Question

所以这是导致此时头疼的代码。得到此错误：

致命错误：调用未定义的方法mysqli_result :: execute（）

不确定我在哪里做错了：

$mysqli_load = new mysqli(HOST, USER, PASS, DB);
$query = 'SELECT `id`, `call_taker`, `call_time`, `call_type`, `call_priority`, `call_location`, `call_city`, `reporter_name`, `reporter_num`, `call_info` FROM `calls`';
$sql = mysqli_query($mysqli_load, $query) or die($mysqli_load->error);
$sql->execute() or die($mysqli_load->error);
$sql->bind_result($c_id, $c_taker, $c_time, $c_type, $c_priority, $c_location, $c_city, $r_name, $r_num, $c_info) or die($mysqli_load->error);
while($row = mysqli_fetch_assoc($sql)){

Answer 1

mysqli_query正在返回mysqli_result object，它具有操作查询结果的方法，如fetch，fetch_object等，因为它是结果。此时已经执行了查询。

mysqli::bind_result的使用位置错误，应该在语句中运行。应该完全避免，因为它已在PHP5.4中删除。

如何以这种方式制定它：

$mysqli_load = new mysqli(HOST, USER, PASS, DB);
$query = 'SELECT `id`, `call_taker`, `call_time`, `call_type`, `call_priority`, `call_location`, `call_city`, `reporter_name`, `reporter_num`, `call_info` FROM `calls`';
$sql = mysqli_query($mysqli_load, $query) or die($mysqli_load->error);

while ($row = mysqli_fetch_assoc($sql)) {
    // data manipulation
}

// and possibly close connection

无需执行和绑定任何内容。

为了将来参考，如果您的查询需要一些用户输入，请查看How can I prevent SQL injection in PHP?

致命错误：调用未定义的方法mysqli_result :: execute（）

1 个答案: