AJAX + jQuery变量不会传递给PHP

问题描述

我正在尝试做一些非常简单的事情:将2个文本变量传递给php脚本，并将其插入到MySQL数据库中.但是由于某种原因，我无法传递变量(因此，我在数据库中只得到空记录).

I am trying to do something very simple: Pass 2 text variables to a php script and insert them into a MySQL db. For some reason however I can't get the variables to pass (so I just get empty records in my DB).

function ajaxCall(){

        $.ajax({
            type: "GET",
            url: "http://www.*.be/bubblingAjax.php",
            cache: false,
            data: "colour="+colour+"&size="+size,
        dataType: "html",
            success: onSuccess
        });
        return false;
    };

和PHP:

<?php
    try
    {
        $connection = mysql_connect("#");
        mysql_select_db("#");

        $colour = mysql_real_escape_string($_GET['colour']);
        $size = mysql_real_escape_string($_GET['size']);

        mysql_query("INSERT INTO bubble (colour, size) VALUES ('$colour', '$size')");
        mysql_close($connection);
        echo "SUCCESS";
        echo $colour;
        echo $size;
    }
    catch(Exception $e)
    {
        echo $e->getMessage();
    }
?>

有人愿意快速浏览一下，指出我的-可能是显而易见的-错误吗?一天来一直让我发疯.

Anyone willing to take a quick look at it and point out my -probably obvious- mistake? It's been driving me nuts for over a day.

推荐答案

这必须有效:

<script type="text/javascript">
$(document).ready(function() {
   //you can wrap the code into an event, e.g click()
   var colour=...
   var size=...
   $.post("http://www.website.com/bubblingajax.php", { colour: colour, size: size },
   function(data) {
     alert("Respond: " + data);
   });
});

</script>

和PHP(只有更改才能发布)

and the PHP (only changed get to post)

<?php
    try
    {
        $connection = mysql_connect("#");
        mysql_select_db("#");

        $colour = mysql_real_escape_string($_POST['colour']);
        $size = mysql_real_escape_string($_POST['size']);

        mysql_query("INSERT INTO bubble (colour, size) VALUES ('$colour', '$size')");
        mysql_close($connection);
        echo "SUCCESS";
        echo $colour;
        echo $size;
    }
    catch(Exception $e)
    {
        echo $e->getMessage();
    }
?>

也可以调试，我建议在检查工具中使用firebug或chrome的构建.

Also to debug, I would suggest using firebug or chrome's build in inspect tools.

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