Php MYSQL错误Get Prepared语句

如何获取mysqli_stmt-object的准备语句

如果在执行mysql语句时出现错误，我希望返回该语句

$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_语句”当然不起作用。那么如何获得完整的查询？在本例中：

"SELECT * FROM database WHERE id = 89c483c8"

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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());
}

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您应该引用十六进制值。没有这样的机会。只需使用

error

属性即可。类似的：

echo $stmt->error