[SOLVED] Warning: mysql_fetch_array():

[brandon99919]

Here's the error that shows up:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in F:\hshome\brandon9\pcprpg.com\menu_user.php on line 28

 

Here's the code itself:

<?php
			   $query = mysql_query("SELECT * FROM pokemon WHERE owner_id = '$userid' and p_slot = '1'");
			   $fetch = mysql_fetch_array($query);
			   $p_icon1 = $fetch['p_icon'];
			   $p_name1 = $fetch['p_name'];
			   
			   if (empty($p_name1)) {
						  echo "<img src='{$p_icon}' alt='{$p_name1}' />";
						  } else {
							  echo "<img src='pokemonball.gif' alt='slot 1' />";
						  }
						  ?>

 

 

and line 28 is the mysql_fetch_array function

[gevans]

You have to execture the query first using mysql_query($query);

 

<?php
               $query = mysql_query("SELECT * FROM pokemon WHERE owner_id = '$userid' and p_slot = '1'");
$result = mysql_query($query);
               $fetch = mysql_fetch_array($result);
               $p_icon1 = $fetch['p_icon'];
               $p_name1 = $fetch['p_name'];
               
               if (empty($p_name1)) {
                       echo "<img src='{$p_icon}' alt='{$p_name1}' />";
                       } else {
                          echo "<img src='pokemonball.gif' alt='slot 1' />";
                       }
                       ?>

[gevans]

sorry, little typo;

 

<?php
               $query = mysql_query("SELECT * FROM pokemon WHERE owner_id = '$userid' and p_slot = '1'");
$result = mysql_result($query);
               $fetch = mysql_fetch_array($result);
               $p_icon1 = $fetch['p_icon'];
               $p_name1 = $fetch['p_name'];
               
               if (empty($p_name1)) {
                       echo "<img src='{$p_icon}' alt='{$p_name1}' />";
                       } else {
                          echo "<img src='pokemonball.gif' alt='slot 1' />";
                       }
                       ?>

[brandon99919]

it still gives the same error

[gevans]

Are you successfully connection to your db?

Is $userid set, and is it what you think it should be.

Output the query before running it to check it.

Do some error checking on the query, is it successful, does it fail?

[brandon99919]

yea the database connection is good and when I tested the query it got back an error saying that the owner_id column doesn't exist. Turns out the column is actually called p_owner, so I changed it to that  and ran the query and it was successful. And yea, the userid variable is set. I still get the same error even tho I fixed the problem with the where clause.  ???

[gevans]

So is this solved?

[brandon99919]

nahh the same error still shows up

[gevans]

If the query was successfuly you wouldn't be getting that error. Try this,

 

<?php
$query = mysql_query("SELECT * FROM `pokemon` WHERE `p_owner`=$userid AND `p_slot`= 1");
$result = mysql_query($query) or trigger_error('Query failed: ' . mysql_error($db), E_USER_ERROR);
$fetch = mysql_fetch_array($result);
$p_icon1 = $fetch['p_icon'];
$p_name1 = $fetch['p_name'];

if (empty($p_name1)) {
echo "<img src='{$p_icon}' alt='{$p_name1}' />";
} else {
echo "<img src='pokemonball.gif' alt='slot 1' />";
}
?>

[brandon99919]

now it gives this error:

 

Warning: mysql_error(): supplied argument is not a valid MySQL-Link resource in F:\hshome\brandon9\pcprpg.com\menu_user.php on line 28

 

Fatal error: Query failed: in F:\hshome\brandon9\pcprpg.com\menu_user.php on line 28

[gevans]

You have to forgive me, I've been working on stuff today that has led me to do the oppsite than help.

 

Fixed Script bellow;

 

<?php
$query = mysql_query("SELECT * FROM `pokemon` WHERE `p_owner`=$userid AND `p_slot`= 1") or trigger_error('Query failed: ' . mysql_error($db), E_USER_ERROR);
$fetch = mysql_fetch_array($query);
$p_icon1 = $fetch['p_icon'];
$p_name1 = $fetch['p_name'];

if (empty($p_name1)) {
   echo "<img src='{$p_icon}' alt='{$p_name1}' />";
} else {
   echo "<img src='pokemonball.gif' alt='slot 1' />";
}
?>

[brandon99919]

now it gives the same error:

 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in F:\hshome\brandon9\pcprpg.com\menu_user.php on line 29

[brandon99919]

my bad for the double post, but yeah it still gives the same error =[

[PFMaBiSmAd]

It's likely that the code you are posting is part of a loop and that the actual line with the error is not the posted code. Post your code from the start of your file through the end of any loop that is involved in what you are doing.

 

The $db variable in the mysql_error() is also not helping and should be removed because it is not relevant to the OP's code.

[brandon99919]

here's the full code:

 

		       <a onClick="subMenu('general_submenu', this);">General</a>
			   <div id='general_submenu' class='sub_menu'>
			        <a href='/index.php?page=news'>News</a>
                        <a href='/index.php?page=promo'>Promo</a>
                        <a href='/index.php?page=staff'>Staff</a>
                        <a href='/index.php?page=logout'>Logout</a>
			   </div>
			   
			   <a onClick="subMenu('myAccount_submenu', this);">My Account</a>
			   <div id='myAccount_submenu' class='sub_menu'>
			        <a href='/index.php?page=profile'>My Profile</a>
			   </div>
                   <a onClick="subMenu('myPokemon_submenu', this);">My Pokémon</a>
			   <div id='myPokemon_submenu' class='sub_menu'>
			        <a href='/index.php?page=stats'>Starter Stats</a>
			   </div>
                   <?php if($user_level == "Admin") { ?>
                   <a onClick="subMenu('adminCP_submenu', this);">Admin CP</a>
                   <div id='adminCP_submenu' class='sub_menu'>
			        <a href='/index.php?page=gen'>Promo</a>
			   </div><?php }?>
			   
			   <a href='#'>Master Quest</a>
                   <div align='center'>
                   <?php
			   include('includes/basefile.php');
			   $query = mysql_query("SELECT * FROM pokemon WHERE p_owner = '$userid' and p_slot = '1'");
			   $result = mysql_query($query);
			   $fetch = mysql_fetch_array($result);
			   $p_icon1 = $fetch['p_icon'];
			   $p_name1 = $fetch['p_name'];
			   
			   if (empty($p_name1)) {
						  echo "<img src='{$p_icon}' alt='{$p_name1}' />";
						  } else {
							  echo "<img src='pokemonball.gif' alt='slot 1' />";
						  }
						  ?>
						  <?php
			   $query = mysql_query("SELECT * FROM pokemon WHERE p_owner = '$userid' and p_slot = '2'");
			   $result = mysql_query($query);
			   $fetch = mysql_fetch_array($result);
			   $p_icon1 = $fetch['p_icon'];
			   $p_name1 = $fetch['p_name'];
			   
			   if (empty($p_name1)) {
						  echo "<img src='{$p_icon}' alt='{$p_name1}' />";
						  } else {
							  echo "<img src='pokemonball.gif' alt='slot 2' />";
						  }
						  ?>
						  <?php
			   $query = mysql_query("SELECT * FROM pokemon WHERE p_owner = '$userid' and p_slot = '3'");
			   $result = mysql_query($query);
			   $fetch = mysql_fetch_array($result);
			   $p_icon1 = $fetch['p_icon'];
			   $p_name1 = $fetch['p_name'];
			   
			   if (empty($p_name1)) {
						  echo "<img src='{$p_icon}' alt='{$p_name1}' />";
						  } else {
							  echo "<img src='pokemonball.gif' alt='slot 3' />";
						  }
						  ?><br />
						  <?php
			   $query = mysql_query("SELECT * FROM pokemon WHERE p_owner = '$userid' and p_slot = '4'");
			   $result = mysql_query($query);
			   $fetch = mysql_fetch_array($result);
			   $p_icon1 = $fetch['p_icon'];
			   $p_name1 = $fetch['p_name'];
			   
			   if (empty($p_name1)) {
						  echo "<img src='{$p_icon}' alt='{$p_name1}' />";
						  } else {
							  echo "<img src='pokemonball.gif' alt='slot 4' />";
						  }
						  ?>
						  <?php
			   $query = mysql_query("SELECT * FROM pokemon WHERE p_owner = '$userid' and p_slot = '5'");
			   $result = mysql_query($query);
			   $fetch = mysql_fetch_array($result);
			   $p_icon1 = $fetch['p_icon'];
			   $p_name1 = $fetch['p_name'];
			   
			   if (empty($p_name1)) {
						  echo "<img src='{$p_icon}' alt='{$p_name1}' />";
						  } else {
							  echo "<img src='pokemonball.gif' alt='slot 5' />";
						  }
						  ?>
						  <?php
			   $query = mysql_query("SELECT * FROM pokemon WHERE p_owner = '$userid' and p_slot = '6'");
			   $result = mysql_query($query);
			   $fetch = mysql_fetch_array($result);
			   $p_icon1 = $fetch['p_icon'];
			   $p_name1 = $fetch['p_name'];
			   
			   if (empty($p_name1)) {
						  echo "<img src='{$p_icon}' alt='{$p_name1}' />";
						  } else {
							  echo "<img src='pokemonball.gif' alt='slot 6' />";
						  }
						  ?>

[PFMaBiSmAd]

And the php error message from that code is?

 

There is no way that code can work, because there are double mysql_query() statements in it.

 

The easiest fix would be to change each of the leading mysql_query() statements so that the line just sets the $query variable with the query statement -

 

$query = "SELECT * FROM pokemon WHERE p_owner = '$userid' and p_slot = '1'";

 

And you should be executing a single query that returns all the rows you want as a set, then you loop through them. Duplicating code usually wastes time due to the typo's made while fixing up the different values in it.

 

 

[brandon99919]

here's the errors:

 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in F:\hshome\brandon9\pcprpg.com\menu_user.php on line 29

 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in F:\hshome\brandon9\pcprpg.com\menu_user.php on line 55

 

 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in F:\hshome\brandon9\pcprpg.com\menu_user.php on line 68

 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in F:\hshome\brandon9\pcprpg.com\menu_user.php on line 81

 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in F:\hshome\brandon9\pcprpg.com\menu_user.php on line 94

 

[PFMaBiSmAd]

Back to ground zero. All that means is your query is failing and no one knows why. For troubleshooting purposes only, replace the lines making up your first query with the following -

 

$query = "SELECT * FROM pokemon WHERE p_owner = '$userid' and p_slot = '1'";
$result = mysql_query($query) or die("Query failed: $query, " . mysql_error());

[brandon99919]

the errors are gone  thanks for the help

 

solved 

 

[PFMaBiSmAd]

I'm guessing the original problem was the column name.

 

You can replace all that php code with the following single query and a loop -

 

Untested, but should work -

<?php
include('includes/basefile.php');
$query = "SELECT * FROM pokemon WHERE p_owner = '$userid' and p_slot IN('1','2','3','4','5','6') ORDER BY p_slot";
$result = mysql_query($query);
$i = 1;
while($row = mysql_fetch_array($result)){
$p_icon = $row['p_icon'];
$p_name = $row['p_name'];
if(!empty($p_name)){
	echo "<img src='{$p_icon}' alt='{$p_name}' />";
} else {
	echo "<img src='pokemonball.gif' alt='slot {$i}' />";
}
$i++;
}
?>

[brandon99919]

it only displays one image