jQuery CodeIgniter/Ajax - 将 post 值发送到控制器
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CodeIgniter/Ajax - Send post values to controller
提问by cabaret
I'm working on a simple login form. When the user clicks on the Login button, I want to send the post values to my controller, validate them against my database (using a model) and return a value. Based on that value, I want to either send the user to the member area, or display an error message on my form. This error message has a class 'error' and is hidden by default. I want to use jQuery's show() to show it when the credentials are wrong.
我正在处理一个简单的登录表单。当用户单击登录按钮时，我想将帖子值发送到我的控制器，根据我的数据库（使用模型）验证它们并返回一个值。根据该值，我想将用户发送到会员区，或者在我的表单上显示错误消息。此错误消息有一个“错误”类，默认情况下是隐藏的。当凭据错误时，我想使用 jQuery 的 show() 来显示它。
My form has two textfields, one for the email address, other one for the password and a submit button. However, I have never really worked with Ajax except for VERY basic things like a simple loader, so I'm not sure what to do next.
我的表单有两个文本字段，一个用于电子邮件地址，另一个用于密码和一个提交按钮。但是，我从来没有真正使用过 Ajax，除了非常基本的东西，比如一个简单的加载器，所以我不确定接下来要做什么。
$("#btn_login").click(
function()
{
// get values
var email_address = $("#email_address").val();
var password = $("#password").val();
// post values? and ajax call?
//stop submit btn from submitting
return(false);
}
);
I assume I have to use jQuery's ajax() method here, so I was thinking of working off this example:
我假设我必须在这里使用 jQuery 的 ajax() 方法，所以我想解决这个例子：
$.ajax({
type: "POST",
url: "some.php",
data: "name=John&location=Boston",
success: function(msg){
alert( "Data Saved: " + msg );
}
});
However, I'm not really sure how to get my post values (those two vars) in the data: thingy.. Or should I take another road here? This seems like something I'll never forget how to do once I learn it, so if someone can help me out I'd be grateful. Thanks a lot.
但是，我不太确定如何在数据中获取我的帖子值（这两个变量）：thingy .. 或者我应该在这里走另一条路？这似乎是我一旦学会了就永远不会忘记怎么做的事情，所以如果有人能帮助我，我将不胜感激。非常感谢。
回答by JohnP
All you need to do is create a key/value pair with the values and assign it to the data parameter and jQuery will do the rest.
您需要做的就是使用值创建一个键/值对并将其分配给数据参数，jQuery 将完成剩下的工作。
//create a js object with key values for your post data
var postData = {
'email' : email_address,
'password' : password
};
$.ajax({
type: "POST",
url: "some.php",
data: postData , //assign the var here
success: function(msg){
alert( "Data Saved: " + msg );
}
});
With this, you should be able to access the data with Codeigniters input
有了这个，您应该能够使用 Codeigniters 访问数据 input
EDIT
编辑
I've set up a test fiddle here : http://jsfiddle.net/WVpwy/2/
我在这里设置了一个测试小提琴：http: //jsfiddle.net/WVpwy/2/
$('#dostuff').click(function(){
var email_address = $("#email_address").val();
var password = $("#password").val();
var postData = {
'email' : email_address,
'password' : password,
'html' : 'PASS'
};
$.post('/echo/html/', postData, function(data){
//This should be JSON preferably. but can't get it to work on jsfiddle
//Depending on what your controller returns you can change the action
if (data == 'PASS') {
alert('login pased');
} else {
alert('login failed');
}
});
});
I just prefer .post, but what you used is an equivalent.
我只是更喜欢 .post，但你使用的是等效的。
Basically your controller should echoout data. Not return. You need to send a string representation of your data back so your script can (evaluate if json) and act on it
基本上你的控制器应该回显数据。不回。您需要将数据的字符串表示发送回，以便您的脚本可以（评估是否为 json）并对其进行操作
Here's a good example as a starting point : http://www.ibm.com/developerworks/opensource/library/os-php-jquery-ajax/index.html
这是一个很好的例子作为起点：http: //www.ibm.com/developerworks/opensource/library/os-php-jquery-ajax/index.html
回答by user1484668
just modifiy a little bit the url of your ajax :
只需稍微修改一下您的 ajax 网址：
$.ajax({
type: "POST",
url: "some/myfunction",
data: "name=John&location=Boston",
success: function(msg){
alert( "Data Saved: " + msg );
}
});
Make sure your url point to the function you want inside your controller. For example: "myfunction" inside the controller some (in the file Some.php)
确保您的 url 指向您想要在控制器中使用的功能。例如：控制器中的“myfunction” some（在文件 Some.php 中）
To access to your post variables inside the controller function do not try to put parameters to myfunction but :
要访问控制器函数内的 post 变量，请不要尝试将参数放入 myfunction，而是：
class Some extends CI_Controller
{
......
public function myfunction()
{
$name = $this->input->post('name');
$location = $this->input->post('location');
}
......
}
回答by Harold Smith
You can definitely use your method, but a shorthand method is this:
您绝对可以使用您的方法，但速记方法是这样的：
$.post("some.php",{name:"John",location:"Boston",email:email_address,password:password},function(data,textStatus) { //response from some.php is now contained in data });
