如何向NSString插入空格。
我需要在索引5处添加一个空格:
NString * dir = @"abcdefghijklmno";
要获得此结果:
abcde fghijklmno
使用:
NSLOG (@"%@", dir);
答案 0 :(得分:85)
您需要使用NSMutableString
NSMutableString *mu = [NSMutableString stringWithString:dir];
[mu insertString:@" " atIndex:5];
或者您可以使用这些方法来拆分字符串:
- substringFromIndex:
- substringWithRange:
- substringToIndex:
并用
重新组合它们- stringByAppendingFormat:
- stringByAppendingString:
- stringByPaddingToLength:withString:startingAtIndex:
但这种方式更值得一试。而且由于NSString
是不可变的,你会打赌很多对象都是无用的。
NSString *s = @"abcdefghijklmnop";
NSMutableString *mu = [NSMutableString stringWithString:s];
[mu insertString:@" || " atIndex:5];
// This is one option
s = [mu copy];
//[(id)s insertString:@"er" atIndex:7]; This will crash your app because s is not mutable
// This is an other option
s = [NSString stringWithString:mu];
// The Following code is not good
s = mu;
[mu replaceCharactersInRange:NSMakeRange(0, [mu length]) withString:@"Changed string!!!"];
NSLog(@" s == %@ : while mu == %@ ", s, mu);
// ----> Not good because the output is the following line
// s == Changed string!!! : while mu == Changed string!!!
这可能导致难以调试的问题。
这就是为什么@property
字符串通常被定义为copy
的原因所以如果你得到一个NSMutableString
,通过复制你肯定它不会因为其他一些意外的代码而改变
我倾向于选择s = [NSString stringWithString:mu];
,因为你不会混淆复制一个可变对象并拥有一个不可变对象。
答案 1 :(得分:4)
在这里,例如,如何每隔3个字符插入一个空格......
NSMutableString *mutableString = [NSMutableString new];
[mutableString setString:@"abcdefghijklmnop"];
for (int p = 0; p < [mutableString length]; p++) {
if (p%4 == 0) {
[mutableString insertString:@" " atIndex:p];
}
}
NSLog(@"%@", mutableString);
结果:abc def ghi jkl mno p
答案 2 :(得分:2)
NSMutableString *liS=[[NSMutableString alloc]init];
for (int i=0; i < [dir length]; i++)
{
NSString *ichar = [NSString stringWithFormat:@"%c", [lStr characterAtIndex:i]];
[l1S appendString:ichar];
if (i==5)
{
[l1S appendString:@" "];
}
}
dir=l1S;
NSLog(@"updated string is %@",dir);
试试这个会帮助你
答案 3 :(得分:1)
对于一项简单的任务,您需要一个简单的解决方案:
NString * dir = @"abcdefghijklmno";
NSUInteger index = 5;
dir = [@[[dir substringToIndex:index],[dir substringFromIndex:index]]componentsJoinedByString:@" "];
答案 4 :(得分:0)
在C ++上,我发现更容易操纵NSString
,然后将其转换为std::string text = "abcdefghijklmno";
text.insert(text.begin()+5, ' ');
NSString* result = [NSString stringWithUTF8String:text.c_str()];
。 E.g:
PNUM = Sheets("Packing Slip").Cells(15, 5)
答案 5 :(得分:0)
有一个不使用可变字符串的解决方案:替换长度为0的范围内的字符:
NSString * test = @"1234";
// Insert before "3", which is at index 2.
NSRange range = NSMakeRange(2, 0);
NSString * inserted = [test stringByReplacingCharactersInRange:range
withString:@"abc"]
NSLog(@"%@", inserted); // 12abc34