我正在尝试使用 javascript 访问对象数组列表,数组列表是:
membersList:{
"kind":"admin#directory#users",
"users":[{"orgUnitPath":"/",
"isMailboxSetup":true,
"id":"1076823423424234",
"isAdmin":false,
"suspended":false,
"isDelegatedAdmin":false,
"isEnforcedIn2Sv":false,
"etag":"\"npJcgsdfsadfsfsff\"",
"ipWhitelisted":false,"changePasswordAtNextLogin":true,
"customerId":"C01looera",
"includeInGlobalAddressList":true,
"lastLoginTime":"1970-01-01T00:00:00.000Z",
"primaryEmail":"[email protected]",
"isEnrolledIn2Sv":false,"kind":"admin#directory#user",
"name":{"givenName":"ilda",
"familyName":"donofrio",
"fullName":"ilda donofrio"},
"creationTime":"2018-06-10T11:56:45.000Z",
"emails":[{"address":"[email protected]",
"primary":true}],
"agreedToTerms":true
}],
"etag":"\"npJcgsdfsadfsfsff/npJcgsdfsadfsfsff\"
}
我正在尝试primary email
从我为其编写代码的列表中访问
for (var j in membersList) {
var member = membersList[j];
Logger.log('member Email:' + member);
}
这只是返回我
[18-06-12 15:39:55:982 EDT] member Email:admin#directory#users
[18-06-12 15:39:55:982 EDT] member Email:{"orgUnitPath":"/","isMailboxSetup":true,"id":"1076823423424234","isAdmin":false,"suspended":false,"isDelegatedAdmin":false,"isEnforcedIn2Sv":false,"etag":"\"npJcgsdfsadfsfsff\"","ipWhitelisted":false,"changePasswordAtNextLogin":false,"customerId":"C01looera","includeInGlobalAddressList":true,"lastLoginTime":"1970-01-01T00:00:00.000Z","primaryEmail":"[email protected]","isEnrolledIn2Sv":false,"kind":"admin#directory#user","name":{"givenName":"ilda","familyName":"donofrio","fullName":"ilda donofrio"},"creationTime":"2018-06-10T11:56:45.000Z","emails":[{"address":"[email protected]","primary":true}],"agreedToTerms":true}
[18-06-12 15:39:55:983 EDT] member Email:"npJcgsdfsadfsfsff/npJcgsdfsadfsfsff\"
当我尝试:
Logger.log('member Email:' + member.primaryEmail);
它的投掷:
[18-06-12 16:02:53:630 EDT] member Email:undefined
[18-06-12 16:02:53:630 EDT] member Email:undefined
[18-06-12 16:02:53:631 EDT] member Email:undefined
有没有办法只提取primaryEmail
使用脚本的值?我是 javascript 和 google 脚本的新手,所以如果这是一个愚蠢的问题,请原谅我。
在谷歌应用程序脚本中对我有用的唯一答案是:
for (var j in membersList) {
var member = membersList[j];
for (var k in member){
var user = member[k];
var memberEmail = user.primaryEmail;
var lastLoginTime = user.lastLoginTime;
}
}
Logger.log('member Email:' + memberEmail);
Logger.log('last Login Time:' + lastLoginTime);
我尝试了这里发布的所有方法,但都没有成功,但正如@AllenG 引用的那样,我只需要遍历users
成员列表对象中的所有 。非常感谢您的帮助。
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我来说两句