blocks|key|154732|text|AnyObject?是可选的，因为字典可能包含也可能不包含"uuid“键的值。要获取optional的值，您必须对其进行解包。请参阅文档中的Optionals。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|154733|处理可选的最安全的方法是把它放在一个条件语句中。|154734|if+let+uuidString+=+regionToMonitor["uuid"]+{
++++//+do+something+with+uuidString
}|code-block|syntax|javascript|154735|如果您绝对确定字典将始终包含此键/值对，则可以使用隐式展开选项(+!后缀)：|154736|println("UUID:+\(regionToMonitor["uuid"]!)")+++++++|154737|在这种情况下，如果密钥没有值，你的应用程序就会崩溃。|154738|如果你经常使用!，看起来就像你一直在大喊大叫……这可能有助于解释为什么你应该谨慎使用它，如果有的话。:)|154739|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.apple.com/library/prerelease/ios/documentation/Swift/Conceptual/Swift_Programming_Language/TheBasics.html#//apple_ref/doc/uid/TP40014097-CH5-XID_428^0|0|A|1Y|9|0|0|0|0|X|1|0|0|0|7|1|0^^$0|@$1|2|3|4|5|6|7|12|8|@$9|13|A|14|B|C]]|D|@$9|15|A|16|1|17]]|E|$]]|$1|F|3|G|5|6|7|18|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|19|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|1A|8|@$9|1B|A|1C|B|C]]|D|@]|E|$]]|$1|O|3|P|5|J|7|1D|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|1E|8|@]|D|@]|E|$]]|$1|S|3|T|5|6|7|1F|8|@$9|1G|A|1H|B|C]]|D|@]|E|$]]|$1|U|3|-4|5|6|7|1I|8|@]|D|@]|E|$]]]|V|$W|$5|X|Y|Z|E|$10|11]]]]

<code>AnyObject?</code> is an optional, because the dictionary may or may not contain a value for the "uuid" key. To get at an optional's value, you have to unwrap it. See <a href="https://developer.apple.com/library/prerelease/ios/documentation/Swift/Conceptual/Swift_Programming_Language/TheBasics.html#//apple_ref/doc/uid/TP40014097-CH5-XID_428" rel="nofollow">Optionals</a> in the documentation.

The safest way to deal with an optional is to put it in a conditional statement.

<pre><code>if let uuidString = regionToMonitor["uuid"] {
 // do something with uuidString
}
</code></pre>

If you're absolutely positively sure the dictionary will always contain this key/value pair, you can use an implicitly unwrapped optional (the <code>!</code> suffix):

<pre><code>println("UUID: \(regionToMonitor["uuid"]!)") 
</code></pre>

In this case, if there's no value for the key your app will crash.

If you use <code>!</code> a lot, it looks like you're yelling all the time... which might help illustrate why you should use it sparingly, if at all. :)

blocks|key|2954204|text|我已经找到了一个有效的解决方案，它在编译时没有任何警告，比如：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2954205|var+regions+=+NSBundle.mainBundle().infoDictionary["Regions"]+as+Array<Dictionary<String,+AnyObject>>

for+region+in+regions+{
++++let+dict:+NSDictionary+=+region
++++var+uuid+=+dict["uuidString"]+as+String
}|code-block|syntax|javascript|2954206|来自NSBundle的infoDictionary返回NSArray和NSDictionary，而不是Swift.Array或Swift.Dictionary。虽然，它们应该是可互换的，但也许它们并不像我们想的那样。|offset|length|style|CODE|2954207|entityMap^0|0|0|2|8|B|E|R|7|Z|C|1F|B|1R|G|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]|$I|V|J|W|K|L]|$I|X|J|Y|K|L]|$I|Z|J|10|K|L]|$I|11|J|12|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|13|8|@]|9|@]|A|$]]]|N|$]]

I've found a working solution, which compiles without warnings and such:

<pre><code>var regions = NSBundle.mainBundle().infoDictionary["Regions"] as Array&lt;Dictionary&lt;String, AnyObject&gt;&gt;

for region in regions {
 let dict: NSDictionary = region
 var uuid = dict["uuidString"] as String
}
</code></pre>

The <code>infoDictionary</code> from the <code>NSBundle</code> returns an <code>NSArray</code> and <code>NSDictionary</code>, not a <code>Swift.Array</code> or <code>Swift.Dictionary</code>. Though, they should be interchangeable, but maybe they aren't as we though.

blocks|key|2741456|text|我最后说了一句难看的话：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2741457|var+uuidString:String+=+regionToMonitor["uuid"]+as!+String|code-block|syntax|javascript|2741458|无警告、无错误、无运行时错误|2741459|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I ended up with the ugly line:

<pre><code>var uuidString:String = regionToMonitor["uuid"] as! String
</code></pre>

no warnings, no errors, no runtime error

blocks|key|2741527|text|您还可以使用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2741528|var+uuidString+=+regionToMonitor["uuid"]?+as+String|code-block|syntax|javascript|2741529|它与你正在做的事情有相同的结果，但在意图上我更清楚。无论如何，as操作符都会强制展开，因此在后面加上感叹号感觉很多余。将问号放在字典下标后面表明您链接的是可选的。|offset|length|style|CODE|2741530|entityMap^0|0|0|V|2|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

You can also use

<pre><code>var uuidString = regionToMonitor["uuid"]? as String
</code></pre>

It has the same results as what you are doing, but is IMHO more clear in intent. The <code>as</code> operator force unwraps anyway, so putting the exclamation mark behind it feels redundant. Putting the question mark behind the dictionary subscript makes it clear you are chaining an optional.

blocks|key|2954274|text|我发现这对我很有效|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2954275|var+uuidString:+String?+=+regionToMonitor["uuid"]+as+AnyObject?+as?+String|code-block|syntax|javascript|2954276|编辑:这是一个旧的swift版本的答案|2954277|请使用接受的答案。|2954278|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I found this to work for me

<pre><code>var uuidString: String? = regionToMonitor["uuid"] as AnyObject? as? String
</code></pre>

EDIT: this was the answer for an older swift version

Please use the accepted answer.

blocks|key|2741545|text|如果您确定需要取消包装的值，则可以使用以下任意一种方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2741546|var+uuidString:String!+=+regionToMonitor["uuid"]
var+uuidString+=+regionToMonitor["uuid"]+as+String!|code-block|syntax|javascript|2741547|或者甚至是这样：|2741548|if+var+uuidString+=+regionToMonitor["uuid"]+{
++++println("\(uuidString)+has+been+unwrapped")
}|2741549|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

If you are sure you want the unwrapped value you can use any of these:

<pre><code>var uuidString:String! = regionToMonitor["uuid"]
var uuidString = regionToMonitor["uuid"] as String!
</code></pre>

or even this:

<pre><code>if var uuidString = regionToMonitor["uuid"] {
 println("\(uuidString) has been unwrapped")
}
</code></pre>

blocks|key|154816|text|我不确定我的解决方案是否有效，但它就在这里。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|154817|var+uuidVar+=+regionToMonitor["uuid"]
var+uuidString:String+=+"\(uuidVar)"|code-block|syntax|javascript|154818|希望能有所帮助。|154819|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I am not sure my solution is effective of not but here it is.

<pre><code>var uuidVar = regionToMonitor["uuid"]
var uuidString:String = "\(uuidVar)"
</code></pre>

Hope it helps.

blocks|key|156130|text|保持简单：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|156131|let+uuidString+=+"\(regionToMonitor["uuid"])"|code-block|syntax|javascript|156132|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Keep it simple:

<pre><code>let uuidString = "\(regionToMonitor["uuid"])"
</code></pre>

I am reading a plist key (NSArray with n NSDictionaries):

<pre><code> let regionsToMonitor = NSBundle.mainBundle().infoDictionary["Regions"] as Array&lt;Dictionary&lt;String,AnyObject&gt;&gt;
</code></pre>

now I iterate over it:

<pre><code> for regionToMonitor in regionsToMonitor {
</code></pre>

and now I want to to get uuidString of the regionToMonitor

in ObjC: <code>NSString *uuidString = regionToMonitor[@"uuidString"];</code>

in swift I try: <code>let uuidString = regionToMonitor["uuid"]!.stringValue;</code>

the above does compile but the string is always <code>nil</code> in swift. <code>regionToMonitor["uuid"]</code> when used without !.stringValue works fine in <code>println</code>

how do I get a valid Swift.String here?

I am trying to pass it to NSUUID!

<hr>

I also tried

<code>let uuidString:String = regionToMonitor["uuid"]</code> 
=> AnyObject isn't convertible to String

<code>let uuidString = regionToMonitor["uuid"] as String</code> 
=> Could not find an overload for 'subscript' that accepts the supplied arguments

<code>let uuidString = regionToMonitor["uuid"];</code> 
=> 'AnyObject?' cannot be implicitly downcast to 'String'; did you mean to use 'as' to force downcast?

Get from AnyObject(NSString) to String

Swift

我正在读取一个plist键(带有n个NSDictionaries的NSArray)： let regionsToMonitor = NSBundle.mainBundle().infoDictionary["Regions"] as Array<Dictionary<String,AnyObject>>现在我对它...

问从AnyObject(NSString)获取字符串
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