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

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;
}

其他回答

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:@""];

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];
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."];

当处理字符串时,我经常发现使源文件objc++更容易,然后我可以使用问题中显示的第二个方法连接std::字符串。

std::string stdstr = [nsstr UTF8String];

//easier to read and more portable string manipulation goes here...

NSString* nsstr = [NSString stringWithUTF8String:stdstr.c_str()];