php statements fail to work when wrapped in a function

Question

The code below works fine when its not in wrapped in a function

$opt = 'logo_img';
$sql="SELECT option_value FROM r0_options WHERE option_name='".$opt."'";
$result=mysqli_query($db,$sql);
$row=mysqli_fetch_array($result,MYSQLI_NUM);
var_dump($row);

However when I do as follows, and call the function, it gives NULL.

function get_result($opt){
    $sql="SELECT option_value FROM r0_options WHERE option_name='".$opt."'";
    $result=mysqli_query($db,$sql);
    $row=mysqli_fetch_array($result,MYSQLI_NUM);
    var_dump($row);
}

get_result('logo_img');

ever heard of variable scope? *well, there you go; now you have* and that's what it is. — Funk Forty Niner, Dec 14 '15 at 17:53

score 1 · Accepted Answer · answered Dec 14 '15 at 17:53

1

It's because you are not passing $db variable, either pass it to function or do the following:

function get_result($opt){
    global $db;

    $sql="SELECT option_value FROM r0_options WHERE option_name='".$opt."'";
    $result=mysqli_query($db,$sql);
    $row=mysqli_fetch_array($result,MYSQLI_NUM);
    var_dump($row);
}

get_result('logo_img');

answered Dec 14 '15 at 17:53

divix

1,265
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oh god. I've spent too much time w/o coffee. Thanks – MWaheed Dec 14 '15 at 17:55

score 0 · Answer 2 · answered Dec 14 '15 at 17:56

0

You are forgotten to pass he '$db' to your function:

function get_result($db, $opt){

answered Dec 14 '15 at 17:56

Franco

2,309
1
11
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php statements fail to work when wrapped in a function

2 Answers2