我遇到一个问题,我要使用jQuery附加到表中,还要附加到<div class="jobEntity" id="bestTable">
中。出于某种原因,我不想这样附加表格。有没有一种方法可以解决我的jQuery代码以正确追加表的问题?感谢您的帮助。
var cache = {};
$("#searchTextField").autocomplete({
minLength: 2,
focus: function(event, ui) {
event.preventDefault();
},
source: function(request, response) {
var term = request.term;
if (request.term in cache) {
response(cache[request.term]);
return;
}
$.ajax({
url: "#{request.contextPath}/JobSearchItem.xhtml",
type: "GET",
data: {
term: request.term
},
dataType: "json",
success: function(data) {
response(data);
}
});
},
select: function select(event, ui) {
event.preventDefault();
var url = '#{request.contextPath}/index.xhtml';
var searchValue = ui.item.value;
var data = new FormData();
data.append('searchValue', searchValue);
$.ajax({
url: url,
data: data,
method: "POST",
processData: false,
contentType: false,
cache: false
}).done(function(text) {
$('#results').append($(text).find('#bestTable'));
$('#results').append($(text).find('#textTable'));
$('#results').append($(text).find('table'));
$("#clearone").show();
});
},
response: function response(event, ui, data) {
if (!ui.content.length) {
var message = {
value: "",
label: "NOTHING HAS BEEN FOUND"
};
ui.content.push(message);
}
}
});
答案 0 :(得分:1)
好吧,你可以做这样的事情:
$.ajax({
url: url,
data: data,
method: "POST",
processData: false,
contentType: false,
cache: false
}).done(function(text) {
let append = $(text).find('#bestTable').text() + $(text).find('#textTable').text() + $(text).find('table').text();
$('#results').append(append);
$("#clearone").show();
});
首先,请准备附加内容,然后立即附加。