ios Cocoa/Objective-C - 验证 NSDictionary 上是否存在密钥
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Cocoa/Objective-C - Verify if a key exist on a NSDictionary
提问by fabio santos
How do I verify if a key exists on a NSDictionary?
如何验证密钥是否存在于NSDictionary?
I know how to verify if it has some content, but I want to very if it is there, because it's dynamic and I have to prevent it. Like in some cases it could happen to have a key with the "name" and is value, but in another cases it could happen that this pair of value don't exists.
我知道如何验证它是否有一些内容，但我很想知道它是否存在，因为它是动态的，我必须阻止它。就像在某些情况下，它可能碰巧有一个带有“名称”和值的键，但在另一种情况下，这对值可能不存在。
回答by CRD
The simplest way is:
最简单的方法是：
[dictionary objectForKey:@"key"] != nil
as dictionaries return nilfor non-existant keys (and you cannot therefore store a nil in a dictionary, for that you use NSNull).
因为字典返回nil不存在的键（因此您不能在字典中存储 nil，因为您使用NSNull）。
Edit: Answer to comment on Bradley's answer
编辑：回答评论布拉德利的回答
You further ask:
你进一步问：
Is there a way to verify if this: [[[contactDetailsDictionary objectForKey:@"professional"] objectForKey:@"CurrentJob"] objectForKey:@"Role"] exists? Not a single key, because is a really giant dictionary, so it could exist in another category.
有没有办法验证： [[[contactDetailsDictionary objectForKey:@"professional"] objectForKey:@"CurrentJob"] objectForKey:@"Role"] 是否存在？不是一个键，因为它是一个非常巨大的字典，所以它可以存在于另一个类别中。
In Objective-C you can send a message to nil, it is not an error and returns nil, so expanding the simple method above you just write:
在 Objective-C 中，你可以向 发送消息nil，它不是错误并返回nil，所以扩展上面的简单方法你只需写：
[[[contactDetailsDictionary objectForKey:@"professional"]
objectForKey:@"CurrentJob"]
objectForKey:@"Role"] != nil
as if any part of the key-sequence doesn't exist the LHS returns nil
好像关键序列的任何部分不存在 LHS 返回 nil
回答by Bradley M Handy
NSDictionaryreturns all of the keys as an NSArrayand then use containsObjecton the array.
NSDictionary将所有键作为 an 返回NSArray，然后containsObject在数组上使用。
NSDictionary* dictionary = [NSDictionary dictionaryWithObjectsAndKeys:@"object", @"key"];
if ([[dictionary allKeys] containsObject:@"key"]) {
NSLog(@"'key' exists.");
}
