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"];
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:@", "];
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];
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;
}
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()];
