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
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];
}
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:@""];
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;
}
使c = [a stringByAppendingString: b]更短的唯一方法是在st点附近使用自动补全。+运算符是C的一部分，它不知道Objective-C对象。
你可以使用NSArray as
NSString *string1=@"This"
NSString *string2=@"is just"
NSString *string3=@"a test"
NSArray *myStrings = [[NSArray alloc] initWithObjects:string1, string2, string3,nil];
NSString *fullLengthString = [myStrings componentsJoinedByString:@" "];
or
你可以使用
NSString *imageFullName=[NSString stringWithFormat:@"%@ %@ %@.", string1,string2,string3];
当为web服务构建请求时，我发现像下面这样做非常容易，并且在Xcode中使连接可读:
NSString* postBody = {
@"<?xml version=\"1.0\" encoding=\"utf-8\"?>"
@"<soap:Envelope xmlns:xsi=\"http://www.w3.org/2001/XMLSchema-instance\" xmlns:xsd=\"http://www.w3.org/2001/XMLSchema\" xmlns:soap=\"http://schemas.xmlsoap.org/soap/envelope/\">"
@" <soap:Body>"
@" <WebServiceMethod xmlns=\"\">"
@" <parameter>test</parameter>"
@" </WebServiceMethod>"
@" </soap:Body>"
@"</soap:Envelope>"
};
