notice: array to string conversion in php

Question

<?php       
 $player[] = array();
    $team_id = $_SESSION['tid'];

    $team_pids = $con->prepare("SELECT p_id FROM players_to_team WHERE t_id = ?");

    $team_pids->bindParam(1,$team_id);

    $team_pids->execute();

    while($info = $team_pids->fetch(PDO::FETCH_ASSOC))
    {
            $player[] = $info['p_id'];
            echo $info['p_id'];
    }
    $pl_1 = $player[0];
    .
        .
        .
    $pl_10 = $player[9];

    echo $player[0]; //notice here
    echo $pl_1;      //notice here
?>
<table>

$query = $con->prepare("SELECT role,name,value FROM players WHERE p_id = '".$pl_1."'");
// notice here
               $query->execute();

               while($result = $query->fetch(PDO::FETCH_ASSOC))
               {
                     echo "<tr>";  
                     echo "<td>".$result['role']."</td>";
                     echo "<td>".$result['name']."</td>";
                     echo "<td>".$result['value']."</td>";
            }
?>
</tr>
</table>

when i echo $info array it works fine, but when i echo $player array or $pl_1 variable or $result array values Notice appears...Array to string conversion and o/p doesn't show. why?

because both are arrays not a string.Instead of echo use print_r($player[0]); and print_r($pl_1); to see the array. — 웃웃웃웃웃
– 웃웃웃웃웃, Commented Oct 24, 2013 at 11:50
you can use var_dump on any variable to see the variable TYPE as well as it's content to get a better understand of your variable assignments through your code. — Latheesan
– Latheesan, Commented Oct 24, 2013 at 11:53
possible duplicate of Reference - What does this error mean in PHP? — ɴᴀᴛᴇ ᴅᴇᴠ
– ɴᴀᴛᴇ ᴅᴇᴠ, Commented Oct 24, 2013 at 12:01

Guillaume Chevalier · Accepted Answer · 2013-10-24 13:15:07Z

11

Try replacing $player[] = array(); by $player = array(); at the beginning (line 2).

This is because that you declare an array at the index 0 of this variable which is told to be an array because of the []. You therefore try to place an array in your array, making it multidimensional.

edited Oct 24, 2013 at 13:15

answered Oct 24, 2013 at 11:55

Guillaume Chevalier

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Comments

deceze · Accepted Answer · 2013-10-24 11:55:29Z

9

You cannot simply echo an array. echo can only output strings. echo 'foo' is simple, it's outputting a string. What is echo supposed to do exactly in the case of echo array('foo' => 'bar')? In order for echo to output anything here, PHP will convert array('foo' => 'bar') to a string, which is always the string "Array". And because PHP knows this is probably not what you want, it notifies you about it.

The problem is you're trying to treat an array like a string. Fix that.

answered Oct 24, 2013 at 11:55

deceze♦

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1 Comment

Anthony Over a year ago

One does not simply echo an array.

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