in_array giving Warning Message - php

Question

All,
I have the following code:
$qry = "Select * from vendor_options order by vendor_option_name ASC";
$result = mysql_query($qry);
while($resultset = mysql_fetch_array($result)){
if(isset($_SESSION['pav_choosen_vendor_categories'])){
for($z=0;$z<$_SESSION['pav_choosen_vendor_categories'];$z++){
$sVendorId = $_SESSION['pav_vendor_categories_' . $z];
if($sVendorId==$resultset['vendor_option_id']){
$vendor_cats_choosen[] = $sVendorId;
}
}
if(in_array($resultset['vendor_option_id'],$vendor_cats_choosen)){
?>
<input type="checkbox" value="<?php echo $resultset['vendor_option_id']; ?>" class="select_vendor" name="vendor_categories[]" checked><?php echo $resultset['vendor_option_name']; ?><br>
<?php
}else{
?>
<input type="checkbox" value="<?php echo $resultset['vendor_option_id']; ?>" class="select_vendor" name="vendor_categories[]"><?php echo $resultset['vendor_option_name']; ?><br>
<?php
}
}
}
I'm trying to check to see if the value returned in the mysql_fetch_array is already in my array. Say the first value it finds in the array is in the fourth iteration of the while loop. I'll get the following error:
Warning: in_array() expects parameter 2 to be array, null
Once it gets to a value that is in the array the rest of them work fine. Why does itgive an error for the first couple? Thanks.

Answer 1

It looks like you have not initialized $vendor_cats_chosen to be an array, and so if the condition if($sVendorId==$resultset['vendor_option_id']) is not true, no elements will be appended to it, turning it implicitly into an array.
Initialize it before the while loop. You should just about always initialize arrays before use.
// Initialize the array
$vendor_cats_chosen = array();
while($resultset = mysql_fetch_array($result)){
....
Now, when your in_array() statement executes, the array may be empty, but will be a valid array.
// $vendor_cats_chosen might be an empty array, or may have elements.
if(in_array($resultset['vendor_option_id'],$vendor_cats_choosen)){

Answer 2

The problem is that your array not always creates. To fix this issue just add
$vendor_cats_choosen = array();
somewhere before while in your code.

Answer 3

You only ever populate the array $vendor_cats_choosen inside an if statement, which means it can potentially contain no values. You also do not declare it before you start the loop which populates it - which you should do anyway, because adding a value to an undeclared array will emit an E_NOTICE.
Add the line
$vendor_cats_choosen = array();
...at the top of the script and the error will disappear. If you think this array should contain values, you may need to examine the logic in your if statement.

Answer 4

In order to eliminate the warning message, two approaches can be followed:
1) use of # before the statement which is generating the warning message ( However, this is not an advisable engineering approach )
2) prior to using the array object, it can be filtered out in an if-else.
For example, in your case, you can add this line
if( $vendor_cats_choosen ){
if(in_array($resultset['vendor_option_id'],$vendor_cats_choosen)){
?>
<input type="checkbox" value="<?php echo $resultset['vendor_option_id']; ?>" class="select_vendor" name="vendor_categories[]" checked><?php echo $resultset['vendor_option_name']; ?><br>
<?php
}else{
?>
<input type="checkbox" value="<?php echo $resultset['vendor_option_id']; ?>" class="select_vendor" name="vendor_categories[]"><?php echo $resultset['vendor_option_name']; ?><br>
<?php
}
} ?>