我是 Bootstrap JS 新手。我有一个用 Bootsrap JS 制作的表,其数据来自 Json 文件。这是代码-
<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);
});
}
});
现在这个有效负载是我的 json 的名称,它来 self 调用数据库的 servlet。
现在我们有 3 个标签。但有些行有 2 个标签。因此,当我将值放入 json 时,Json 看起来像 -
{"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"}]}
现在您可以看到,对于 UID=U02,有 2 个标签。输出看起来如所附图片。如何删除空白值或空值???请任何人帮助我...
最佳答案
我想,你是在说标签栏中额外的 ,
...一个困惑的解决方案是
$.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所说
var tags = [value.tag1, value.tag2, value.tag3].filter(function (value) {
return value !== undefined;
});
console.log(tags)
关于javascript - 如何从 json 中删除 null 对象,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/26602758/