CODEWITHSUNDEEP
phpmysqlsqlfunction
Marcello Infoweb
Marcello Infoweb

Reputation: 89

How to declare a variable

I have problem undefined variable. The thing is, I found a tutorial to make a type micro blog Twitter, had some problems, I decided, but this is taking me patience.

Error Message:

Notice: Undefined variable: extra in caminho\do\arquivo\functions.php on line 58

function:

function show_users($user_id=0){

    if ($user_id > 0){
        $follow = array();
        $fsql = "SELECT user_id FROM following WHERE follower_id='$user_id'";
        $fresult = mysql_query($fsql);

        while($f = mysql_fetch_object($fresult)){
            //array_push($follow, $f->user_id);
            $follow[] = $f->user_id;
        }
        if (count($follow)){
            $id_string = implode(',', $follow);
            $extra =  " AND id IN ($id_string)";
        }else{
            return array();
        }
    }
    $users = array();
    $sql = "SELECT id, username FROM users WHERE status='active' $extra ORDER BY username";
    $result = mysql_query($sql);

    while ($data = mysql_fetch_object($result)){
        $users[$data->id] = $data->username;
    }
    return $users;
}

The line in question is:

$sql = "SELECT id, username FROM users WHERE status='active' $extra ORDER BY username";

Any suggestions?

Upvotes: 0

Views: 119

Answers (4)

M.RK
M.RK

Reputation: 21

function show_users($user_id=0,extra='') // use this
        {    
    if ($user_id > 0){
        $follow = array();
        $fsql = "SELECT user_id FROM following WHERE follower_id='$user_id'";
        $fresult = mysql_query($fsql);

        while($f = mysql_fetch_object($fresult)){
            //array_push($follow, $f->user_id);
            $follow[] = $f->user_id;
        }
        if (count($follow)){
            $id_string = implode(',', $follow);
            $extra =  " AND id IN ($id_string)";
        }else{
            return array();
        }
    }

Upvotes: 0

Lab
Lab

Reputation: 1063

There are two ways :

1. function show_users($user_id=0){
   $extra='';

2. function show_users($user_id=0,$extra=''){

Actually you are declaring $extra only when you have $user_id > 0 BUT you must be having $user_id equals to or less than 0 for the current situation and thus your query couldn't find $extra.

I hope you get an idea.

Upvotes: 1

Marc B
Marc B

Reputation: 360592

function show_users(...) {
   $extra = ''; // <--just add this line
   blah blah blah blah blah
}

In your code, $extra gets defined ONLY if both if() tests succeed. Obviously they're not, so you end up with an undefined $extra in your sql string.

Upvotes: 0

Sam
Sam

Reputation: 2970

function show_users($user_id=0){
 $extra = ''; //if no extra, you'll get just an empty string
 //etc...
}

Upvotes: 0

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