Javascript Can';t仅从通过php页面获取的json数据返回用户名
我能够从外部php页面获取json数据,并在JavaScript控制台上打印它。但这些都是以对象的形式出现的。以下是我在控制台上接收到的数据:Javascript Can';t仅从通过php页面获取的json数据返回用户名,javascript,php,jquery,ajax,json,Javascript,Php,Jquery,Ajax,Json,我能够从外部php页面获取json数据,并在JavaScript控制台上打印它。但这些都是以对象的形式出现的。以下是我在控制台上接收到的数据: [{"id":"1","username":"iammuneeb","password":"4297f44b13955235245b2497399d7a93","name":"Mirza Muneeb"},{"id":"2","username":"iamfaiz","password":"4297f44b13955235245b2497399d7a93
[{"id":"1","username":"iammuneeb","password":"4297f44b13955235245b2497399d7a93","name":"Mirza Muneeb"},{"id":"2","username":"iamfaiz","password":"4297f44b13955235245b2497399d7a93","name":"Faiz"}]
如何提取用户名并将其转换为有序列表。(其他职等)
这就是我到目前为止所做的:
$(document).ready(function (e) {
$('#delete').click(function (e) {
var jsonData = getResultsInJson(username);
jsonData.success(function (data) {
var output = "<ol>";
for (var i in data) {
output += "<li>" + data.username + "</li>";
}
output += "</ol>";
$('#placeholder').html(data);
console.log(data.username);
});
});
});
使用(y中的x)格式时,x是键,y是数组或对象。因为我只是钥匙,所以您需要将其作为一个:
for (var i in data) {
output += "<li>" + data[i].username + "</li>";
}
for(数据中的变量i){
输出+=“”+数据[i]。用户名+“ ”;
}
您需要按字母顺序排列吗
试着这样做:
$(document).ready(function() {
var json = [{"id":"1","username":"iammuneeb","password":"4297f44b13955235245b2497399d7a93","name":"Mirza Muneeb"},{"id":"2","username":"iamfaiz","password":"4297f44b13955235245b2497399d7a93","name":"Faiz"}];
var usernames = new Array();
$.each(json, function(index, object) {
usernames[index] = object.username;
});
var sorted = usernames.sort();
var output = '';
$.each(sorted, function(index, value) {
output += '<li>' + value + '</li>';
});
$('#placeholder').html('<ol>' + output + '</ol>');
});
$(文档).ready(函数(){
var json=[{“id”:“1”,“用户名”:“iammuneeb”,“密码”:“4297f44b13955235245b2497399d7a93”,“名称”:“Mirza Muneeb”},{“id”:“2”,“用户名”:“iamfaiz”,“密码”:“4297f44b13955235245b2497399d7a93”,“名称”:“Faiz”};
var usernames=新数组();
$.each(json、函数(索引、对象){
用户名[索引]=object.username;
});
var sorted=usernames.sort();
var输出=“”;
$.each(已排序,函数(索引,值){
输出+=''+值+' ';
});
$('#占位符').html(''+output+'');
});
您是否尝试过i.username
??然后不要在PHP中返回密码等。这更像是一个PHP问题,而不是JavaScription问题。这里的基本问题是您正在执行:$(“#占位符”).html(数据)
它应该在哪里$('#占位符').html(输出)代码>仍然只有一个用户名,而有两个用户名?
$(document).ready(function() {
var json = [{"id":"1","username":"iammuneeb","password":"4297f44b13955235245b2497399d7a93","name":"Mirza Muneeb"},{"id":"2","username":"iamfaiz","password":"4297f44b13955235245b2497399d7a93","name":"Faiz"}];
var usernames = new Array();
$.each(json, function(index, object) {
usernames[index] = object.username;
});
var sorted = usernames.sort();
var output = '';
$.each(sorted, function(index, value) {
output += '<li>' + value + '</li>';
});
$('#placeholder').html('<ol>' + output + '</ol>');
});