PHP: Can't fetch an array when using a php variable

Question

I'm trying to fetch an array from a MySQL database but I can't when using a php variable ($final_date) instead of a string (2015-02-16). Can anyone help me?

Here is the code:

$date = new DateTime(date("Y-m-d"));
$date->modify('-1 day');
$prev_date = $date->format('Y-m-d');
$final_date = (string)$prev_date;

function GetPrevUsers(){
    $Adname = "Error";
    $query = mysql_query("SELECT users FROM stats WHERE date = '".$final_date."'");
    while($row = mysql_fetch_array($query)){
        $Adname  = $row['users'];
    }
    return $Adname ;
}

echo GetPrevUsers();

For some reason it the function works when I use the actual date instead of the php variable like this:

$query = mysql_query("SELECT users FROM stats WHERE date = '2015-02-16'");

Try echoing your query to make sure it is correct. Also, it is better to send $final_date as a parameter with you GetPrevUsers() function. — Jerodev
– Jerodev, Commented Feb 17, 2015 at 14:59
The query was right, but as you said I had to send $final_date as a parameter. Thanks! — suchanoob
– suchanoob, Commented Feb 17, 2015 at 15:06

Tacticus · Accepted Answer · 2015-02-17 14:59:44Z

2

It doesn't work as the scopes are different; you try to access a variable from within your function. You could either pass the variable to the function [e.g. : function GetPrevUsers($final_date){ yourCode }] , or add the variable $final_date to the $GLOBALS array.

The first approach is generally the preferred one; only add variables to the $GLOBALS array when it's really neccessary.

answered Feb 17, 2015 at 14:59

Tacticus

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1 Comment

suchanoob Over a year ago

Solved! Thanks a bunch! I will mark your answer as correct when the time limit is over.

Jigar · Accepted Answer · 2015-02-17 15:01:53Z

1

Keep $final_date as global, check function:

function GetPrevUsers(){
    global $final_date;
    $Adname = "Error";
    $query = mysql_query("SELECT users FROM stats WHERE date = '".$final_date."'");
    while($row = mysql_fetch_array($query)){
        $Adname  = $row['users'];
    }
    return $Adname ;
}

answered Feb 17, 2015 at 15:01

Jigar

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