Basic PHP/MySQL Error with mysql_fetch_array()

Question

The error I get:

...mysql_fetch_array() expects parameter 1 to be resource, boolean given...

awayid is in the address bar properly. I can print it out just fine, but for some reason the following code gives me the above error.

$result = mysql_query("select * from team where id=" . $_GET['awayid']);
$row = mysql_fetch_array($result);

EDIT Tried the mysql_error(). It seems I forgot to select a database... however, even why I use mysql_select_db('gamelydb'); I still get the mysql error No database selected

That error means that you query failed. You can use mysql_error to find out why. The mysql_query function returns false (a boolean) when it fails. — Francois Deschenes
– Francois Deschenes, Commented Jun 27, 2011 at 16:56
This error means "$result" did not contain a valid result set. Their must be some error in your query which most probably is that $_GET['awayid'] is empty. Try debugging the query by echoing it and then running directly in MySQL or using mysql_error() function. — Abhay
– Abhay, Commented Jun 27, 2011 at 17:00
possible duplicate of mysql_fetch_array() expects parameter 1 to be resource problem — RobertPitt
– RobertPitt, Commented Jun 27, 2011 at 17:01
possible duplicate of mysql_fetch_array() expects parameter 1 to be resource, boolean given in select — Jürgen Thelen
– Jürgen Thelen, Commented Aug 8, 2012 at 20:30

Andrew Moore · Accepted Answer · 2011-06-27 16:56:56Z

3

Your query is failing... Therefore $result is set to false.

$result = mysql_query("select * from team where id=" . $_GET['awayid']);
var_dump($result); // bool(false)

Call mysql_error() to get the error message for your query:

echo mysql_error();

answered Jun 27, 2011 at 16:56

Andrew Moore

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Marc B · Accepted Answer · 2011-06-27 16:56:59Z

2

Your query is failing and returning a boolean FALSE. Try this:

$result = mysql_query("select ...") or die(mysql_error());
                                   ^^^^^^^^^^^^^^^^^^^^^^---- add this

This will kill the script and show you the exact reason the query is failing.

answered Jun 27, 2011 at 16:56

Marc B

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phant0m · Accepted Answer · 2011-06-27 16:57:01Z

1

mysql_query() returns false if the query is unsuccessful, i.e. an error occured. That is why you need to check $result for being false first.

Use mysql_error() to output the error.

answered Jun 27, 2011 at 16:57

phant0m

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Tarek · Accepted Answer · 2011-06-27 16:57:29Z

1

You need to be sure there is results from your query :

while ($row = mysql_fetch_array($result)) {
// echo $row[] ... ;   
}

answered Jun 27, 2011 at 16:57

Tarek

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Dan Simon · Accepted Answer · 2011-06-27 16:57:34Z

1

First of all, your query is very open to SQL injection attacks. Do not directly insert anything from $_GET or $_POST (or really anywhere) into your query. At the minimum, use mysql_real_escape_string on the variable.

mysql_query is returning false becuase there is something wrong with the query. You can use mysql_error to see what the last reported error is.

answered Jun 27, 2011 at 16:57

Dan Simon

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Sascha Galley · Accepted Answer · 2011-06-27 16:58:23Z

1

if ($result = mysql_query("select * from team where id='" . $_GET['awayid']) . "'") {
    $row = mysql_fetch_array($result);
}
else {
    echo mysql_error();
}

Anyway...you know that writing a $_GET parameter right into the SQL query is very very bad? Try it with PHP Data Objects.

answered Jun 27, 2011 at 16:58

Sascha Galley

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RobertPitt · Accepted Answer · 2011-06-27 16:58:51Z

1

Did you try and search around first Tory, we answer these questions over and over again, next time please search around.

The reason why this error occurs is because your running a query with mysql_query that fails, because it fails it returns false, you then pass the value of false to mysql_fetch_array, it's like doing mysql_fetch_array(false)

You need to make sure that mysql_query is successful:

try something like this:

if(false !== ($result = mysql_query("select * from team where id=" . $_GET['awayid'])))
{
    $row = mysql_fetch_array($result);
}else
{
    die("Query has failed: " . mysql_error())
}

answered Jun 27, 2011 at 16:58

RobertPitt

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