CODEWITHSUNDEEP
phpmysqliprepared-statement
user10588397
user10588397

Reputation:

Unsure why I am getting: Number of variables doesn't match number of parameters in prepared statement

I am trying to do a simple test but am experiencing errors. I am trying to add the parameters and I have done some research, none of which has helped me understand where I have gone wrong.

I have already tried looking on the PHP website and Stackoverflow. This is for a test project.

$stmt = $con->prepare('SELECT username, rank, id, steamid, avatar FROM users WHERE id="$uid"');

$stmt->bind_param('i', $uid);
$stmt->execute();
$stmt->bind_result($username, $rank, $id, $steamid, $avatar);
$stmt->fetch();
$stmt->close();

My expected result is for it to select only the rows specified with the "WHERE" call. My actual result is this error:

mysqli_stmt::bind_param(): Number of variables doesn't match number of parameters in prepared statement

Upvotes: 0

Views: 885

Answers (2)

Marios Nikolaou
Marios Nikolaou

Reputation: 1336

You get error because you don't use column name = ? in the prepared statement. Then using bind_param you bind values to parameters, set parameters and execute.

$stmt = $con->prepare('SELECT username, rank, id, steamid, avatar FROM users WHERE id = ?');

$stmt->bind_param('i', $uid);
//set parameters and execute
$uid = 'value here';
$stmt->execute();

$stmt->close();

Upvotes: 0

Bill Karwin
Bill Karwin

Reputation: 562681

You are binding one variable, but you have zero parameters in your prepared statement. Interpolating a PHP variable into a string is not a parameter.

WRONG: Zero parameters, just string interpolation:

$stmt = $con->prepare('SELECT username, rank, id, steamid, avatar FROM users 
  WHERE id="$uid"');

RIGHT: One parameter:

$stmt = $con->prepare('SELECT username, rank, id, steamid, avatar FROM users
  WHERE id=?');

Upvotes: 6

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