Question

我需要能够使用名为$ topid的变量从我的PHP脚本发送一个值，该变量需要将该id发送到getData.php脚本，所以我尝试了

$(document).ready(function(){

    // Load more data
    $('.load-more').click(function(){
        var topid = Number($('#topid').val());
        var row = Number($('#row').val());
        var allcount = Number($('#all').val());
        row = row + 3;

        if(row <= allcount){
            $("#row").val(row);

            $.ajax({
                url: 'getData.php',
                type: 'post',
                data: {row:row},
                beforeSend:function(){
                    $(".load-more").text("Loading...");
                },
                success: function(response){

                    // Setting little delay while displaying new content
                    setTimeout(function() {
                        // appending posts after last post with class="post"
                        $(".post:last").after(response).show().fadeIn("slow");

                        var rowno = row + 3;

                        // checking row value is greater than allcount or not
                        if(rowno > allcount){

                            // Change the text and background
                            $('.load-more').text("Hide");
                            $('.load-more').css("background","darkorchid");
                        }else{
                            $(".load-more").text("Load more");
                        }
                    }, 2000);


            }
        });
    }else{
        $('.load-more').text("Loading...");

        // Setting little delay while removing contents
        setTimeout(function() {

            // When row is greater than allcount then remove all class='post' element after 3 element
            $('.post:nth-child(3)').nextAll('.post').remove().fadeIn("slow");

            // Reset the value of row
            $("#row").val(0);

            // Change the text and background
            $('.load-more').text("Load more");
            $('.load-more').css("background","#15a9ce");

        }, 2000);


    }

});

}）;

但不知怎的，它似乎没有工作我需要的是id被发送到getData.php脚本然后我通过post来获得它像$ topid = $ _POST ['topid'];

Answer 1

如果您需要将topid发送到getData更改

data: {row:row}

类似

data: {topid : topid}

如何通过post为这个ajax脚本添加一个新变量

1 个答案: