斗士狗 2016-02-14 12:39 采纳率: 0%
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Ajax和jQuery与PHP

I have this code which posts data using jQuery and ajax to itself.

<input type="button" id="butt"  value="button11111111111111" >
<script>
$("#butt").on('click',function(e)
    {
        $.ajax(
            {
                type:'POST',
                url:'test.php',
                data:product_type:"cake"
            });
    });
</script>

<?php
if(isset($_POST['product_type']))
         $abc=$_POST['product_type'];

 if(isset($abc))
      echo $abc; 
  ?>

Now when I try to run this code.I do get the ok status inside the Network Console of Chrome but this code doesn't echo the output.

I simply want to display the result that has been passed by the ajax method.

I am new to ajax and jquery so i don't know much about how they work exactly but is it possible? If yes, then how could i achieve that without actually refreshing the page?

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1条回答 默认 最新

  • weixin_33713707 2016-02-14 12:54
    关注

    Here is what I think you want, i.e. test.php is the source page AND the ajax page!

    <?php
    if(isset($_POST['product_type'])) {
        Header("Content-type: text/plain; charset=utf-8");
        $abc=$_POST['product_type'];
        if(isset($abc)) {
            echo $abc;
        }
    }
    else {
?>
<html>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.0/jquery.min.js">   </script>
<input type="button" id="butt"  value="button11111111111111" >
<script>
    $("#butt").on('click',function(e) {
        $.ajax({
            type:'POST',
            url:'test.php',
            data: {
                product_type:"cake"
            },
            success: function(data) {
                alert('got ' + data);
            }
        });

    });
</script>
</html>
<?php
    }
?>
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