从JavaScript函数调用PHP类方法
问题说明
可能重复：
从javascript调用php函数
我知道php是服务器端和JavaScript是客户端。但我想知道在调用JavaScript函数时如何运行PHP方法。下面是我的代码，我知道错误是但我怎么能执行php方法？
I understand that php is server side and JavaScript is client side. But I would like to know how to run a PHP method when a JavaScript function is called. Below is my code, I know the error is but how can I perform the php method?
<script type="text/javascript">
function toggle()
{
var ele = document.getElementById("addCatTextBox");
var text = document.getElementById("addCatButtonText");
if(ele.style.display == "block") {
ele.style.display = "block";
text.innerHTML = "Save category";
<?php Category::addCategory($inCatName)?>
}
else {
ele.style.display = "none";
text.innerHTML = "Add new category";
}
}
</script>
感谢您的帮助。
正确答案
使用Prototype库（www.prototypejs.org）：
Using the Prototype library (www.prototypejs.org):
Javascript：
Javascript:
<script type="text/javascript">
function toggle()
{
var ele = document.getElementById("addCatTextBox");
var text = document.getElementById("addCatButtonText");
if(ele.style.display == "block") {
ele.style.display = "block";
text.innerHTML = "Save category";
var options={
method: 'get',
parameters: 'inCatName=' ele.value,
onSuccess: function(xhr) {
// TODO: Whatever needs to happen on success
alert('it worked');
},
onFailure: function(xhr) {
// TODO: Whatever needs to happen on failure
alert('it failed');
}
};
new Ajax.Request('addCategory.php', options);
}
else {
ele.style.display = "none";
text.innerHTML = "Add new category";
}
}
</script>
addCategory.php：
addCategory.php:
<?php
$inCatName=isset($_REQUEST["inCatName"]) ? $_REQUEST["inCatName"] : null;
Category::addCategory($inCatName);
?>
我们的想法是Javascript向addCategory发送GET（或者可能是POST）请求。幕后的php页面，传递它创建类别所需的任何信息。
The idea is that the Javascript sends a GET (or it could be POST) request to the addCategory.php page behind the scenes, passing it whatever info it needs to create the category.
希望这足以让你前进。我的代码中缺少很多东西 - 您需要验证变量addCategory.php接收并执行任何其他相关的安全检查，然后再将其放在数据库附近。 addCategory.php还需要任何包含文件等，以便它知道你的Category类。最后，addCategory.php应该将某种形式的变量返回到调用它的Javascript代码，以便它知道结果是什么。
Hopefully this is enough to get you going. There's a lot missing from my code - you'll need to validate the variables addCategory.php receives and perform any other pertinent security checks before letting it anywhere near the database. addCategory.php will also require any include files, etc so that it knows about your Category class. Finally, addCategory.php should really return some form of variable back to the Javascript code that called it so that it knows what the outcome was.
这篇好文章是转载于：学新通技术网
