如何將NSArray輸出連接到以逗號分隔的NSString

Question

我使用以下代碼嘗試將數組輸出連接到NSString中。

NSArray *array1 = [NSArray arrayWithObjects:[item objectForKey:@"id"], nil]; 
NSString *joinedString = [array1 componentsJoinedByString:@","]; 

NSLog(@"joinedString is %@", joinedString); 

我想這個輸出連接字符串：加入字符串是55,56,57,66,88 ......等等......此刻的輸出是：

2013-03-05 13:13:17.052 [63705:907] joinedString is 55 
2013-03-05 13:13:17.056 [63705:907] joinedString is 56 
2013-03-05 13:13:17.060 [63705:907] joinedString is 57 
2013-03-05 13:13:17.064 [63705:907] joinedString is 66 

Answer 1

您可能正在運行一個循環內的連接方法。

我想這就是你想要的。

NSMutableArray * array1 = [NSMutableArray array]; // create a Mutable array 

for(id item in items){ 
     [array1 addObject:[item objectForKey:@"id"]]; // Add the values to this created mutable array 
} 

NSString *joinedString = [array1 componentsJoinedByString:@","]; 

NSLog(@"joinedString is %@", joinedString); 

Answer 2

你可以做到這一點的，

就拿

NSArray *array=@[@"A",@"B",@"C"]; 
NSString *string=[array componentsJoinedByString:@","]; 
NSLog(@"%@",string); 

輸出是：

A,B,C 

Answer 3

編輯：

首先檢查[item objectForKey:@"id"]是正確與否？

，然後使用下面的代碼：

NSArray *array1 = [NSArray arrayWithObjects:[item objectForKey:@"id"], nil]; 
NSString *commaSpStr; 

commaSpStr = [array1 componentsJoinedByString:@", "]; 

NSLog(@"%@", commaSpStr); 

Answer 4

什麼都你正在寫在[item objectForKey:@"id"]一個正確，也可能是問題一經查實這一個以外的所有的罰款。

NSMutableArray *array = [[NSMutableArray alloc] 
          initWithObjects:@"55",@"56",@"57",@"58", nil]; 

    NSString *joinedString = [array componentsJoinedByString:@","]; 
     NSLog(@"%@",joinedString); 

Answer 5

您重建array1每次。創建array1的實例變量，將值插入[item objectForKey:@"id"]，您將看到joinedString將被更新。

Answer 6

我一直在評論這裏的幾個答案，並發現大多數答案只是給出的代碼作爲解決此代碼的答案，原因是因爲代碼提供了（請參閱提供的代碼）完美無瑕。

（通過提問題的人提供）

NSArray *array1 = [NSArray arrayWithObjects:[item objectForKey:@"id"], nil]; 
NSString *joinedString = [array1 componentsJoinedByString:@","]; 

NSLog(@"joinedString is %@", joinedString); 

由於用戶沒有提供如何itemNSDictionary創建我假設一個NSArray已創建其中包含一些NSDictionaries

NSArray *array = [[NSArray alloc] initWithObjects:[NSDictionary dictionaryWithObjectsAndKeys:@"55", @"id", nil], 
        [NSDictionary dictionaryWithObjectsAndKeys:@"65", @"id", nil], 
        [NSDictionary dictionaryWithObjectsAndKeys:@"75", @"id", nil], 
        [NSDictionary dictionaryWithObjectsAndKeys:@"65", @"id", nil], 
        nil]; 

問題是沒有提供的代碼，因爲我們知道item是NSDictionary我們知道[item objectForKey:@"id"]不會返回個別項目它返回ids的NSArray。所以根據它是否爲NSArray它會記錄類似joinedString is (55, 56, 57...)"。我們也知道，它不能只是一個字符串，因爲我們也只有一個值，所以它會記錄下這樣的事情，並且這並不是我們想要的。讓已經提供什麼的唯一方法是有這樣的事情

for(NSDictionary *item in array) { 

    NSArray *array1 = [NSArray arrayWithObjects:[item objectForKey:@"id"], nil]; 
    NSString *joinedString = [array1 componentsJoinedByString:@","]; 

    NSLog(@"joinedString is %@", joinedString); 
} 

因此，如果這是不是辦法的情況下解決這個是做

NSMutableArray *array1 = [NSMutableArray array]; 
    for(NSDictionary *item in array) { 

    [array1 addObject:[item objectForKey:@"id"]]; 

    } 
    // Note that this doesn't need to be in a for loop `componentsJoinedByString:` only needs to run once. 
    NSString *joinedString = [array1 componentsJoinedByString:@","]; 
    NSLog(@"joinedString is %@", joinedString); 

這樣做的輸出會（如用戶想）

joinedString is 55,65,75,65 

一旦問題提問者提供缺少的代碼，我會糾正他的回答基於存在的代碼，但在那之前我假設。

Answer 7

NSMutableArray *arr = [[NSMutableArray alloc] init]; 

for (NSDictionary *item in array) { 
    [arr addObject:[item objectForKey:@"id"]]; 
} 

NSString *joinedStr = [arr componentsJoinedByString:@","];