在Objective-C中有没有(stringByAppendingString:)字符串连接的快捷方式，或者一般使用NSString的快捷方式?
例如，我想做:
NSString *myString = @"This";
NSString *test = [myString stringByAppendingString:@" is just a test"];
更像是:
string myString = "This";
string test = myString + " is just a test";
当前回答
我能想到两个答案……这两种方法都没有使用连接运算符那么令人愉快。
首先，使用一个NSMutableString，它有一个appendString方法，消除了对额外临时字符串的一些需求。
其次，使用NSArray通过componentsJoinedByString方法进行连接。
其他回答
如果你有两个NSString字面量，你也可以这样做:
NSString *joinedFromLiterals = @"ONE " @"MILLION " @"YEARS " @"DUNGEON!!!";
这对于加入#定义也很有用:
#define STRINGA @"Also, I don't know "
#define STRINGB @"where food comes from."
#define JOINED STRINGA STRINGB
享受。
NSString *label1 = @"Process Name: ";
NSString *label2 = @"Process Id: ";
NSString *processName = [[NSProcessInfo processInfo] processName];
NSString *processID = [NSString stringWithFormat:@"%d", [[NSProcessInfo processInfo] processIdentifier]];
NSString *testConcat = [NSString stringWithFormat:@"%@ %@ %@ %@", label1, processName, label2, processID];
这样使用stringByAppendingString:
NSString *string1, *string2, *result;
string1 = @"This is ";
string2 = @"my string.";
result = [result stringByAppendingString:string1];
result = [result stringByAppendingString:string2];
OR
result = [result stringByAppendingString:@"This is "];
result = [result stringByAppendingString:@"my string."];
假设你不知道这里有多少根弦。
NSMutableArray *arrForStrings = [[NSMutableArray alloc] init];
for (int i=0; i<[allMyStrings count]; i++) {
NSString *str = [allMyStrings objectAtIndex:i];
[arrForStrings addObject:str];
}
NSString *readyString = [[arrForStrings mutableCopy] componentsJoinedByString:@", "];
NSString *result=[NSString stringWithFormat:@"%@ %@", @"Hello", @"World"];
