如何从json中删除null对象
问题描述:
我是Bootstrap JS中的新成员。我有一张使用Bootsrap JS制作的表格,其数据来自Json文件。这是代码-如何从json中删除null对象
<div class="container-fluid">
<div class="row">
<div class="col-md-8">
<div class="fixedscroll">
<table id="user_table" class="table table-hover table-bordered table-striped responsive" style="margin-bottom: 0;" class="display">
<thead>
<tr>
<th>UID</th>
<th>Name</th>
<th>Address</th>
<th>Tags</th>
<th>Edit tags</th>
</tr>
</thead>
</table>
</div>
</div>
</div>
function showAll(){
$.ajax({
url: "showAll",
dataType:"json",
success:function(data){
$('#user_table tr:gt(0)').remove();
jQuery.each(data['Payload'], function(index, value) {
var row = '<tr>'
+ '<td id="tduid">'+ value['uid']+ '</td>'
+ '<td>'+ value['name']+ '</td>'
+ '<td>'+ value['address']+ '</td>'
+ '<td>'+ value['tag1']+ ',' + value['tag2']+ ',' + value['tag3']+'</td>' + '<td>'+ '<button class="deleteUser btn btn-danger" type="submit">Edit</button>' + '</td></tr>';
$('#user_table').append(row);
});
}
});
现在这个Payload是我的json的名称,它来自我称之为数据库的servlet。
现在让我们看看有3个标签。但有些行有2个标签。所以,当我把值JSON将JSON看起来喜欢 -
{"Payload":[{"uid":"u01","name":"Subho","address":"Dumdum","tag1":"aircel","tag2":"vodafone","tag3":"airtel"},{"uid":"u02","name":"Jeet","address":"Baruipur","tag1":"airtel","tag2":"","tag3":"aircel"},{"uid":"u03","name":"Diba","address":"Jadavpur","tag1":"vodafone","tag2":"aircel","tag3":"airtel"},{"uid":"u04","name":"Tommy","address":"Baguihati","tag1":"aircel","tag2":"vodafone","tag3":""},{"uid":"u05","name":"Jonty","address":"Rahara","tag1":"","tag2":"vodafone","tag3":"airtel"},{"uid":"u06","name":"Gourav","address":"Tripura","tag1":"aircel","tag2":"vodafone","tag3":"airtel"}]}
现在你可以看到,对于UID = U02有2个标签。输出看起来像附图。我如何删除空白值或空值?请谁能帮我...
我想,你说的有关在标签栏额外,
...一个凌乱的解决方案是
$.ajax({
url: "showAll",
dataType: "json",
success: function (data) {
$('#user_table tr:gt(0)').remove();
jQuery.each(data['Payload'], function (index, value) {
var tags = $.map([value.tag1, value.tag2, value.tag3], function (value) {
return value || undefined;
});
var row = '<tr>' + '<td id="tduid">' + value['uid'] + '</td>' + '<td>' + value['name'] + '</td>' + '<td>' + value['address'] + '</td>' + '<td>' + tags + '</td>' + '<td>' + '<button class="deleteUser btn btn-danger" type="submit">Edit</button>' + '</td></tr>';
$('#user_table').append(row);
});
}
});
正如@Alnitak说
var tags = [value.tag1, value.tag2, value.tag3].filter(function (value) {
return value !== undefined;
});
console.log(tags)
答
我想,你说的有关在标签栏额外,
...一个凌乱的解决方案是
$.ajax({
url: "showAll",
dataType: "json",
success: function (data) {
$('#user_table tr:gt(0)').remove();
jQuery.each(data['Payload'], function (index, value) {
var tags = $.map([value.tag1, value.tag2, value.tag3], function (value) {
return value || undefined;
});
var row = '<tr>' + '<td id="tduid">' + value['uid'] + '</td>' + '<td>' + value['name'] + '</td>' + '<td>' + value['address'] + '</td>' + '<td>' + tags + '</td>' + '<td>' + '<button class="deleteUser btn btn-danger" type="submit">Edit</button>' + '</td></tr>';
$('#user_table').append(row);
});
}
});
正如@Alnitak说
var tags = [value.tag1, value.tag2, value.tag3].filter(function (value) {
return value !== undefined;
});
console.log(tags)
okk ..谢谢..我会这样做的。 – Subho 2014-10-28 07:26:41
回答
我想,你说的有关在标签栏额外
,
...一个凌乱的解决方案是正如@Alnitak说
优秀!!!!非常感谢你@Arun P Jhony .. – Subho 2014-10-28 07:25:55
我认为你需要'.join(',')'标签在一起。此外,“更多ES5”解决方案将使用'.filter'而不是'$ .map'。与来自其他语言的'map'函数相比,jQuery的'$ .map'的行为有点,呃,奇怪。 – Alnitak 2014-10-28 08:17:20
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