Question

在我的php中我在while循环中有这样的代码......

$result = mysql_query("SELECT * FROM wallpost ORDER BY wallpostid DESC");
while($row = mysql_fetch_assoc($result)){
$rate = "<div class=\"example-".$id." \" data-postid=\"".$id."\"></div></br>Post id: <span class=\"example-rating-".$id."\">".$id."</span>";
}

jquery是......

$(document).ready(function() {
$('[class^="example-"]').not('[class^="example-rating-"]').ratings(3).bind('ratingchanged', function (event, data) {
var child = $(this).find('[class^="example-rating-"]');
child.text(data.rating);

$.post('count.php', {
     rate: data.rating,
     wallpostid: jQuery(this).data("postid")
}, function (data) {
    alert(data);
});
});

表示值A我得到空值，但是如果我替换

var a = $('.example-rating-50').html();  //let say the wallpostid is 50

它只能将值50传递给count.php 如果现在让我说我有2个wallpostid，它是22和50（循环时循环）如果我评价wallpostid是22然后我想从PHP传递$ id = 22的值到jquery和$ .post到count.php。如果我对wallpostid = 50进行评分，请执行此操作。

Answer 1

闭包变量i

是一个问题

由于你在回调中使用i，它将具有来自循环的最后一个值102，这就是它失败的原因

$(document).ready(function () {
    for (var i = 1; i < 101; i++) {
        (function(idx){
            $('.example-' + idx + '').ratings(3).bind('ratingchanged', function (event, data) {
                $('.example-rating-' + idx + '').text(data.rating);
                var a = $('.example-rating-' + idx + '').html();

                $.post('count.php', {
                    rate: data.rating,
                    wallpostid: a
                }, function (data) {
                    alert(data);
                });
            });
        })(i)
    }
});

更好的解决方案可能是

$rate = "<div class=\"examples example-".$id." \" data-idx=\"".$id"\"></div></br>Post id: <span class=\"example-rating-".$id."\">".$id."</span>";

然后

$(document).ready(function () {

    $('.examples').ratings(3).bind('ratingchanged', function (event, data) {
        var i = $(this).data('idx')
        var a = $('.example-rating-' + i + '').text(data.rating);

        $.post('count.php', {
            rate: data.rating,
            wallpostid: data.rating
        }, function (data) {
            alert(data);
        });
    });

});

Answer 2

你在javascript中不需要for循环，有选择器可以处理这个，例如class^=

<强> PHP

$rate = "<div class=\"example-".$id." \" data-postid=\"".$id."\"></div></br>Post id: <span class=\"example-rating-".$id."\">".$id."</span>";

<强> JS

//Find elements that have class starting with 'example-' but Not example-rating- 
$('[class^="example-"]').not('[class^="example-rating-"]').ratings(3).bind('ratingchanged', function (event, data) {
    //Find Child element that has class starting with 'example-rating-'
    var child = $(this).find('[class^="example-rating-"]');
    child.text(data.rating);

    $.post('count.php', {
         rate: data.rating,
         wallpostid: jQuery(this).data("postid")
    }, function (data) {
        alert(data);
    });
});

Answer 3

我不想动态构建元素选择器。

var $examples = $('[class^=example-]:not([class^=example-rating])');
var $ratings = $('[class^=example-rating]');

$examples.ratings(3).bind('ratingchanged', function (event, data) {
  var index = $examples.index(this);
  var $rating = $ratings.eq(index);
  $rating.text(data.rating);
  var a = $rating.attr('class').split('-')[2];

  $.post('count.php', {
    rate: data.rating,
    wallpostid: a
  }, function (data) {
    alert(data);
  });
});

用于在jquery中循环

3 个答案: