是否可以在没有HTML表单的情况下发布ajax帖子？如果是，我应该怎么做？用什么PHP变量来获取变量？PHP在提取的文件中。我没有使用任何框架。
function ajax(instruction, push, url, callback){
var xmlhttp; // the object for the httprequest
if (window.XMLHttpRequest) {
// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp = new XMLHttpRequest();
} else {
// code for IE6, IE5
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange = function() { // every time the readystate changes
ajaxLoad(xmlhttp.readyState); // Calls function with the ready state each time it uppdates
if (xmlhttp.readyState == 4) { // status 200 = sucessfull page! NOT 404! // 1 2 3 4 are states of the request (4 is when it's done)
// When load bar is complete
if(xmlhttp.status == 200){
callback(xmlhttp.responseText); // goes to the callback function (from the argument "callback") and then passes the xmlhttp
}
else if(xmlhttp.status == 404){ // Could not find file
ajaxError() // Function that will call the ajax but with the error file
}else{}
ajaxDone(); // activates all the nessesary js to check what to do with some parts of the site
}
else{}
};
xmlhttp.open(instruction,url, true); // sends a the var q to the next php file
if(instruction === "GET"){
xmlhttp.send(''); // Sends the request
}
else if(instruction === "POST"){
xmlhttp.send(url); // Sends the request
}
else{
console.log("This ajax does not support " + instruction + " requests.");
}
if(push == true){ // Change the link to the url of the ajax with
var urlPath = window.location.protocol + "//" + window.location.host + window.location.pathname; // where the host is on
if(url == "home.php"){ // If it's the starting page REMOVE THE ?p= !!
var urlPath = window.location.protocol + "//" + window.location.host + window.location.pathname;
window.history.pushState({path:urlPath},'','./'); // an empty url push (!REMOVE THE DOT WHEN THE SITE IS HOSTED PROPERLY)
return; // exit's the function
}else{}
var newLink = "?p=" + url; // Gives us the link we want except that we don't want the .php
newLink = newLink.substring(0, newLink.indexOf('.')); // makes a new string with character 0 to the dot! Will not include the ending of the file
window.history.pushState({path:urlPath},'',newLink); // the push
}
else{}
}<a href="?p=page1" onclick="ajax('POST', true, 'page1.php', function(content){ document.getElementById('content_holder').innerHTML = content;}); return false;">To page 1</a>您可以在这里找到一些关于如何进行Vanilla JS Ajax调用的答案：http://www.sitepoint.com/guide-vanilla-ajax-without-jquery
即将在没有表格的情况下发送，您已经在此处收到了回复：使用XMLHttpRequest 发送POST数据
您可以使用全局变量$_get["your_param_name"]和$_POST["your_param_name"]来获取params服务器端（php），它们是数组，所以我认为您知道如何使用它们。
当然，您可以在纯js中进行AJAX请求，甚至jquery也可以在后面用纯js处理AJAX请求。
JavaScript:
var ajax = {};
ajax.x = function () {
var xhr;
if (window.XMLHttpRequest) {
xhr = new XMLHttpRequest();
} else {
xhr = new ActiveXObject("Microsoft.XMLHTTP");
}
return xhr;
};
ajax.send = function (url, callback, method, data, async) {
if (async === undefined) {
async = true;
}
var x = ajax.x();
x.open(method, url, async);
x.onreadystatechange = function () {
if (x.readyState == 4) {
callback(x.responseText)
}
};
if (method == 'POST') {
x.setRequestHeader('Content-type', 'application/x-www-form-urlencoded');
}
x.send(data)
};
ajax.get = function (url, data, callback, async) {
var query = [];
for (var key in data) {
query.push(encodeURIComponent(key) + '=' + encodeURIComponent(data[key]));
}
ajax.send(url + (query.length ? '?' + query.join('&') : ''), callback, 'GET', null, async)
};
ajax.post = function (url, data, callback, async) {
var query = [];
for (var key in data) {
query.push(encodeURIComponent(key) + '=' + encodeURIComponent(data[key]));
}
ajax.send(url, callback, 'POST', query.join('&'), async)
};
调用Ajax方法：我建议您不要在onclick中使用它
ajax.get('ajax.php',{DATA_TO_PASS},function(response) {
//Do something with response
console.log(response);
},true);
CCD_ 1接收ajax数据；或：
ajax.post('ajax.php',{DATA_TO_PASS},function(response) {
//Do something with response
console.log(response);
},true);
CCD_ 2接收ajax数据；
使用ajax post不需要表单。
$.post( "test.php", { 'choices[]': [ "Jon", "Susan" ] } );
与使用表单的方式相同，您可以使用$_POST 从php获取值