我似乎不知道如何使用ajax来发布。 我做了一个愚蠢的形式来尝试它,即使把它一直削减到只有两个值,仍然无法获得任何东西。 我的 html 是这样的:
<html>
<head>
<script type="text/javascript" src="j.js"></script>
<title>Test this<
<body>/title>
</head>
<form name="testForm" onsubmit="postStuff()" method="post">
First Name: <input type="text" name="fname" id="fname" /><br />
Last Name: <input type="text" name="lname" id="lname" /><br />
<input type="submit" value="Submit Form" />
</form>
<div id="status"></div>
</body>
</html>
然后,到目前为止,我的外部javascript只是一个函数:
function postStuff(){
// Create our XMLHttpRequest object
var hr = new XMLHttpRequest();
// Create some variables we need to send to our PHP file
var url = "processForm.php";
var fn = document.getElementById("fname").value;
var ln = document.getElementById("lname").value;
var vars = "firstname="+fn+"&lastname="+ln;
hr.open("POST", url, true);
hr.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
// Access the onreadystatechange event for the XMLHttpRequest object
hr.onreadystatechange = function() {
if(hr.readyState == 4 && hr.status == 200) {
var return_data = hr.responseText;
document.getElementById("status").innerHTML = return_data;
}
}
// Send the data to PHP now... and wait for response to update the status div
hr.send(vars); // Actually execute the request
document.getElementById("status").innerHTML = "processing...";
}
虽然我的 php 只是回响了这些东西:
<?php
$firstname = $_POST['firstname'];
$lastname = $_POST['lastname'];
echo $firstname ." - ". $lastname ."<br />";
?>
我在Firebug或chrome的工具中找不到任何错误。谁能说我做错了什么?
整个问题是由您同时提交表单和执行 AJAX 调用的事实引起的! status
肯定是更新的,但在同一时刻页面刷新(请注意,<input>
值消失了)
只需通过更改标记来避免表单提交,
<form name="testForm" action="" method="">
First Name: <input type="text" name="fname" id="fname" /><br />
Last Name: <input type="text" name="lname" id="lname" /><br />
<input type="button" value="Submit Form" onclick="postStuff();" />
你的代码有效。或者根本不使用表单。无论如何,当您使用 AJAXing 时,它是没有用的。
更新
我在回答之前重现了整个场景:
xhr.html
<html>
<head>
<title>Test this</title>
</head>
<body>
<form name="testForm" action="" method="">
First Name: <input type="text" name="fname" id="fname" /><br />
Last Name: <input type="text" name="lname" id="lname" /><br />
<input type="button" value="Submit Form" onclick="postStuff();" />
</form>
<div id="status"></div>
<script>
function postStuff(){
// Create our XMLHttpRequest object
var hr = new XMLHttpRequest();
// Create some variables we need to send to our PHP file
var url = "xhr.php";
var fn = document.getElementById("fname").value;
var ln = document.getElementById("lname").value;
var vars = "firstname="+fn+"&lastname="+ln;
hr.open("POST", url, true);
hr.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
// Access the onreadystatechange event for the XMLHttpRequest object
hr.onreadystatechange = function() {
console.log(hr);
if(hr.readyState == 4 && hr.status == 200) {
var return_data = hr.responseText;
document.getElementById("status").innerHTML = return_data;
}
}
// Send the data to PHP now... and wait for response to update the status div
hr.send(vars); // Actually execute the request
document.getElementById("status").innerHTML = "processing...";
}
</script>
</body>
</html>
xhr.php
<?php
$firstname = $_POST['firstname'];
$lastname = $_POST['lastname'];
echo $firstname ." - ". $lastname ."<br />";
?>
制作:
<form name="testForm" onsubmit="postStuff()" method="post">
First Name: <input type="text" name="fname" id="fname" /> <br />
Last Name: <input type="text" name="lname" id="lname" /> <br />
<input type="submit" value="Submit Form" />
</form>
到按钮标签中:
<form name="testForm">
First Name: <input type="text" name="fname" id="fname" /> <br />
Last Name: <input type="text" name="lname" id="lname" /> <br />
<button type="button" onclick="postStuff();">Submit Form!</button>
</form>
据我所知,页面从表单提交刷新。如果您使用的是 ajax,则不需要使用表单。
另请阅读:为什么在HTML中使用onClick()是一种不好的做法?因为无论如何你都会将帖子包含在一个函数中。
编辑:我刚刚注意到你的标题和头标签在你放置的来源中被破坏了。
这是我的做法:
在您的 html 文件中放入<SCRIPT type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.8.3/jquery.js"></SCRIPT>
然后你可以调用这个函数,它将调用(在我的例子中)queryDB.php
脚本。
function queryDB(db,query,doAfter){
$.ajax({
type: 'POST',
data: { host: "localhost",
port: "5432",
db: db,
usr: "guest",
pass: "guest",
statemnt: query
},
url: 'scripts/php/queryDB.php',
dataType: 'json',
async: false,
success: function(result){
// call the function that handles the response/results
doAfterQuery_maps(result,doAfter);
},
error: function(){
window.alert("Wrong query 'queryDB.php': " + query);
}
});
};
将帖子发送到测试.php 在同一层次结构中并接受 html 变量中的结果
$.ajax(
{
type: "POST",
url: "test.php",
data: {'test': test, 'name': 0, 'asdf': 'asdf'},
success: function(html)
{
alert(html);
}
});
在收件人的 PHP 中,按如下方式指定
<?php
echo "come here";
echo $_POST['test'];
?>
目录结构
$ tree
.
├── a.php
└── test.php
参考https://off.tokyo/blog/ajax%E3%81%A7post%E3%82%92%E5%8F%97%E3%81%91%E5%8F%96%E3%82%8B%E6%96%B9%E6%B3%95/
也许最好使用像jquery这样的库,然后你可以做类似的事情: $('form').submit(function(){$.post('detinatnion', $('form').serialize());});
但是要回答你的问题,因为你有理由使用纯JS,那么:
<form method="post" action="pathToFileForJsFallback.">
First name: <input type="text" id="fname" name="fname" /> <br />
last name: <input type="text" id="lname" name="lname" /> <br />
<input type="submit" value="Submit Form" />
<div id="status"></div>
</form>
.JS:
function postStuff(){
var activexmodes=["Msxml2.XMLHTTP", "Microsoft.XMLHTTP"] //activeX versions to check for in IE
if (window.ActiveXObject){ //Test for support for ActiveXObject in IE first (as XMLHttpRequest in IE7 is broken)
for (var i=0; i<activexmodes.length; i++){
try{
mypostrequest = new ActiveXObject(activexmodes[i]);
}
catch(e){
//suppress error
}
}
}
else if (window.XMLHttpRequest) // if Mozilla, Safari etc
mypostrequest = new XMLHttpRequest();
else
return false;
mypostrequest.onreadystatechange=function(){
if (mypostrequest.readyState==4){
if (mypostrequest.status==200 || window.location.href.indexOf("http")==-1){
document.getElementById("result").innerHTML=mypostrequest.responseText;
}
else{
alert("An error has occured making the request");
}
}
}
var fname=encodeURIComponent(document.getElementById("fname").value);
var lname=encodeURIComponent(document.getElementById("lname").value);
var parameters="fname="+fname+"&lname="+lname;
mypostrequest.open("POST", "destination.php", true);
mypostrequest.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
mypostrequest.send(parameters);
}
我再次建议你使用像jquery这样的库来学习js,因为当你学会如何做这些东西时,这些库,硬件和所有东西都会如此之快,以至于像这样的JavaScript代码对于日常使用的实际使用将变得毫无用处。
你需要在函数末尾返回 false。
function postStuff(){
// Create our XMLHttpRequest object
var hr = new XMLHttpRequest();
// Create some variables we need to send to our PHP file
var url = "processForm.php";
var fn = document.getElementById("fname").value;
var ln = document.getElementById("lname").value;
var vars = "firstname="+fn+"&lastname="+ln;
hr.open("POST", url, true);
hr.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
// Access the onreadystatechange event for the XMLHttpRequest object
hr.onreadystatechange = function() {
if(hr.readyState == 4 && hr.status == 200) {
var return_data = hr.responseText;
document.getElementById("status").innerHTML = return_data;
}
}
// Send the data to PHP now... and wait for response to update the status div
hr.send(vars); // Actually execute the request
document.getElementById("status").innerHTML = "processing...";
return false;
}