douyu9433 2014-03-21 15:07
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Jquery没有将值传递给php页面

I am coding a php album to preview the images and comments i am using Jquery below is code

php

$photo .="<div class=\"imagearray\"><span>".$imagerealname ."</span><br/><div class=\"imagecontent\"><image src='uploads/thumb_".$query['imagesrc']."' alt='image' id='".$query['ImageSN']."' /></div>share delete</div>";

To call images from Database

then click on any image jquery function executes

// display photo preview ajaxy
    $('.imagearray .imagecontent img').click(function (event) {
        if (event.preventDefault) event.preventDefault();

        getPhotoPreviewAjx($(this).attr('id'));
    });
})

getPhotoPreviewAjx is

function getPhotoPreviewAjx(id) {

        var id = id;
        alert(id);  
    $.post('commentblock.php',
            { action: 'get_info', Id: id },
        function(data){
            $('#photo_preview .pleft').html(data.data1);
            $('#photo_preview .pright').html(data.data2);
            $('#photo_preview').show();
        }, "json"
    );
};

this code is working till alert(id); and next i dont think it is passing values to commentblock.php

starting of commentblock.php is

if( $_POST['action'] == "get_info" &&  isset($_POST['Id']))
{
echo $_POST['id'];

and it is not working need some guideness struck here :(

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2条回答 默认 最新

  • dsfdsf8888 2014-03-21 15:12
    关注

    (After debugging)

    The javascript works correctly. There is an error or something stopping it working correctly in server side code.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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