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
listOfCatalogIDs =[@[@"id[]=",listOfCatalogIDs] componentsJoinedByString:@""];
其他回答
这里有一个简单的方法，使用新的数组文字语法:
NSString * s = [@[@"one ", @"two ", @"three"] componentsJoinedByString:@""];
^^^^^^^ create array ^^^^^
^^^^^^^ concatenate ^^^^^
尝试stringWithFormat:
NSString *myString = [NSString stringWithFormat:@"%@ %@ %@ %d", "The", "Answer", "Is", 42];
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];
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];
正在尝试在lldb窗格中执行以下操作
[NSString stringWithFormat:@"%@/%@/%@", three, two, one];
这错误。
而是使用alloc和initWithFormat方法:
[[NSString alloc] initWithFormat:@"%@/%@/%@", @"three", @"two", @"one"];
