CODEWITHSUNDEEP
phparraysconditional-statements
user3759862
user3759862

Reputation: 57

Array search dont know search

i have this array: 

include("config.php");
$start = "2014-06-20 08:00:00";
$data = mysql_query ("select * from evenement WHERE start = '$start'");
$zaznam = mysql_fetch_array ($data);
while($zaznam = mysql_fetch_array ($data))
{
 $arr2[] = $zaznam["resourceId"];   //store query values in second array

}

If i echo $arr2 i get this:

Array ( [0] => STK1 )

now i make condition for array_search:

if (array_search('STK1', $arr2)) {
    echo "Arr2 contains STK1 <br>";
}
else {
    echo "Arr2 not contains STK1 <br>";
}

but i get this Arr2 not contains STK1 how it is possible? What im doing wrong?

Upvotes: 0

Views: 28

Answers (2)

Jim
Jim

Reputation: 22656

From the array_search documentation:

This function may return Boolean FALSE, but may also return a non-Boolean value which evaluates to FALSE. Please read the section on Booleans for more information. Use the === operator for testing the return value of this function.

array_search is returning 0 because it found a match at index 0 of the array. This is evaluating to false.

Instead try:

if (array_search('STK1', $arr2) !== false) {
    echo "Arr2 contains STK1 <br>";
}
else {
    echo "Arr2 not contains STK1 <br>";
}

Upvotes: 1

Jimmy T.
Jimmy T.

Reputation: 4190

That is totally correct behaviour for PHP.

The documention for the return value says:

Returns the key for needle if it is found in the array, FALSE otherwise.

In your case you are getting 0 which also evaluates to false in an if.

You have to check if the value is not false using the !== operator.

if (array_search('STK1', $arr2) !== false) {
    echo "Arr2 contains STK1 <br>";
}
else {
    echo "Arr2 not contains STK1 <br>";
}

Upvotes: 2

Related Questions