PHP Checking if something exists in database

Question

I'm making an iOS app that sends a username string to this PHP file and then the PHP file checks to see if their username exists in a database, in a table called "members". I got this code online and modified it a little to fit my needs. This is the code:

// Main method to redeem a code function redeem() { // Check for required parameters if (isset($_POST["username"])) { // Put parameters into local variables $code = $_POST["username"]; echo $code; // Look up code in database $user_id = 0; echo "userid"; $stmt = $this->db->prepare('SELECT username FROM members WHERE username=', $code); echo "dbprepare"; $stmt->bind_param("is", $code); echo "bindparam"; $stmt->execute(); echo "execute"; $stmt->bind_result($id, $code); echo "bindresult"; while ($stmt->fetch()) { break; } $stmt->close();

The code is tripping up on bind_param, it only gets to echo "dbprepare". Am I doing something incorrectly? How do I check for the username?

None of us have any idea what bind_param or bind_result mean. What API / framework are you using? — Calvin Froedge
– Calvin Froedge, Commented Dec 25, 2011 at 1:50

Kashif Khan · Accepted Answer · 2011-12-25 01:50:24Z

try this code

 $stmt = $this->db->prepare('SELECT username FROM members WHERE username=?'); echo "dbprepare"; $stmt->bind_param("s", $code);

mario · Accepted Answer · 2011-12-25 01:50:44Z

I would guess you do miss an actual placeholder here:

$stmt = $this->db->prepare('SELECT username FROM members WHERE username=?', $code);

See the added ?. The prepare call does not just append the value. You do need to tell it where it belongs. (If your class implements prepare/bind as in mysqli or PDO, and as commonly understood.)

See also what Kashif wrote. If it's really mysqli, then you pass the variable separately, and with maybe just 's' as type specifier.

Lion · Accepted Answer · 2011-12-25 01:53:04Z

You forgot to add a ? in your SQL.

$stmt = $this->db->prepare('SELECT username FROM members WHERE username=?', $code); echo "dbprepare"; $stmt->bind_param("is", $code);

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