将javascript变量传递给php mysql选择查询 [英] pass javascript variable to php mysql select query

问题描述

我在php中运行mysql select查询这样

I am running a mysql select query in php like this

<?php
  $getvalue="SELECT id,name from table1 WHERE column1='$var1' and column2='$var2'";
  $result=mysql_query($getvalue) or die(mysql_error());

  while($row=mysql_fetch_array($result)){
       extract($row);
       echo $name;
  }
?>

var1和var2是同一页面上的javascript变量。我知道客户端变量不能传递给服务器端。但是有变通方法是在同一页面吗？

var1 and var2 are javascript variables on the same page. I know client side variable cannot be passed to server side. But is there any workaround as the variables are in the same page.

推荐答案

为了让你做到这一点 - 我相信 - 你必须使用AJAX。

In order for you to make this happen - I believe - you have to use AJAX.

代码如下所示：

        $.ajax({
            url: 'your_script.php',
            type: 'POST',
            data: {var1: javascript_var_1, var2: javascript_var_2},
            success: function(data) {
                console.log("success");
            }
        });

您的PHP看起来与此类似（不记得JSON编码：

Your PHP will look similar to this (without keeping in mind the JSON encode:

<?php

$var1 = $_POST['var1'];
$var2 = $_POST['var2'];

  $getvalue="SELECT id,name from table1 WHERE column1='$var1' and column2='$var2'";
  $result=mysql_query($getvalue) or die(mysql_error());

  while($row=mysql_fetch_array($result)){
       extract($row);
       echo $name;
  }
?>

然后你可以对结果进行JSON编码成功输出它们。你的PHP脚本 - 但是 - 必须存在于另一个php文件中。

Then you can JSON encode the results and pretty much output them on the success. Your php script - however - must live on another php file.

另外，转义你的数据。使用预备语句。

Also, escape your data. Use prepared statements.

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