Php function in_array() not returning expecting result [duplicate]

Question

I'm trying to get whether name exists in my array that I have acquired from my database table. The array contains the whole name field of the table members However,on using in_array() function, the result for any search is always false

Here is how the array looks like

    Array ( [0] => Array ( [name] => Jane ) 
            [1] => Array ( [name] => Kelvin ) 
            [2] => Array ( [name] => James ) 
            [3] => Array ( [name] => Leon ) 
            [4] => Array ( [name] => Steve ) 
            [5] => Array ( [name] => Jenifer ) 
           )

And here is my php code to check if a name exists

<?php

    require_once 'databaseconn.php'; 
    #To get connection
    $mysqli = new mysqli($db_hostname, $db_username, $db_password,$db_database);
    #To get array of names from my database
    $stmt = $mysqli-> prepare('SELECT name FROM members WHERE id');
    $stmt->execute();
    $result = $stmt->get_result();
    #To store result as array in variable $application_names
    while ($row = $result->fetch_assoc()) {
        $member_names[] = $row;
    }
    #To test if name exists
    $name = "Jane";

    if (in_array($name, $member_names))
    {
    echo "Match found";
    }
    else
    {
    echo "Match not found";
    }

?>

Hint: if (in_array($name, array_coumn($member_names, "Name"))) — dWinder
– dWinder, Commented May 22, 2019 at 10:28

Jimmix · Accepted Answer · 2019-05-22 10:29:15Z

0

<?php

    require_once 'databaseconn.php'; 
    #To get connection
    $mysqli = new mysqli($db_hostname, $db_username, $db_password,$db_database);
    #To get array of names from my database
    $stmt = $mysqli-> prepare('SELECT name FROM members WHERE id');
    $stmt->execute();
    $result = $stmt->get_result();
    #To store result as array in variable $application_names
    while ($row = $result->fetch_assoc()) {
        $application_names[] = $row;
    }
    #To test if name exists
    $name = "Jane";

    $flatArr = [];

    foreach ($member_names as $memberArr) {
        if(isset($memberArr['name'])) {
            $flatArr[] = $memberArr['name'];
        }
    }

    if (in_array($name,  $flatArr))
    {
    echo "Match found for: " . $name;
    }
    else
    {
    echo "Match not found for: " . $name;
    }

?>

answered May 22, 2019 at 10:29

Jimmix

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