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。

我一直在看这篇文章，最后总是整理答案，找到一个简单的解决方案，可以根据需要处理尽可能多的变量:

[NSString stringWithFormat:@"%@/%@/%@", three, two, one];

例如:

NSString *urlForHttpGet = [NSString stringWithFormat:@"http://example.com/login/username/%@/userid/%i", userName, userId];

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。

宏:

// stringConcat(...)
//     A shortcut for concatenating strings (or objects' string representations).
//     Input: Any number of non-nil NSObjects.
//     Output: All arguments concatenated together into a single NSString.

#define stringConcat(...) \
    [@[__VA_ARGS__] componentsJoinedByString:@""]

测试用例:

- (void)testStringConcat {
    NSString *actual;

    actual = stringConcat(); //might not make sense, but it's still a valid expression.
    STAssertEqualObjects(@"", actual, @"stringConcat");

    actual = stringConcat(@"A");
    STAssertEqualObjects(@"A", actual, @"stringConcat");

    actual = stringConcat(@"A", @"B");
    STAssertEqualObjects(@"AB", actual, @"stringConcat");

    actual = stringConcat(@"A", @"B", @"C");
    STAssertEqualObjects(@"ABC", actual, @"stringConcat");

    // works on all NSObjects (not just strings):
    actual = stringConcat(@1, @" ", @2, @" ", @3);
    STAssertEqualObjects(@"1 2 3", actual, @"stringConcat");
}

备用宏:(如果你想强制一个最小数量的参数)

// stringConcat(...)
//     A shortcut for concatenating strings (or objects' string representations).
//     Input: Two or more non-nil NSObjects.
//     Output: All arguments concatenated together into a single NSString.

#define stringConcat(str1, str2, ...) \
    [@[ str1, str2, ##__VA_ARGS__] componentsJoinedByString:@""];

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:@""];

一个选项:

[NSString stringWithFormat:@"%@/%@/%@", one, two, three];

另一个选择:

我猜你不满意多个追加(a+b+c+d)，在这种情况下，你可以这样做:

NSLog(@"%@", [Util append:one, @" ", two, nil]); // "one two"
NSLog(@"%@", [Util append:three, @"/", two, @"/", one, nil]); // three/two/one

使用类似于

+ (NSString *) append:(id) first, ...
{
    NSString * result = @"";
    id eachArg;
    va_list alist;
    if(first)
    {
        result = [result stringByAppendingString:first];
        va_start(alist, first);
        while (eachArg = va_arg(alist, id)) 
        result = [result stringByAppendingString:eachArg];
        va_end(alist);
    }
    return result;
}