CODEWITHSUNDEEP
phpmysqlarraysimplode
user2848253
user2848253

Reputation:

Implode a single column of values from a query result set

I'm trying to put an array into a query but I doens't work. I'm tying it with implode() but then it gives me " Array to string conversion in ... on line 26". Why? With json_encode it worked out ...

$sql = mysql_query("SELECT follows
            FROM follow 
             WHERE follower LIKE '".$id."'") or die (mysql_error());

if (mysql_num_rows($sql) < 1) {
    echo "<br/>";
    echo "Follow someone";
} else {
    //Put all the id's of the users the user is following in an array. 
    $i = 0;
    $user_follows = array();
    while ( $row = mysql_fetch_assoc($sql) ) {
        $user_follows[$i] = $row;
        $i++;
    }

    $user_follows = implode(" , ", $user_follows);
    echo $user_follows;
}

Upvotes: 0

Views: 113

Answers (1)

Barmar
Barmar

Reputation: 781059

The second argument to implode must be an array of strings. But you're doing:

$user_follows[$i] = $row;

Since $row is an array, you're making an array of arrays (a 2-dimensional array), not an array of strings. That should be:

$user_follows[] = $row['follows'];

You don't need the $i variable, assigning to $array[] appends a new element to the array.

Upvotes: 2

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