dou6577 2016-05-13 05:24
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如何将JavaScript变量的数据传递给另一个php页面?

I have two pages: test.php and backend.php.

test.php has three images which when clicked takes the user to backend.php. I want the src of the image clicked to be displayed on backend.php. I am trying to achieve this task via JavaScript and Ajax but the src of the image doesn't appear on backend.php. Code for the same are given below:

test.php:

<?php
    session_start();
?>

<html>
    <head>
        <script src="jquery-1.9.1.js">
        </script>

        <script>
            function clickIt(data){
                $.ajax({
                           url: 'backend.php',
                           data: {"imgsrc":data},
                           type: 'post',
                           success:function(res){
                              alert(res.message);
                           }
                });
            }
        </script>
    </head>

    <body>
        <a href="backend.php">
            <img onclick="clickIt(this.src)" src="img1.jpg"/>
        </a>
        <a href="backend.php">
            <img onclick="changeIt(this.src)" src="img2.jpg"/>
        </a>
        <a href="backend.php">
            <img onclick="changeIt(this.src)" src="img3.jpg"/>
        </a>
    </body>
</html>

backend.php:

<?php
    session_start();
?>

<?php
    echo $_POST['imgsrc'];
    echo '<a href="test.php">Back</a>';
?>

What am I possibly doing wrong?

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2条回答 默认 最新

  • dtfo55908 2016-05-13 05:37
    关注

    Try this;

    HTML:

    <body>
        <a href="javascript:void(0);">
            <img onclick="clickIt(this.src)" src="img1.jpg"/>
        </a>
        <a href="javascript:void(0);">
            <img onclick="changeIt(this.src)" src="img2.jpg"/>
        </a>
        <a href="javascript:void(0);">
            <img onclick="changeIt(this.src)" src="img3.jpg"/>
        </a>
    </body>
    

    Jquery:

    function clickIt(data){
        window.location.href = 'backend.php?imgsrc='+data; // redirect to backend.php
    }
    

    In backend.php

    <?php
        echo $_GET['imgsrc']; // use $_GET
        echo '<a href="test.php">Back</a>';
    ?>
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论
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