mysqli :: get_result有什么问题? [英] What's wrong with mysqli::get_result?
本文介绍了mysqli :: get_result有什么问题?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我有以下代码:
$postId = $_GET['postId'];
$mysqli = new mysqli('localhost', 'username', 'database', 'name_db');
mysqli_report(MYSQLI_REPORT_ALL);
$stmt = $mysqli->stmt_init();
$stmt->prepare("
SELECT *
FROM posts
WHERE postId = ?
");
$stmt->bind_param('i', $postId);
$stmt->execute();
$result = $stmt->get_result();//Call to undefined method
$info = $result->fetch_array(MYSQLI_ASSOC);
echo json_encode($info);
我在上面得到了一些错误标记.我做错了什么?
And I get some error marked above. What have i done wrong?
将fecth_array()更改为fetch_array()
changed fecth_array() to fetch_array()
推荐答案
正如其他人所述,它仅在最新的PHP中可用.您可以执行以下操作(回答类似问题):
As others have stated, it is only available in bleeding edge PHP. You could do something like this (answer to a similar question):
function bind_array($stmt, &$row) {
$md = $stmt->result_metadata();
$params = array();
while($field = $md->fetch_field()) {
$params[] = &$row[$field->name];
}
call_user_func_array(array($stmt, 'bind_result'), $params);
}
// ....
bind_array($stmt, $info);
$stmt->fetch();
echo json_encode($info);
或使用 mysqli :: query 进行简单的查询没有参数-请勿将其与动态生成的SQL语句一起使用.
Or use mysqli::query if you have a simple query with no parameters - don't use it with dynamically generated SQL-statements.
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