Javascript 如何在codeigniter中将数据从视图发送到控制器
我是新的代码点火器。在我看来,我添加此代码是为了在我的网站中启用google登录Javascript 如何在codeigniter中将数据从视图发送到控制器,javascript,php,database,codeigniter,Javascript,Php,Database,Codeigniter,我是新的代码点火器。在我看来,我添加此代码是为了在我的网站中启用google登录 <html lang="en"> <head> <meta name="google-signin-scope" content="profile email"> <meta name="google-signin-client_id" content="720765566138-5c6jreo4r7ma6cm0hdblj5cmjtdiruk4.apps.googleuse
<html lang="en">
<head>
<meta name="google-signin-scope" content="profile email">
<meta name="google-signin-client_id" content="720765566138-5c6jreo4r7ma6cm0hdblj5cmjtdiruk4.apps.googleusercontent.com">
<script src="https://apis.google.com/js/platform.js" async defer> </script>
</head>
<body>
<div class="g-signin2" data-onsuccess="onSignIn" data-onfailure="onFailure" data-theme="dark"></div>
<script>
function onFailure(msg){ console.log(msg); }
function onSignIn(googleUser) {
console.log("onSignIn");
var profile = googleUser.getBasicProfile();
var user_name = profile.getName();
var id = googleUser.getAuthResponse().id;
};
</script>
</body>
</html>
函数onFailure(msg){console.log(msg);}
函数onSignIn(谷歌用户){
控制台日志(“onSignIn”);
var profile=googleUser.getBasicProfile();
var user_name=profile.getName();
var id=googleUser.getAuthResponse().id;
};
如何将变量user_name和id发送到控制器,以便调用模型将值插入数据库。您必须执行控制器的操作,并且在此过程中,您必须按如下方式传递变量:
function test{
$user_name = "XYZ";
$this->load->view('view_name', $user_name);
}
如果要传递多个变量,请使用以下命令:
function test{
$user_name = "XYZ";
$data = 123;
$this->load->view('view_name', array("user_name"=>$user_name,"data"=>$data));
}
May this will help you :)
如果您的方法是
post
$this->input->post();
默认情况下,此方法包含所有其他数据。get方法
$this->input->get();
让您的表单
如下所示
<form action="baseurl/controller_name/method">
add hidden field of user_id then sumbmit
</form>
您只需在获得信息后使用ajax发布信息(使用下面的jquery):
函数onSignIn(谷歌用户)
{
var profile=googleUser.getBasicProfile();
var user_name=profile.getName();
var id=googleUser.getAuthResponse().id;
$.ajax({
'url':'the/path/to/controller',
'type':'POST',
'data':{'user\u name':user\u name,'id':id},
'success':function(){},//由您决定
“错误”:函数()//由您决定
})
};
我想将JavaScript变量user\u name从视图发送到控制器。然后将JavaScript值传递到隐藏字段,并以动作url的形式传递该变量。。。。然后您可以通过$this->input->get()获得该变量;控制器中。可能存在的副本
function YOUR_METHOD()
{
print_r($this->input->get())//By default get method
}
<script>
function onSignIn(googleUser)
{
var profile = googleUser.getBasicProfile();
var user_name = profile.getName();
var id = googleUser.getAuthResponse().id;
$.ajax({
'url': 'the/path/to/controller',
'type': 'POST',
'data':{'user_name': user_name, 'id':id},
'success': function(){}, //up to you
'error': function() // up to you
})
};
</script>