PHP json编码数组并将返回值存储在mysql中

Question

我想在 mysql 中将 php 数组存储为 JSON 。为此，我有 type='longtext' 的 meeting_point_json 列。

这是数组:

Array
(
    [1] => Array
        (
            [date] => 23/4/2012
            [meeting_time] => 23:04
            [meeting_place] => town hall
            [venue] => London
            [opponents] => Tigers
            [official_incharge] => Mr Putin
        )

    [2] => Array
        (
            [date] => 23/4/2050
            [meeting_time] => 13:04
            [meeting_place] => chief office
            [venue] => Kenya
            [opponents] => Peococks
            [official_incharge] => Mr Black
        )

    [3] => Array
        (
            [date] => dsad
            [meeting_time] => sadas
            [meeting_place] => jjjjj
            [venue] => jjjj
            [opponents] => dasds
            [official_incharge] => asad
        )

)

这是 php 代码:

$data = json_encode($_POST['team_meeting_pt']);
    $sql = "UPDATE yami_sub_team set meeting_point_json = $data where id = $subteam_id";
    if(mysql_query($sql)){
        exit("Done!");
    }else{
        die('Something went wrong, changes not saved. Error details: ' . mysql_error());
    }

它应该可以工作，但相反，我收到一个错误:

Something went wrong, changes not saved. Error details: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '"1":{"date":"23\/4\/2012","meeting_time":"23:04","meeting_place":"town hall","ve' at line 1

知道我在这里做错了什么吗？

Answer 1

参见:$data 变量更改为 ".$data."和$subteam_id 变量，带有 ".$subteam_id."

记住 sql 查询仅作为字符串发送，而不是内部变量......并按照您编写 PHP 的所有方式进行操作。始终用“..”和“..”包裹变量

“..”和“..”取决于您的 strip 包装器:例如:如果看起来像这样$string = mysql_query("从“.$variable.”中选择数据。”);所以，你必须使用“..”包装!如果像这样$string = mysql_query('从'.$variable.'中选择数据。'');因此，您必须使用 '..' 包装器!

祝 friend 好运