in array function php

Question

I use in_array() to check whether a value exists in an array like below, But does not work as expected:

$projectId = $_GET['projectId'];
$userId = $_GET['userId'];

$qr2 = mysql_query("SELECT `userId`,`status` FROM `fndn_ProjectBacker` WHERE `projectId`='$projectId' && `status`='1'");

for($i = 1; $row2 = mysql_fetch_assoc($qr2); $i++) {
    $arr1[$i] = array($row2);
}

if (in_array($userId, $arr1)) {
    echo "True";
}

plz help me!

Community · Accepted Answer · 2017-05-23 11:58:14Z

2

It doesn't work as you want it, because you create a two dimensional array with this:

$arr1[$i]=array($row2);
        //^^^^^^     ^ See here

Remove the array declaration here and you can also change out your for loop with a while loop, e.g.

while($row2 = mysql_fetch_assoc($qr2)){
    $arr1[] = $row2["userId"];
}

if (in_array($userId, $arr1)) {
    echo "True";
}

I also want to welcome you in 2015 and highly recommend to use mysqli_* prepared statements or PDO, they are much safer. (Also see: Why shouldn't I use mysql_* functions in PHP?)

edited May 23, 2017 at 11:58

CommunityBot

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answered Jul 26, 2015 at 11:25

Rizier123

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3 Comments

Rizier123 Over a year ago

@FatemehNamkhah What does not work mean? Do you get any errors? Add error reporting to the top of your file(s): <?php ini_set("display_errors", 1); error_reporting(E_ALL); ?> Also what is the output of: print_r($arr1); after the while loop?

Fatemeh Namkhah Over a year ago

Array ( [0] => Array ( [userId] => 0 [status] => 1 ) [1] => Array ( [userId] => 5 [status] => 1 ) [2] => Array ( [userId] => 25 [status] => 1 )...

Rizier123 Over a year ago

@FatemehNamkhah updated my answer. Just reload the page and try the code again. You just need to add the userid to your array.

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