Getting an error message when I try to take values from an array to insert into MySQL

Question

For some reason I am getting an error this error:

mysql_fetch_array() expects parameter 1 to be resource, array

Here's the relevant code - I am trying to to take all of the values from an array ($culture), and insert each into a MySQL query individually.

if (isset($_POST["orgid"])  && ($_POST['orgid'] !== '')) {
$orgid = $_POST['orgid'];
    $query = mysql_query("UPDATE organization SET org_name='".$orgname."', add_1='".$add1."', add_2='".$add2."', city='".$city."', state='".$state."', zip='".$zip."', url='".$url."', email='".$email."', phone='".$phone."', contact='".$contact."', hours='".$hours."', notes='".$notes."', description='".$description."' WHERE org_id='".$orgid."'");   

while ($cultrow = mysql_fetch_array($culture)) {
    $query = mysql_query("update org_cult_xref set org_id='".$orgid."', cult_id='".$cultrow."'");    
}
}

Here's the form info:

<select name="culture[]" multiple="multiple"><?php
    while ($cultrow = mysql_fetch_array($rescult)) {
        ECHO '<option name="culture[]" value="'. stripslashes($cultrow['cult_id']) .'">'. stripslashes($cultrow['cult_desc']) .'</option>';
    }
    ?>
        </select></fieldset>

$culture is set from all the variables of a picklist. Let me show you. — Steven Matthews
– Steven Matthews, Commented Nov 28, 2011 at 13:53
mysql_fetch_array() is used for retrieving data from a query and not for populating an update statement. — liquorvicar
– liquorvicar, Commented Nov 28, 2011 at 13:54

LostMohican · Accepted Answer · 2011-11-28 14:04:42Z

2

if $culture is not some sort of resource returned from a mysql_query function, you should supply the $query to mysql_fetch_array function. Of course if you want to use the result from the first update query.

answered Nov 28, 2011 at 14:04

LostMohican

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Marijn van Vliet · Accepted Answer · 2011-11-28 14:05:07Z

0

Perhaps you are looking for the foreach structure?

foreach($culture as $cultid) {
    $query = mysql_query("update org_cult_xref set org_id='".$orgid."', cult_id='".$cultid."'");    
}

answered Nov 28, 2011 at 14:05

Marijn van Vliet

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2 Comments

Steven Matthews Over a year ago

Ah, this makes more sense - I am so used to getting arrays from MySQL! Sorry, still new to PHP. Also, I like your handle, used a similar one for years.

Steven Matthews Over a year ago

Hrm, this does not appear to update org_cult_xref. Neither does an "insert into" - why would this be?

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