php warning mysql_fetch_assoc

Question

I am trying to access some information from mysql, but am getting the warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource for the second line of code below, any help would be much appreciated.

$musicfiles=getmusicfiles($records['m_id']);
$mus=mysql_fetch_assoc($musicfiles);
for($j=0;$j<2;$j++)
{
 if(file_exists($mus['musicpath']))
 {
  echo '<a href="'.$mus['musicpath'].'">'.$mus['musicname'].'</a>';       
 }
 else
 {
  echo 'Hello world';     
 }
}

function getmusicfiles($m_id)
{
$music="select * from music WHERE itemid=".$s_id;
$result=getQuery($music,$l);
return $result;
}

And what's the definition of getQuery()? Is there a call to mysql_query in there, somewhere? If not, you will get the warning, as it should be. — Piskvor left the building
– Piskvor left the building, Commented Dec 14, 2009 at 15:31
yeah there is, and it works fine since I used it quite a bit in my site — death the kid
– death the kid, Commented Dec 14, 2009 at 15:35
@death the kid: If it worked fine, you wouldn't be asking this question ;) Can we see it? — Piskvor left the building
– Piskvor left the building, Commented Dec 14, 2009 at 15:38
Note that your parameter ($m_id) is not the same as the value you're trying to append to your SQL querty ($s_id) - perhaps this is your problem? — Dexter
– Dexter, Commented Dec 14, 2009 at 15:41

Piskvor left the building · Accepted Answer · 2009-12-14 15:58:33Z

Generally, the mysql_* functions are used as follows:

$id = 1234;
$query = 'SELECT name, genre FROM sometable WHERE id=' . $id;
// $query is a string with the MySQL query
$resource = mysql_query($query);
// $resource is a *MySQL result resource* - a mere link to the result set
while ($row = mysql_fetch_assoc($resource)) { 
    // $row is an associative array from the result set
    print_r($row);
    // do something with $row
}

If you pass something to mysql_fetch_assoc that is not a MySQL result resource (whether it's a string, an object, or a boolean), the function will complain that it doesn't know what to do with the parameter; which is exactly what you are seeing.

A common gotcha: you get this warning if you pass something (other than a valid query string) to mysql_query:

$id = null;
$query = 'SELECT name, genre FROM sometable WHERE id=' . $id;
$res = mysql_query($query); 
// $res === FALSE because the query was invalid
// ( "SELECT name, genre FROM sometable WHERE id=" is not a valid query )
mysql_fetch_assoc($res); 
// Warning: don't know what to do with FALSE, as it's not a MySQL result resource

philjohn · Accepted Answer · 2009-12-14 15:15:03Z

1

Without seeing the code of getmusicfiles there's not a lot we can really help you with. You should be returning a valid mysql resource in that function.

answered Dec 14, 2009 at 15:15

philjohn

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Comments

Dexter · Accepted Answer · 2009-12-14 15:20:31Z

1

As others have noted, you need to return a valid mysql resource into the mysql_fetch_assoc function to retrieve the next row. For example:

$sql = "select * from table";

$resultSet = mysql_query($sql) or die("Couldn't query the database.");
echo "Num Rows: " . mysql_num_rows($resultSet);

while ($resultRowArr = mysql_fetch_assoc($resultSet)) {
    ...
}

answered Dec 14, 2009 at 15:20

Dexter

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Comments

Digital Craft Studios · Accepted Answer · 2009-12-14 15:42:23Z

0

I think you need to specify what the function getQuery()

$result=getQuery($music,$l);

does

answered Dec 14, 2009 at 15:42

Digital Craft Studios

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Comments

Jason Plank · Accepted Answer · 2011-10-17 20:07:05Z

0

It depends on what exactly getmusicfiles() does. It must return a result of mysql_query() function call, then it will be a "valid MySQL result".

And you most probably wanted to put the line $mus=mysql_fetch_assoc($musicfiles) inside of the for cycle to fetch several rows one after another.

edited Oct 17, 2011 at 20:07

Jason Plank

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answered Dec 14, 2009 at 15:16

alemjerus

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Comments

Jason Plank · Accepted Answer · 2011-10-17 20:10:56Z

0

function getmusicfiles($m_id) {
$music="select * from music WHERE itemid=".$s_id;

$m_id != $s_id ?

edited Oct 17, 2011 at 20:10

Jason Plank

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answered Mar 29, 2010 at 2:38

jaysivo

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