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];

其他回答

尝试stringWithFormat:

NSString *myString = [NSString stringWithFormat:@"%@ %@ %@ %d", "The", "Answer", "Is", 42];

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"];
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];
listOfCatalogIDs =[@[@"id[]=",listOfCatalogIDs] componentsJoinedByString:@""];

我试过这个代码。这对我很有效。

NSMutableString * myString=[[NSMutableString alloc]init];
myString=[myString stringByAppendingString:@"first value"];
myString=[myString stringByAppendingString:@"second string"];