How can I view the contents of a prepared statement?

Question

I\'m working on learning to use prepared statements with mysqli in PHP and usually, if I\'m having a problem with a query I just echo it to the screen to see what it looks l

Answer 1

• You can use PDOStatement->debugDumpParams to get some informations about the prepared statement (in case you're using pdo).
• Prepared statements are logged in MySQL's general log:

For prepared statements that are executed with the mysql_stmt_prepare() and mysql_stmt_execute() C API functions, the server writes Prepare and Execute lines to the general query log so that you can tell when statements are prepared and executed.
[...] the server writes the following lines to the general query log:
Prepare [1] SELECT ?
Execute [1] SELECT 3

So for debugging purposes active the general log and keep an eye on that file.

edit: oh, the question has a [mysqli] tag... completely overlooked that.
If the statement isn't executed at all have you (double/tripple) checked that no error occurred along the way?

echo "
Debug: start
\n";

$mysqli = new mysqli('localhost', 'localonly', 'localonly', 'test');
if ($mysqli->connect_error) {
  die('Connect Error (' . $mysqli->connect_errno . ') ' . $mysqli->connect_error);
}

$result = $mysqli->query('CREATE TEMPORARY TABLE foo (id int auto_increment, x int, primary key(id))');
if ( false=== $result) { 
 die('error : '. $mysqli->error);
}

$stmt = $mysqli->prepare('INSERT INTO foo (x) VALUES (?)');
if ( false===$stmt ) {
  die ('prepare() failed: ' . $mysqli->error);
}

$result = $stmt->bind_param('i', $x);
if ( false===$result ) {
  die('bind_param() failed');
}

$x = 1;
$result = $stmt->execute();
if ( false===$result ) {
  die('execute() failed: '.$stmt->error);
}

echo "
Debug: end
\n";