undefined index invalid argument for loop error

Question

Please point me in the right direction here Im trying to do the following:

Multiple select boxes are generated for each eventid
User choose who they think will win each event
The name of each selectBox is the event id assigned to variable $id
At end of while loop I want to extract the array $id value in For loop, however im getting error "undefined offset & invalid argument" on my for loop...

Here is my form

enter image description here

    $i=0;//counter
    while($row=mysql_fetch_array($result)){

    $team1 = $row['team1'];
    $team2 = $row['team2'];
    $id[$i]= $row['event_id']; 

    echo'<h3>'.$team1.' VS '.$team2.'</h3>';
    echo'<select name="'.$id[$i].'">';
            echo'<option value="'.$row['team1'].'">'.$team1.'</option>';
            echo'<option value="'.$row['team2'].'">'.$team2.'</option>';
            echo'</select>';    
        $i++;
    }//while

Here is my for loop giving error, I suspect problem is in the $_POST['$id']...

if(isset($_POST['submit'])){    

    foreach($_POST[$id] as $eventId => $winner){
     echo'<h3>'.$eventId.'</h3>';
}//for loop
}//end isset

Any help will be greatly appreciated

@Lashane is this what you are asking $id[$i]= $row['event_id']; WHERE $row['event_id']; is retrieved from database — Marilee, Apr 17 '15 at 17:59
possible duplicate of [PHP: "Notice: Undefined variable" and "Notice: Undefined index"](http://stackoverflow.com/questions/4261133/php-notice-undefined-variable-and-notice-undefined-index) — chris85, Apr 17 '15 at 18:00
What is the exact error message and to what line does it relate? — ficuscr, Apr 17 '15 at 18:01
I think you need to change $id to just id in your code. You are using a variable when you are actually trying to use a constant. — Maximus2012, Apr 17 '15 at 18:01
no, before `isset` you have to define scalar (not array) variable `$id` - how and where it is defined? note - this variable has nothing to do with $id array which is used in first part of your code — Iłya Bursov, Apr 17 '15 at 18:02
@ficuscr the message is Illegal offset type and Invalid argument supplied for foreach() — Marilee, Apr 17 '15 at 18:04
Then thinking its as Maximus2012 said then... see: http://stackoverflow.com/questions/6628/php-array-indexing-arrayindex-vs-arrayindex-vs-arrayindex — ficuscr, Apr 17 '15 at 18:08
@Maximus2012 that works but returns the index of the array and does not give me the event_id im looking for — Marilee, Apr 17 '15 at 18:08
@Maximus2012 its pointing me in the correct directions im playing around with the idea but doesn't quite work — Marilee, Apr 17 '15 at 18:18
What part is not working? On a different note, could you use radio buttons in place of dropdown list since your logic needs to choose one of the two options. — Maximus2012, Apr 17 '15 at 18:19
@Maximus2012 I got it working thank you so much for the help, ill change to radiobuttons ;-) — Marilee, Apr 17 '15 at 18:28

score 2 · Accepted Answer · answered Apr 17 '15 at 18:06

$id is defined as an array here$id[$i]= $row['event_id'];. Arrays can not be used as a key in a foreach array or otherwise. This is what is causing you an error on this line

foreach($_POST[$id] as $eventId => $winner){ //$id is an array of values
    echo'<h3>'.$eventId.'</h3>';
}//for loop

You have to make a second foreach statement for the $id then use the id values in your current foreach statement.

foreach( $id as $key => $val ) {
foreach( $_POST[$val] as $eventId => $winner){

thank you very much for your help, I tweaked it a bit but it works like a bomb — Marilee, Apr 17 '15 at 18:27

score 0 · Answer 2 · answered Apr 17 '15 at 18:24

You don't know what the event_id values are on the action page, so I think you'll have to query for those again something like:

if(isset($_POST['submit'])){
    /* do your query here */
    $data = array();
    while($row=mysql_fetch_array($result)){
     $data[] = $_POST[$row['event_id']];
    }

    foreach($data as $key => $value){
        print_r($_POST[$value]);
        echo "<br>";
    }
}

undefined index invalid argument for loop error

2 Answers2