Question

在快速使用Firebase时，我遇到了一个非常奇怪的问题。似乎脚本可以继续执行而无需等待完成块完成。这就是我的意思：

我有一个功能：

func create_user(email:String, password:String) -> Bool{

    var create_suc = false

    Auth.auth().createUser(withEmail: email, password: password) { (user, error) in
        if let error = error{
            // Error creating new user.
            create_suc = false
            print("In fail closure: \(create_suc)")
        }else{
            // Create successfully.
            create_suc = true
            print("In success closure: \(create_suc)")
        }
    }
    print("Before return: \(create_suc)")
    return create_suc
}

以及调用此函数的上下文：

 @IBAction func sign_up(_ sender: UIButton) {
        let result = create_user(email: acc_email.text!, password: acc_password.text!)

        print("Create successfully: \(result)")

我得到的结果很奇怪：

**Before return: false**
**Create successfully: false**
2018-08-26 16:41:55.313457-0700 chat_app[251:9301] TIC Read Status [2:0x0]: 1:57
2018-08-26 16:41:55.313611-0700 chat_app[251:9301] TIC Read Status [2:0x0]: 1:57
2018-08-26 16:41:56.013564-0700 chat_app[251:9300] TIC Read Status [3:0x0]: 1:57
2018-08-26 16:41:56.013721-0700 chat_app[251:9300] TIC Read Status [3:0x0]: 1:57
**In success closure: true**

Answer 1

任务异步并不奇怪。

string url = "EntityDefinitions(LogicalName='contact')/Attributes(LogicalName='new_country')/Microsoft.Dynamics.CRM.PicklistAttributeMetadata? $select=LogicalName&$expand=OptionSet($select=Options)";

var countrylist = await client .FindEntriesAsync(url);

Swift Firebase createUser函数完成块

1 个答案: