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:@", "];

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."];

通过创建AppendString宏的快捷方式…

#define AS(A,B)    [(A) stringByAppendingString:(B)]

NSString *myString = @"This";

NSString *test = AS(myString,@" is just a test");

注意:

如果使用宏，当然只需使用可变参数，请参阅ethb的答案。

我喜欢的方法是:

NSString *firstString = @"foo";
NSString *secondString = @"bar";
NSString *thirdString = @"baz";

NSString *joinedString = [@[firstString, secondString, thirdString] join];

你可以通过在NSArray中添加join方法来实现:

#import "NSArray+Join.h"
@implementation NSArray (Join)
-(NSString *)join
{
    return [self componentsJoinedByString:@""];
}
@end

@[]是NSArray的简短定义，我认为这是连接字符串最快的方法。

如果你不想使用类别，直接使用componentsJoinedByString:方法:

NSString *joinedString = [@[firstString, secondString, thirdString] componentsJoinedByString:@""];

如何缩短stringByAppendingString和使用#define:

#define and stringByAppendingString

因此你可以使用:

NSString* myString = [@"Hello " and @"world"];

问题是它只适用于两个字符串，你需要包装额外的括号更多的追加:

NSString* myString = [[@"Hello" and: @" world"] and: @" again"];

对于所有Objective C爱好者，在ui测试中需要这个:

-(void) clearTextField:(XCUIElement*) textField{

    NSString* currentInput = (NSString*) textField.value;
    NSMutableString* deleteString = [NSMutableString new];

    for(int i = 0; i < currentInput.length; ++i) {
        [deleteString appendString: [NSString stringWithFormat:@"%c", 8]];
    }
    [textField typeText:deleteString];
}