in_array problem (wrong data type for second argument)

[poleposters]

Hi,

 

I'm using the in_array function to check if a value exists in an array.

 

Abridged code is below.

 

Basically I'm retrieving an array of records from the data base and checking if the string "massage" is in it. That way I can check a checkbox.

 

I'm still unsure about arrays, so I'm not even sure if I've vreated it properly in order to use the in_array function.

 

$bid=$_SESSION['business_id'];
$query="SELECT * FROM specialties WHERE business_id=$bid";
$result=mysql_query($query);
while($check=mysql_fetch_array($result)){
$specialty=$check['specialty'];


if (in_array("massage", $specialty)) {
    echo "checked=\"checked\"";

}

 

I get the following error

 

in_array() [function.in-array]: Wrong datatype for second argument

 

Can anyone help? Or suggest another way to see if checkbox values exist in a database so that they can be checked.

 

Thanks

[vbnullchar]

are you sure $specialty is an array?

[poleposters]

I'm not sure. I'm quite new at using arrays. I think the problem might lie with the array rather than the in_array function. But I've been reading tutorials and I can't seem to get my head around it.

 

Is there a better(correct) way to create an array from the records in the database?

[redarrow]

 

I think it not in the array sorry.........

 

 

OFF topic....

 

This is for those dont understand in_array....................

 

 

 

<?php


$find="luke";


$a=array("john","luke","paul");


if(in_array("$find",$a)){


if(preg_match('/^[a-z]/',$find)){


echo "$find is in the array";

}else{

echo " didnt find $find in this array";

}

}else{

	echo " sorry $find is not a  ((a to z))  letter!";
}

?>

[sKunKbad]

line 5 try

 

$specialty[]=$check['specialty'];