php - 无法访问 php 数组进行 m​​ysql 查询

Question

foreach($res as $key) {
    echo $key['id'].'<br>';
    $sql= "INSERT INTO Post VALUES('".$key['id']."')";

    if (mysql_query($sql) == TRUE) {
        echo "New record created successfully";
    } else {
        echo "Error: " . $sql . "<br>" . $link->error;
    }
}

对于第一个条目，它可以工作并将其添加到数据库中，但之后会出现以下错误: 尝试获取非对象的属性

print_r($res) 给出以下输出:

Array
(
    [0] => Array
        (
            [message] =>*******
            [created_time] => DateTime Object
                (
                    [date] => 2016-01-13 20:14:32.000000
                    [timezone_type] => 1
                    [timezone] => +00:00
                )

            [id] => 1413366785545501_1686416964907147
        )

    [1] => Array
        (
            [message] =>*****
            [created_time] => DateTime Object
                (
                    [date] => 2015-11-27 15:51:16.000000
                    [timezone_type] => 1
                    [timezone] => +00:00
                )

            [id] => 1413366785545501_1674032329478944
        )

    [2] => Array
        (
            [message] => The 'AD- MAD' competition :)
            [story] => KRITA // Chalk your art. added 4 new photos.
            [created_time] => DateTime Object
                (
                    [date] => 2015-10-07 19:02:12.000000
                    [timezone_type] => 1
                    [timezone] => +00:00
                )

            [id] => 1413366785545501_1662706990611478
        )

    [3] => Array
        (
            [message] => *****
            [story] => KRITA // Chalk your art. created a poll.
            [created_time] => DateTime Object
                (
                    [date] => 2015-10-02 21:06:05.000000
                    [timezone_type] => 1
                    [timezone] => +00:00
                )

            [id] => 1413366785545501_1661659277382916
        )

我只想将 id 存储在数据库中

Answer 1

mysql_connect()不返回对象，因此当您尝试使用 $link->error 时，您会收到给定的错误消息。将其替换为 mysql_error() :

echo "Error: " . $sql . "<br>" . mysql_error();

您将收到一条清晰的错误消息，说明 SQL 查询失败的原因。

但是我必须重复评论中的注释:mysql_* 函数已被弃用，它们已从 PHP 7 中删除，当您升级到该版本时，您的代码将停止工作。您不应该使用它们编写新代码，请使用 mysqli_* or PDO相反。