我需要在for循环中创建一个小延迟:
for (i = 1; i <= cloneIndex; i++) {
var myElem = document.getElementById('form' + i);
if (myElem != null) {
function postData() {
return {
udd: document.getElementById('udd').value,
data: date_in,
hora_ini: hour_in,
hora_fim: hour_out,
cat: $('#form' + i).find('select[id="cat"]').val(),
m1: $('#form' + i).find('select[id="q1"]').val(),
m2: $('#form' + i).find('select[id="q2"]').val(),
m3: $('#form' + i).find('select[id="q3"]').val(),
m4: $('#form' + i).find('select[id="q4"]').val(),
m5: $('#form' + i).find('select[id="q5"]').val()
}
}
var newItem = postData();
$2sxc(@Dnn.Module.ModuleID).webApi.post('app/auto/content/audits', {}, newItem);
}
}
在stackoverflow示例之后,我尝试了这个解决方案:
for (i = 1; i <= cloneIndex; i++) {
(function(i){
setTimeout(function(){
var myElem = document.getElementById('form' + i);
if (myElem != null) {
function postData() {
return {
udd: document.getElementById('udd').value,
data: date_in,
hora_ini: hour_in,
hora_fim: hour_out,
cat: $('#form' + i).find('select[id="cat"]').val(),
m1: $('#form' + i).find('select[id="q1"]').val(),
m2: $('#form' + i).find('select[id="q2"]').val(),
m3: $('#form' + i).find('select[id="q3"]').val(),
m4: $('#form' + i).find('select[id="q4"]').val(),
m5: $('#form' + i).find('select[id="q5"]').val()
}
}
var newItem = postData();
$2sxc(Dnn.Module.ModuleID).webApi.post('app/auto/content/audits', {}, newItem);
}
}, 1000 * i);
}(i));
}
但是这打破了里面的功能。似乎myElem现在总是为空。太多“我”?我该如何解决这个问题?
答案 0 :(得分:1)
您需要在闭包内定义变量,使其对每次迭代都是唯一的:
for (var i = 1; i < 10; i++) {
(function() {
var k = i; // <-- k will be different for each iteration, because it was declared inside the closure. i was defined outside the closure.
setTimeout(function() {
console.log("Delayed: ", i, k)
}, i * 1000)
}());
}
&#13;
...或者在闭包定义中包含i
:
for (var i = 1; i < 10; i++) {
(function(i) {
setTimeout(function() {
console.log("Delayed: ", i)
}, i * 1000)
}(i));
}
&#13;
答案 1 :(得分:0)
没关系。代码不起作用的原因很简单。下面的其余代码没有等待延迟循环结束,所以它实际上破坏了函数。
这修复它(放在setTimeout函数内):
k++;
if (k == cloneIndex) {rest of the code that needs the loop to end}