[英]javascript array filter from json
我不好,我刪除了舊的東西
我把Ajax調用簡化了,我為不活動的服務器添加了一個新的ul
,並在其中添加了新的列表項。
https://jsfiddle.net/0m2xs3ah/5/
$.each(response.results,function(index, value){
if(value.status){
$('#active-servers').append('<li>'+value.server+'</li>');
}else{
$('#inactive-servers').append('<li>'+value.server+'</li>');
}
});
var response = { "results": [ { "server": "server04.republic-m.com", "status": true }, { "server": "server06.republic-m.com", "status": true }, { "server": "server06.republic-m.com", "status": true }, { "server": "server17.republic-m.com", "status": false }, { "server": "server18.republic-m.com", "status": true }, { "server": "server20.republic-m.com", "status": true }, { "server": "server21.republic-m.com", "status": false }, { "server": "server25.republic-m.com", "status": true } ] } $.each(response.results,function(index, value){ if(value.status){ $('#active-servers').append('<li>'+value.server+'</li>'); }else{ $('#inactive-servers').append('<li>'+value.server+'</li>'); } }); var servers = response.results; var onlineServers = servers.filter(function(server){ if(server.status){ return (server.status == true); } }); var offlineServers = servers.filter(function(server){ if(!server.status){ return (server.status == false); } }); console.log(offlineServers) console.log(onlineServers)
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <div> <span>Active</span> <ul id= "active-servers"> </ul> </div> <div> <span>Inactive</span> <ul id= "inactive-servers"> </ul> </div>
我認為您只需要一個if語句來選擇將div放入結果中:
$(document).ready( function() {
$.ajax({
type: 'POST',
url: 'servers',
data: {results:'results', status:'status'},
dataType: 'json',
cache: false,
success: function(response) {
$('#active-servers, #inactive-servers').html('');
$.each(response.results,function(index,result){
if (result.status)
$('#active-servers').append('<li>'+result.server+'</li>');
else
$('#inactive-servers').append('<li>'+result.server+'</li>');
});
}
});
});
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