PHP Function to select from MySQL database and pass results to array

Question

I'm trying to create a function that will return an array with all the fields of the query result and then loop through it and echo everything. This is what I have done so far. I tried some approaches from similar questions such as Putting a SQL query in a PHP function but so far I haven't been able to get it to work.

<?php

    $ID = $_POST['ID'];

    function query($ID){
    $servername = "x.x.x.x";
    $username = "xxxxx";
    $password = "xxxxx";
    $dbname = "xxxxx";

    $conn = mysqli_connect($servername, $username, $password, $dbname);
      if (mysqli_connect_errno()){
              echo "Failed to connect to MySQL: " . mysqli_connect_error();
                               }

    $query = "SELECT ID, NAME, POSITION, TELEPHONE_NUMBER, EMAIL FROM GROUP WHERE ID = '$ID'";
    $result = mysqli_query($conn, $sql);

    while($row = mysqli_fetch_array($result)){
    $resArr[] = $row;
    }
    return $resArr;
    }

    $person = query();
    foreach($person as $post) {
    echo $post['ID'] . "<br>";
    echo $post['NAME'] . "<br>";
    echo $post['POSITION'] . "<br>";
    echo $post['TELEPHONE_NUMBER'] . "<br>";
    echo $post['EMAIL'] . "<br>";
    }
    ?>

Other than your gaping wide-open sql injection attack vulnerability, what exactly is wrong with this code? doesn't work doesn't tell us anything. if it worked, you wouldn't be here. — Marc B
– Marc B, Commented May 5, 2015 at 14:55
you define your function with a parameter - function query($ID), but you call it without a parameter - $person = query();. You need to do $person = query($ID); — Sean
– Sean, Commented May 5, 2015 at 14:58
Marc B - User input is limited to integers on the page where POST is getting data from and if I knew what was wrong I still wouldn't be here I would probably fix it. Anyway, thanks Sean that did it and thanks to the rest of you guys! — John
– John, Commented May 6, 2015 at 8:16

imran qasim · Accepted Answer · 2015-05-05 15:19:01Z

1

there are few mistakes has been done for example $ID in your query can not get translated into normal id value so you have to close your query like this if ID is an integer than

 $query = "SELECT ID, NAME, POSITION, TELEPHONE_NUMBER, EMAIL FROM GROUP WHERE ID = ".$ID;

if ID is a varchar or any other than.

 $query = "SELECT ID, NAME, POSITION, TELEPHONE_NUMBER, EMAIL FROM GROUP WHERE ID = '".$ID."'";

also you are getting id via post but you are not assigning it to your function where you are calling it.

   $person = query();

it must be

     $person = query($ID);

try these fixes and than post your result back.

Regards

answered May 5, 2015 at 15:19

imran qasim

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