javascript JQuery、AJAX：如何使用 json 返回值填充数组？
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JQuery, AJAX: How can I populate an array with a json return?
提问by daveomcd
I'm trying to following code but only get "undefined" in the alert box. Anyone know how I can actually populate the array outside of the .ajax?
我正在尝试遵循代码，但在警报框中只显示“未定义”。任何人都知道我如何实际填充 .ajax 之外的数组？
$(document).ready(function() {
var reviewArray = new Array();
getReviews();
alert(reviewArray[0]);
});
function getReviews()
{
$.ajax({
type : 'GET',
url : 'reviewbox.php',
dataType : 'json',
success : function ( data ) {
$.each( data.reviews, function( i, itemData ) {
reviewArray[i] = itemData.review;
});
},
error : function ( XMLHttpRequest, textStatus, errorThrown) {
var err = "An error has occured: " + errorThrown;
$("body").append(err);
}
});
}
回答by Felix Kling
You have two problems:
你有两个问题：
- Ajax calls are asynchronous. When
alertis executed, the array is not filled yet (the Ajax call did not return yet). - The array
reviewArrayis not in any (parent) scope ofgetReviews(i.e. not accessible from that function).
- Ajax 调用是异步的。当
alert被执行时，阵列中不填充尚未（Ajax调用还不回）。 - 该数组
reviewArray不在任何（父）范围内getReviews（即不能从该函数访问）。
Put the alertin the callback:
将 放入alert回调中：
$(document).ready(function() {
getReviews(function(reviewArray) {
alert(reviewArray[0]);
});
});
function getReviews(callback) {
$.ajax({
/*...*/
success : function (data) {
var reviewArray = [];
$.each( data.reviews, function( i, itemData ) {
reviewArray[i] = itemData.review;
});
callback(reviewArray);
},
/*...*/
});
}
If you want to do it with declaring reviewArraybeforehand, you also have to define getReviewsin the readycallback:
如果要reviewArray事先声明，还必须getReviews在ready回调中定义：
$(document).ready(function() {
var reviewArray = [];
getReviews(function() {
alert(reviewArray[0]);
});
function getReviews(callback) {
$.ajax({
/*...*/
success : function (data) {
$.each( data.reviews, function( i, itemData ) {
reviewArray[i] = itemData.review;
});
callback();
},
/*...*/
});
}
});
But this way, the actual flow of your application might be more confusing.
但是这样一来，您的应用程序的实际流程可能会更加混乱。
回答by NickAldwin
Try this:
试试这个：
$(document).ready(function() {
getReviews();
});
function getReviews()
{
var reviewArray = new Array();
$.ajax({
type : 'GET',
url : 'reviewbox.php',
dataType : 'json',
success : function ( data ) {
$.each( data.reviews, function( i, itemData ) {
reviewArray[i] = itemData.review;
});
alert(reviewArray[0]);
},
error : function ( XMLHttpRequest, textStatus, errorThrown) {
var err = "An error has occured: " + errorThrown;
$("body").append(err);
}
});
}
