如何调用谷歌的getBasicProfile（）到谷歌登录只有按钮点击？ [英] how to call getBasicProfile() of google to google signin on only button click?
本文介绍了如何调用谷歌的getBasicProfile（）到谷歌登录只有按钮点击？的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
我已在Google网站上登录Google：
< script src =https://apis.google.com/js/platform.jsasync defer>< / script>
< script>
函数onSignIn（googleUser）{
var profile = googleUser.getBasicProfile（）;
//console.log（'ID：'+ profile.getId（））; //不要发送到你的后端！改用ID标记。
//console.log('Image URL：'+ profile.getImageUrl（））;
//console.log('Name：'+ profile.getName（））;
//console.log('Email：'+ profile.getEmail（））;
var user_uname = profile.getName（）;
var user_email = profile.getEmail（）;
alert（user_uname）;
}
< / script>
以下是登录Google的按钮：
< div class =g-signin2data-onsuccess =onSignIn>< / div>
我想给用户google登录，但问题出在页面加载时onSignIn（）函数被调用自动。
我只需要点击按钮。任何人都可以帮助我？
解决方案
最佳解决方案是仅在用户未登录时呈现登录按钮。
< html>
< head>
< meta name =google-signin-client_idcontent =YOUR_CLIENT_ID>
< / head>
< body>
< script>
函数onSignIn（googleUser）{
var profile = googleUser.getBasicProfile（）;
var user_name = profile.getName（）;
alert（user_name）;
函数onLoad（）{
gapi.load（'auth2，signin2'，function（）{
var auth2 = gapi.auth2.init（） ;
auth2.then（function（）{
//当前值
var isSignedIn = auth2.isSignedIn.get（）;
var currentUser = auth2.currentUser.get（） ;
if（！isSignedIn）{
//渲染g-signin2按钮
gapi.signin2.render（'google-signin-button'，{
'onsuccess'：'onSignIn'
}）;
}
}）;
}）;
}
< / script>
< div id =google-signin-button>< / div>
< script src =https://apis.google.com/js/platform.js?onload=onLoadasync defer>< / script>
< / body>
< / html>
i have take google signin in my website :
<script src="https://apis.google.com/js/platform.js" async defer></script>
<script>
function onSignIn(googleUser) {
var profile = googleUser.getBasicProfile();
//console.log('ID: ' + profile.getId()); // Do not send to your backend! Use an ID token instead.
//console.log('Image URL: ' + profile.getImageUrl());
//console.log('Name: ' + profile.getName());
//console.log('Email: ' + profile.getEmail());
var user_uname = profile.getName();
var user_email = profile.getEmail();
alert(user_uname);
}
</script>
and here is a button to login google:
<div class="g-signin2" data-onsuccess="onSignIn"></div>
i want to give user google signin but the problem is whenever page is load onSignIn() function is called automatically. i want it only on button click. can anybody help me?
解决方案
Best solution is to render sign-in button only when user is not signed in.
<html>
<head>
<meta name="google-signin-client_id" content="YOUR_CLIENT_ID">
</head>
<body>
<script>
function onSignIn(googleUser) {
var profile = googleUser.getBasicProfile();
var user_name = profile.getName();
alert(user_name);
}
function onLoad() {
gapi.load('auth2,signin2', function() {
var auth2 = gapi.auth2.init();
auth2.then(function() {
// Current values
var isSignedIn = auth2.isSignedIn.get();
var currentUser = auth2.currentUser.get();
if (!isSignedIn) {
// Rendering g-signin2 button.
gapi.signin2.render('google-signin-button', {
'onsuccess': 'onSignIn'
});
}
});
});
}
</script>
<div id="google-signin-button"></div>
<script src="https://apis.google.com/js/platform.js?onload=onLoad" async defer></script>
</body>
</html>
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