Question

所以我能够成功地从我的JSON对象中提取数据，我遇到的问题是数据是四重的。有人能告诉我为什么会这样吗？我以前从未见过这个。这是JS：

jQuery(function() {
    jQuery.getJSON(NewConvData, function(data){
        var channelHTML = ''
        jQuery.each(data, function(i){
            console.log(data);
            for (var i in data.results) {
            channelHTML += '<tr><td class="tg-yw4l">' + data.results[i].call_mine_status + '</td><td class="tg-yw4l">' + data.results[i].cdr_source + '</td><td class="tg-yw4l">' + data.results[i].tracking_number + '</td><td class="tg-yw4l">' + data.results[i].disposition + '</td><td class="tg-yw4l">' + data.results[i].duration + '</td><td class="tg-yw4l">' + data.results[i].external_id + '</td><td class="tg-yw4l">' + data.results[i].id + '</td><td class="tg-yw4l">' + data.results[i].is_outbound + '</td><td class="tg-yw4l">' + data.results[i].ouid + '</td><td class="tg-yw4l">' + data.results[i].repeat_call + '</td><td class="tg-yw4l">' + data.results[i].caller_id + '</td><td class="tg-yw4l">' + data.results[i].calldate + '</td><td class="tg-yw4l">' + data.results[i].ringto_number + '</td><td class="tg-yw4l"><a href="' + data.results[i].file_url + '">Recorded Call</a></td></tr>';
            }
        });
        jQuery('#conv_table').append(channelHTML);
    });              
});

以下是Fiddle link

Answer 1

您有两个循环each和for都检查

var channelHTML = ''
            for (var i in data.results) {
            channelHTML += '<tr><td class="tg-yw4l">' + data.results[i].call_mine_status + '</td><td class="tg-yw4l">' + data.results[i].cdr_source + '</td><td class="tg-yw4l">' + data.results[i].tracking_number + '</td><td class="tg-yw4l">' + data.results[i].disposition + '</td><td class="tg-yw4l">' + data.results[i].duration + '</td><td class="tg-yw4l">' + data.results[i].external_id + '</td><td class="tg-yw4l">' + data.results[i].id + '</td><td class="tg-yw4l">' + data.results[i].is_outbound + '</td><td class="tg-yw4l">' + data.results[i].ouid + '</td><td class="tg-yw4l">' + data.results[i].repeat_call + '</td><td class="tg-yw4l">' + data.results[i].caller_id + '</td><td class="tg-yw4l">' + data.results[i].calldate + '</td><td class="tg-yw4l">' + data.results[i].ringto_number + '</td><td class="tg-yw4l"><a href="' + data.results[i].file_url + '">Recorded Call</a></td></tr>';
            }
            jQuery('#conv_table').append(channelHTML);
        });

Bugs in code

Answer 2

不要为同一件事使用两个循环，删除

for (var i in data.results)此循环并按jQuery.each(data, function(i)修改jQuery.each(data.results, function(i)。

“守则”将完美运作，不会重复。

这是小提琴链接Fiddle link

删除重复的JSON对象

2 个答案: