Can't echo single value from array with a function in PHP

Question

I am having trouble with PHP, where I populate an array with results from a MySQL query.

The problem is when I make a function to echo a certain element of the array, it's not working, where as without a function there are no errors.

Establish connection, perform query, store result in variable:

require_once("db.php");

$conn = mysqli_connect($servername, $username, $password, $dbname);
if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());
}

$query = "SELECT * FROM arlista";
$query_result = mysqli_query($conn, $query);

mysqli_close($conn);
$result_array = array();

I pass the query results to an array, then I want to query a single value from the array. The problem is if I use a function like this, this does not work. I can't get the element of the array to display in the browser.

function arlista($attr, $rownum){
    while($row = mysqli_fetch_array($query_result)){
        $result_array[$i] = $row[$attr];
        $i++;
    }
    echo $result_array[$rownum];
}

arlista("ar",1);

However this works if I do not use a function. The browser is displaying the value.

while($row = mysqli_fetch_array($query_result)){
$result_array[$i] = $row["ar"];
$i++;
}

echo $result_array[1];

Could someone explain what is going wrong with the function or how do I fix it to work? Thank you!

The server is running PHP 5.6.19

@cpugourou Undefined variables become 0 when used in an arithmetic context. — Barmar
– Barmar, Commented Apr 1, 2016 at 23:23
Thanks. Never thought of this one. Nevertheless I will continue to properly declare all my variables... kinda old school. Probably because I never use while to go across an array... — cpugourou
– cpugourou, Commented Apr 1, 2016 at 23:27

ak_ · Accepted Answer · 2016-04-01 23:16:56Z

2

Check your variable scoping. Your function has no variable $query_result. Turning on error reporting would also give you notices about the problem.

Using a global would work:

function arlista($attr, $rownum){
    global $query_result, $result_array;
    while($row = mysqli_fetch_array($query_result)){
        $result_array[$i] = $row[$attr];
        $i++;
    }
    echo $result_array[$rownum];
}

arlista("ar",1);

Hope that helps!

answered Apr 1, 2016 at 23:16

ak_

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Comments

user2162942 · Accepted Answer · 2016-04-01 23:20:36Z

2

A function does not bring in the variables that you haven't defined inside the function, or passed in. Since you did not define or pass in $query_result and $result_array, the script would not work. However, without the function, that variable is defined, so the script will work. To make the function work, all you need to do is pass in the variable $query_result, $result_array or define it inside the function.

Edit: As Adam said, you can define in the function to use it as a global variable: global $query_result, $result_array;

edited Apr 1, 2016 at 23:20

answered Apr 1, 2016 at 23:13

user2162942

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Barmar · Accepted Answer · 2016-04-01 23:23:19Z

2

You need to pass $query_result as an argument to the function. You should also initialize the variables that you use within the function. But there's no real need for the $i variable, since you can use [] to push a new element onto an array.

function arlista($attr, $rownum, $query_result){
    $result_array = array();
    while($row = mysqli_fetch_array($query_result)){
        $result_array[] = $row[$attr];
    }
    echo $result_array[$rownum];
}

arlista("ar", 1, $query_result);

You could also use global $query_result, but explicit arguments are generally better programming style.

answered Apr 1, 2016 at 23:23

Barmar

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1 Comment

Zsolt Mester Over a year ago

thank you very much for the explanation, I understand what was missing now. it's working nicely.

GitaarLAB · Accepted Answer · 2016-04-01 23:23:31Z

1

You need to either pass $query_result as a function parameter, or define it as a global variable within the function.

edited Apr 1, 2016 at 23:23

GitaarLAB

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answered Apr 1, 2016 at 23:15

Richard Armerding

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