php - MYSQL错误获取准备语句

Question

如何获取 mysqli_stmt 对象的准备语句？

如果在执行 mysql 语句时出现错误，我想返回该语句。

$id = "89c483c8";
$query = "SELECT * FROM database WHERE id = ?";
if (!($stmt = $database->prepare($query) { ... }
else {
    $stmt->bind_param("s", $id);
    if (!$stmt->execute())
         return $stmt->get_statement; //doesn't exist
}

“$stmt->get_statement”当然不行。那么我如何获得完整的查询呢？在这个例子中:

"SELECT * FROM database WHERE id = 89c483c8"

Answer 1

这是捕获 sql 错误的最佳方法:

try {
    $res = $mysqli_instance->query($query);
}catch (mysqli_sql_exception $e) {
    print "Error Code <br>".$e->getCode();
    print "Error Message <br>".$e->getMessage();
    print "Strack Trace <br>".nl2br($e->getTraceAsString());
}

或者最简单的方法:

echo $stmt->error

http://php.net/manual/en/mysqli.error.php