固定的！代码在第一个ajax函数中^_^

document.getElementById('color').addEventListener('click', color);
 function color(){
 var xhr = new XMLHttpRequest();
 xhr.open('GET', 'setting/color.php', true);
 xhr.onload = function(){
 if(this.status == 200){
 history.pushState(null, null, '?c=color');
 document.getElementById('content').innerHTML = this.responseText;
var colito = document.getElementsByClassName('cols');

var i;
 for(i=0;i < colito.length;i++){
colito[i].addEventListener('click', colorSid);
 function colorSid(e){
e.preventDefault();
var cols = this.value;
var params = "cols="+cols;

 var xhr = new XMLHttpRequest();
 xhr.open('POST', '../phppath/color.php', true);
 xhr.setRequestHeader('Content-type', 'application/x-www-form-urlencoded');
 xhr.onload = function(){
   console.log(this.responseText);
 }
 xhr.send(params);
 }
 }
 } else if(this.status = 404){
document.getElementById('content').innerHTML = 'Not Found';
}
  }
 xhr.onerror = function(){
 console.log('Request Error...');
  }
 xhr.send();
  }

我看到了您的javascript代码。您将在事件列表中调用的函数放入，它必须在javascript for循环之外。下面我为您在标题中提到的问题的答案编写代码。

$.ajax({
     url:"your url",
     type: "GET",
     beforeSend: function(xhr){
         //xhr.setRequestHeader('Access-Control-Allow-Origin', '*');
     },
     success: function(data) {
        console.log(data);
        var resp = JSON.parse(data);

          //call another ajax function here

      }
  });

你必须保持你的html在一个字符串中 color.php 和回声。

$html = '<div class="f-col s1 m1 l1">
            <label class="container f-padding">
              <input class="cols" type="radio" name="cols" value="dark light-hover">
              <span class="checkmark dark"></span>
            </label>
        </div>

        <div class="f-col s1 m1 l1">
            <label class="container f-padding">
              <input class="cols" type="radio" name="cols" value="light dark-hover" checked="checked">
              <span class="checkmark light"></span>
            </label>
        </div>';

echo $html;