error with a very simple query and was bashing my head against the desk until considering your advice. There were no visible spaces in my code but I deleted it anyway and typed it over... that was all it took to remedy the problem. My best guess is that Notepad++ was storing the whitespace incorrectly. Much appreciated :P

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  #21  
Old February 23rd, 2012, 01:32 PM
suki suki is offline
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help

Hi all,

I have been tring to get the data from the database, but I get the same error. I hope some one can help me. here is my code

$sql = mysql_query("select s.accessNum, s.boxNum, b.boxName, s.sampleType, s.organ,e.diagonosis, s.sourceName,s.notes,s.comments from slide s, box b, exam e where b.boxNum=s.boxNum and s.organ like '%eye%' and e.slideId=s.slideId") or die( mysql_error() );
$result = mysql_num_rows($sql)or die(mysql_error());
echo "$result Rows\n";

if ($row = mysql_fetch_array($result))
{ echo $row['accessNum']; }
else mysql_error()

at the line echo "$result Rows\n"; I am getting the number of rows that are retrived. But at the next line I get the error

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in line 110

Please help

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  #22  
Old March 7th, 2012, 09:51 PM
jaldjoelajeoaj jaldjoelajeoaj is offline
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$sql = mysql_query("select s.accessNum, s.boxNum, b.boxName, s.sampleType, s.organ,e.diagonosis, s.sourceName,s.notes,s.comments from slide s, box b, exam e where b.boxNum=s.boxNum and s.organ like '%eye%' and e.slideId=s.slideId") or die( mysql_error() );
$result = mysql_num_rows($sql)or die(mysql_error());
echo "$result Rows\n";URL
URL
URL

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  #23  
Old April 4th, 2012, 08:39 AM
Weaver239 Weaver239 is offline
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Great work ! I will keep following this thread!!

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  #24  
Old May 22nd, 2012, 12:09 PM
sara_zeid sara_zeid is offline
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sql statment errors

hi
i have an sql statement that works on localhost but its shows an error on the server

$result=mysql_query("SELECT U_Name , location , c_name , gender , m_name FROM university , college , major WHERE ( university.U_ID = college.U_ID AND college.C_ID = major.C_ID )AND( university.U_Name ='$uni_name' )AND( major.gender = '$gender'OR major.gender = 'Both' AND major.m_name LIKE \"%$key_name%\" )AND( college.location LIKE \"%$location%\" AND college.c_name LIKE \"%$key_name%\" )");
IF ($result == FALSE ){
ECHO "ERROR";
}
$NUM=0;
while($row=mysql_fetch_array($result))
{
echo "<html>\n";

echo "<h4>".$row['U_Name']. " Has College " .$row['c_name']." located in :".$row['location']." - major :" .$row['m_name']." </h4>";


echo "<br>";
$NUM++;
}
IF($NUM==0){
ECHO "No Result To Display ! ";}

i dont know what the error in my sql I've try every thing
it shows
ERROR
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /homepages/xxxx/xxxxxxx/htdocs/result.php on line 64
No Result To Display !

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