[英]after Ajax get content, I want call another ajax function
我在侧边栏中创建了一个ajax调用,当我在导航中单击“颜色”时,它可以完美运行,并且一切正常。
代码是:
<a id='color'>Color</a>
JavaScript函数:
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;
} else if(this.status = 404){
document.getElementById('content').innerHTML = 'Not Found';
}
}
xhr.onerror = function(){
console.log('Request Error...');
}
xhr.send();
}
之后,我要在“颜色”页面上此代码,它位于color.php中,该代码来自ajax调用。
<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>
JavaScript函数在这里:
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);
}
}
如果我不刷新页面,它将不起作用,任何人都知道是什么问题?
您必须将html保留在color.php
中的字符串中并回显它。
$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;
我看到了您的javascript代码。您放置了在事件列表器中调用的函数,该函数必须在javascript for loop。之外。在下面,我编写代码以回答标题中提到的问题。
$.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
}
});
固定! 代码在第一个ajax函数^ _ ^内部
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();
}
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