如何使用jQuery POST将变量从JavaScript传递给PHP [英] How to pass variable from JavaScript to PHP using jQuery POST

问题描述

我从页面chat.php中的以下Javascript函数传递变量 sessionnum ：

I am passing the variable sessionnum from the following Javascript function in the page chat.php:

$(document).ready(function(){

        timestamp = 0;
        updateMsg();
        $("form#chatform").submit(function(){
            $.post("backend.php",{
                        message: $("#msg").val(),
                        name: author,
                        action: "postmsg",
                        time: timestamp,
                        tablename1: sessionnum
                    }, function(xml) {
                $("#msg").empty();

                addMessages(xml);

                document.forms['chatform'].reset()
                fixScroll();
            });
            return false;
        });
    });

到backend.php中的以下PHP函数：

To the following PHP function in backend.php:

if(@$action == "postmsg") {
    mysql_query("INSERT INTO `$tablename1` (`user`,`msg`,`time`)
                VALUES ('$name','$message',".time().")",$dbconn);
    mysql_query("DELETE FROM `$tablename1` WHERE id <= ".
                (mysql_insert_id($dbconn)-$store_num),$dbconn);
    }

$messages = mysql_query("SELECT user,msg
                         FROM `$tablename1`
                         WHERE time>$time
                         ORDER BY id ASC
                         LIMIT $display_num",$dbconn);

仅当我对诸如 $ tablename1 =的作业进行硬编码时才有效在backend.php中有100 ，即使变量及其值都是整数和相同的值。这个hack是不可接受的，因为我实际上必须传递变量。我的代码中是否有错误？

It only works when I hard-code an assignment such as $tablename1 = 100 in backend.php even though both the variable and its value are integers and the same value. This hack is not acceptable, as I actually have to pass the variable. Is there a bug in my code?

此代码改编自 http://articles.sitepoint.com/article/ajax-jquery/3

感谢任何帮助张贴使用jQuery正确变量。

Thanks for any help POSTING the variable correctly with jQuery.

推荐答案

尝试将POST变量更改为 $ _ POST ['variable_name'] 。您使用的语法依赖于全局变量注册。这是一项功能，a）默认情况下未启用，b）启用时会产生重大安全风险。因此，请尝试将服务器端代码更改为：

Try changing the POST variables to $_POST['variable_name']. You're using a syntax that relies on globals being registered as variables. This is a feature that is a) not enabled by default and b) poses a major security risk when it is enabled. Thus, try changing your server-side code to:

$action = $_POST['action'];
$tablename1 = mysql_real_escape_string($_POST['tablename1']);
$name = mysql_real_escape_string($_POST['name']);
$message = mysql_real_escape_string($_POST['message']);

if(@$action == "postmsg") {
    mysql_query("INSERT INTO `$tablename1` (`user`,`msg`,`time`)
                VALUES ('$name','$message',".time().")",$dbconn);
    mysql_query("DELETE FROM `$tablename1` WHERE id <= ".
                (mysql_insert_id($dbconn)-$store_num),$dbconn);
    }

$messages = mysql_query("SELECT user,msg
                         FROM `$tablename1`
                         WHERE time>$time
                         ORDER BY id ASC
                         LIMIT $display_num",$dbconn);

请注意，为了防止某些SQL注入，您在SQL中使用的变量使用 mysql_real_escape_string对问题（用户可能更改）进行了转义。

Note that, in order to prevent some SQL injections, the variables that you're using in your SQL queries (that the user can potentially change) have been escaped using mysql_real_escape_string.

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