PHP json_encode在MySQL

数组和商店回报，我想一个PHP阵列作为存储在MySQL JSON。为此我有meeting_point_json列，type ='longtext'。PHP json_encode在MySQL

这里是数组：

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 
     ) 

)

这里是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()); 
    }

它应该工作，但取而代之的是，我得到一个错误：

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

任何想法我在这里做错了什么？

来源

2012-11-01 CuriousMind

什么的$数据的全部价值？ –

如果_POST [''team_meeting_pt]是空的或更糟的是NOT SET？ – Svetoslav

@Svetlio或者更糟的是，试图破解他的数据库？ :) –

看到：与".$data." 和 $subteam_id与".$subteam_id."

可变记得SQL查询只发送为字符串，而不是变量中有... 和它做你的所有方式来写PHP $data变量的变化。始终以包裹变量“”和“..”

“”和“..”是取决于你的带包装：例如：如果是这样的 $string = mysql_query("SELECT DATA FROM ".$variable.""); 所以，你必须使用“..”包装！如果这样 $string = mysql_query('SELECT DATA FROM '.$variable.''); 所以，你必须使用“..”包装！

祝你好运的朋友

来源

2012-11-01 13:41:39

它应该是这样的：

$sql = "UPDATE yami_sub_team set meeting_point_json = '" . mysql_escape_string($data) . "'where id = $subteam_id";

你可以阅读更多有关该功能的位置：http://php.net/manual/en/function.mysql-escape-string.php

来源

2012-11-01 12:27:54

试试这个行，而不是你的：

在

$sql = "UPDATE yami_sub_team SET `meeting_point_json`='".$data."' WHERE `id`='".$subteam_id."';";

来源

2012-11-01 12:34:42 anonym

仍然倾向于SQL注入。使用'mysql_real_escape_string'，或者更确切地说，不要使用已弃用的mysql_ *函数，并用预处理语句选择PDO。 –

$data = "'" . mysql_escape_string(json_encode($_POST['team_meeting_pt'])) . "'";

变化1号线

来源

2012-11-01 12:36:58 Svetoslav

首先我会用一个准备语句。并将结果绑定到它以避免所有困难的船只。

$conn = new mysqli($servername, $username, $password, $dbname); 
//replaced $data with ? for security reasons 
$sql = "UPDATE yami_sub_team set meeting_point_json = ? where id = $subteam_id"; 
$stmt->$conn_prepare($sql); 
//binds data to the ? 
// the 's' states string is replacing ? 
$stmt->bind_param('s',$data); 
if($stmt->execute()){ 
{ 
    exit("Done!"); 
}else{ 
    die('Something went wrong....); 
}

来源

2016-01-13 10:21:49 jmodine

PHP json_encode在MySQL

回答

相关问题