dongxi5505 2018-07-25 07:47
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PHP没有捕获ajax请求

I am having request like this:

$(function(){
    $("#novaProfaktura").click(function(){
        $.ajax({
          type: "POST",
          url: "../Pages/Program.php",
          data: { action: "testing" },
          success: function()
          {
              alert("Ajax done"); //Testing purpose
          }
        });
        location.reload();
    });
});

And then in my body I have php like this:

<div id="Content">
    <?php
    echo("Php running <br>");
    if($_SERVER['REQUEST_METHOD'] == 'POST')
    {
        echo("Post running <br>");
        if($_POST['action'] = "testing")
        {
            echo("Action running <br>");
        }
    }

    ?>
</div>

What I am getting from this is alert message so I assume ajax is set up properly and getting Php running but it doesn't pass if($_SERVER['REQUEST_METHOD'] == 'POST'

As you can see jQuery code is running on div click.

What am I doing wrong?

EDIT: I have used function(data) and it seems to work BUT it doesn't return only echoed data but WHOLE page with echo. I used document.write() and it rewrite whole page but how to get particular echo value?

  • 写回答

2条回答 默认 最新

  • duanjiao5723 2018-07-25 12:50
    关注

    Answer for this was to instead of this:

    success: function()
    {
        alert("Ajax done"); //Testing purpose
    }
    

    use this:

    success: function(data)
    {
        alert("Ajax done"); //Testing purpose
        document.getDocumentById('emptyDiv').innerHTML = data;
    }
    

    Now as data you are getting all HTML text from document you are running function from so instead of url: "../Pages/Program.php" I used separated file url: "../Php/Testing.php" and inside that file I have only this code:

    <?php
        echo("Testing");
    ?>
    
    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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