mysqli::get_result 有什么问题?

2021-12-25 00:00:00 php mysqli

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);

我收到了上面标记的一些错误.我做错了什么?

And I get some error marked above. What have i done wrong?

将 fecth_array() 改为 fetch_array()

changed fecth_array() to fetch_array()

推荐答案

正如其他人所说，它仅在前沿 PHP 中可用.你可以做这样的事情(answer to a类似的问题):

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);

或者使用 mysqli::query 如果你有一个简单的查询无参数 - 不要将它与动态生成的 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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