在不可选的情况下获取可选的Firebase数据库值 [英] Getting an optional firebase database value when it shouldn't be optional

问题描述

我正在运行以下查询以从快照获取值，但是该值作为可选返回.

I am running the following query to get a value from the snapshot, but the value is returning as optional.

ref.child("PGroups").observeSingleEventOfType(.Value, withBlock: { (snapshot) in
    let groupName = String(rest.childSnapshotForPath("/GroupName").value)
    print(groupName)
})

并得到以下打印声明:"Optional(Name)" 推荐为:"Name"

and am getting the following printed statement: "Optional(Name)" as appoosed to just: "Name"

推荐答案

只需添加一个感叹号，告诉编译器解开可选内容:

Simply add an exclamation point to tell the compiler to unwrap the optional:

print(groupName！)应该打印名称"

或者，您也可以将代码更改为以下内容:

Alternatively you could also change your code to something like this:

ref.child("PGroups").observeSingleEventOfType(.Value, withBlock: { (snapshot) in
    let snapshotValue = snapshot.value as! [String: Any]
    let groupName = snapshotValue["GroupName"] as! String
    print(groupName)
})

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