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
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"];
其他回答
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"];
尝试stringWithFormat:
NSString *myString = [NSString stringWithFormat:@"%@ %@ %@ %d", "The", "Answer", "Is", 42];
对于所有Objective C爱好者，在ui测试中需要这个:
-(void) clearTextField:(XCUIElement*) textField{
NSString* currentInput = (NSString*) textField.value;
NSMutableString* deleteString = [NSMutableString new];
for(int i = 0; i < currentInput.length; ++i) {
[deleteString appendString: [NSString stringWithFormat:@"%c", 8]];
}
[textField typeText:deleteString];
}
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"];
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;
}
