2019-04-21 20:21
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I have an array and when using in_array to check if object exists it always returns false.

As an example, here is the array:

array:1 [
0 => "name1, name2, name3"

and this is how I check if index exists

if(!in_array('name1', $array)) { return FALSE; } else { return TRUE;}

Name1 exists in array I have however I think I'm missing something small here.

Not sure what it could be or even if this function is the right function to use to check? Expected result is to be true as object exists.


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2条回答 默认 最新

  • dongniechi7825
    dongniechi7825 2019-04-21 20:26

    Since your array only has one item with a long string in_array returns false since there is no exact match.
    What you can do is to use preg_grep which is regex for arrays.

    var_dump(preg_grep("/name1/i", $arr));

    You can use it with bool return like this:

    if(preg_grep("/name1/i", $arr)){
        echo "true";
        echo "false";

    Since preg_grep returns empty if nothing is found.
    And empty array means false in an if().

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  • dongsou0083
    dongsou0083 2019-04-21 20:36

    Your array is one element that is one string, it does not contain multiple elements. If you want to use in_array(), it must be an array - not a string.

    There's a few different ways of achieving this, one us by using explode() on the element to make it into an array and then use in_array().

    You can also return the value of in_array() instead of running a check yourself - the returnvalue of the function is a boolean true/false.

    $array = array("name1, name2, name3");
    $parts = explode(", ", $array[0]);
    return in_array("name1", $parts);
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