PHP - in_array problems

Question

I wrote the following, but its saying that its the 'wrong datatype'.

Could someone tell me where im going wrong.

$newQuery = mysql_query("SELECT * FROM table WHERE left_id='$authId'");
$catid = mysql_fetch_array($newQuery);

if(in_array("99", $catId)){
    // A Director
} else {
    // A Researcher
}

Cheers,

For variable names PHP does impose case sensitivity, if only function names acted in the same manner... — Russell Dias
– Russell Dias, Commented Feb 1, 2011 at 5:56

Dan Grossman · Accepted Answer · 2011-02-01 06:40:04Z

8

You are assigning the row to $catid and later using $catId. Notice the capital "I". You also need to make sure $catId is an array because mysql_fetch_array will return FALSE if there are no rows.

edited Feb 1, 2011 at 6:40

Dan Grossman

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answered Feb 1, 2011 at 5:52

shankhan

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1 Comment

sberry Over a year ago

+1. @Chris, just another crazy thing about PHP... variables are case-sensitive, but function are not... Man this language pisses me off sometimes.

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