的OBJ-C：創建陣列從字符串中的項目在<>

Question

我想一個字符串解析到一個數組中的元素是<之間>例如<this is column 1><this is column 2>等....

幫助將不勝感激。

感謝

Answer 1

NSString *input =@"<one><two><three>"; 
NSString *strippedInput = [input stringByReplacingOccurencesOfString: @">" withString: @""]; //strips all > from input string 
NSArray *array = [strippedInput componentsSeperatedByString:@"<"]; 

注意，[陣列objectAtIndex：0]將是一個空串（ 「」）的噸他當然不會工作，如果其中一個「實際」字符串包含<或>

Answer 2

一種方法可能是使用兩種componentsSeparatedByCharactersInSet或從NSString的componentsSeparatedByString。

NSString *test = @"<one> <two> <three>"; 

NSArray *array1 = [test componentsSeparatedByCharactersInSet:[NSCharacterSet characterSetWithCharactersInString:@"<>"]]; 

NSArray *array2 = [test componentsSeparatedByString:@"<"]; 

你需要做一些清理之後，無論是在數組2或陣列1

Answer 3

東西的情況下，去除空白的字符串的情況下，微調證明：

NSString *string = @"<this is column 1><this is column 2>"; 
NSScanner *scanner = [NSScanner scannerWithString:string]; 

NSMutableArray *array = [NSMutableArray arrayWithCapacity:0]; 

NSString *temp; 

while ([scanner isAtEnd] == NO) 
{ 
    // Disregard the result of the scanner because it returns NO if the 
    // "up to" string is the first one it encounters. 
    // You should still have this in case there are other characters 
    // between the right and left angle brackets. 
    (void) [scanner scanUpToString:@"<" intoString:NULL]; 

    // Scan the left angle bracket to move the scanner location past it. 
    (void) [scanner scanString:@"<" intoString:NULL]; 

    // Attempt to get the string. 
    BOOL success = [scanner scanUpToString:@">" intoString:&temp]; 

    // Scan the right angle bracket to move the scanner location past it. 
    (void) [scanner scanString:@">" intoString:NULL]; 

    if (success == YES) 
    { 
     [array addObject:temp]; 
    } 
} 

NSLog(@"%@", array);