Jquery $ .get()传递变量问题

时间:2021-04-23 10:03:21

I can't pass different values of variable through $.get() function please check this code to learn more about my problem

我无法通过$ .get()函数传递不同的变量值,请检查此代码以了解有关我的问题的更多信息

var addressFieldValues = ['address1', 'address2', 'address3'];

for(i=0; i<addressFieldValues.length; i++) {
 var address = addressFieldValues[i];
 $.get('function.php', address, function(data){
  alert(address); // alerts address1 all time
 });
}

Why is it alerting "address1" these 3 times? since it should alert 3 different addresses at all.

为什么这3次警告“地址1”?因为它应该警告3个不同的地址。

2 个解决方案

#1


2  

One option would be to wrap your get request in a function, and pass in address as an argument. This way, you avoid the asynchronous issues.

一种选择是将get请求包装在一个函数中,并将地址作为参数传递。这样,您就可以避免异步问题。

function get(address) {
    $.get('', address, function(data) {
        alert(address);
    });
}

var addressFieldValues = ['address1', 'address2', 'address3'];
for (i = 0; i < addressFieldValues.length; i++) {
    var address = addressFieldValues[i];
    get(address);
}

#2


0  

Set request in object format, not string:

以对象格式设置请求,而不是字符串:

$.get('function.php', {'address': address}, function(data){
  alert(data);
});

Second problem: alert server data instead.

第二个问题:改为警告服务器数据。

#1


2  

One option would be to wrap your get request in a function, and pass in address as an argument. This way, you avoid the asynchronous issues.

一种选择是将get请求包装在一个函数中,并将地址作为参数传递。这样,您就可以避免异步问题。

function get(address) {
    $.get('', address, function(data) {
        alert(address);
    });
}

var addressFieldValues = ['address1', 'address2', 'address3'];
for (i = 0; i < addressFieldValues.length; i++) {
    var address = addressFieldValues[i];
    get(address);
}

#2


0  

Set request in object format, not string:

以对象格式设置请求,而不是字符串:

$.get('function.php', {'address': address}, function(data){
  alert(data);
});

Second problem: alert server data instead.

第二个问题:改为警告服务器数据。