Fre*_*Bud 7 javascript google-login google-plus
我正在尝试按照Google+指示使用我自己的按钮启动Google+登录流程.
关于回调函数,gapi.auth.signIn引用说(引用):
"全局命名空间中的一个函数,在呈现登录按钮时调用,并在登录流程完成后调用."
出现Google登录对话框,要求我登录,但在与对话框进行任何交互之前,会立即调用两次回调.两次我得到一个类似的authResult,错误="immediate_failed",error_subtype ="access_denied",status.signed_in = false
这是为什么？
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<html>
<head>
<meta http-equiv=Content-Type content="text/html; charset=utf-8" />
<script src="https://apis.google.com/js/client:platform.js?onload=googleRender" async defer></script>
</head>
<body>
<script>
function googleRender() { // executed when Google APIs finish loading
var googleSigninParams = {
'clientid' : '746836915266-a016a0hu45sfiouq7mqu5ps2fqsc20l4.apps.googleusercontent.com',
'cookiepolicy' : 'http://civoke.com',
'callback' : googleSigninCallback ,
'requestvisibleactions' : 'http://schema.org/AddAction',
'scope' : 'https://www.googleapis.com/auth/plus.login'
};
var googleSigninButton = document.getElementById('googleSigninButton');
googleSigninButton.addEventListener('click', function() {
gapi.auth.signIn(googleSigninParams);
});
}
function googleSigninCallback(authResult) {
console.log('googleSigninCallback called: ');
console.dir(authResult);
if (authResult['status']['signed_in']) {
document.getElementById('googleSigninButton').setAttribute('style', 'display: none'); // hide button
console.log('User is signed-in to Google');
} else {
console.log('User is NOT signed-in. Sign-in state: ' + authResult['error']);
}
}
</script>
<button id="googleSigninButton">Sign in with Google</button>
</body>
</html>
当状态发生变化时,不仅在用户登录时调用回调函数.在googleSigninCallback(authResult),您应该首先检查用户是否已登录,然后您应该检查方法值是否为AUTO或PROMPT.PROMPT应该只在用户登录时返回一次,这就是您所需要的.这是代码:
function googleSigninCallback(authResult) {
if (authResult['status']['signed_in'] && authResult['status']['method'] == 'PROMPT') {
// User clicked on the sign in button. Do your staff here.
} else if (authResult['status']['signed_in']) {
// This is called when the status has changed and method is not 'PROMPT'.
} else {
// Update the app to reflect a signed out user
// Possible error values:
// "user_signed_out" - User is signed-out
// "access_denied" - User denied access to your app
// "immediate_failed" - Could not automatically log in the user
console.log('Sign-in state: ' + authResult['error']);
}
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