警告：in_array（）[function.in-array]：第二个参数的数据类型错误

Question

I am a little rusty in PHP after several years without programming with this language.

I am fetching data from an array in mySQL using PHP.

My problem is that at some point I fetch data from a table, using this:

$myArray = mysql_fetch_array( $data ); 

but at some point, the data base may be empty, so $myArray will be null or contain no elements.

Then I have to check if a particular string that is generated on-the-fly is on $myArray. If it is, another string must be generated and verified if it already exists on $myArray.

$myString = generateString();
if (in_array($myString, $myArray)) {
    $myString = generateString();
}

this code gives me this error:

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

To prevent in_array from running when $myArray is empty I did this:

if (count($myArray) > 0) {
    if (in_array($myString, $myArray)) {
        $myString = generateString();
    }
}

but count is giving me 1 when the array is empty (?)...

How do I solve that?

just another question: $myString is being modified inside the IF that is already testing it. Is this possible in PHP?

Answer 1

Well, if your query failed, you'll have... wait for it... no array!

So that in_array call would give you a warning about $myArray not being an array (but false ).

You need to check your query for errors, and better yet:

Please, don't use mysql_* functions in new code . They are no longer maintained and the deprecation process has begun on it. See the red box ? Learn about prepared statements instead, and use PDO or MySQLi - this article will help you decide which. If you choose PDO, here is a good tutorial .