Objective c 有没有办法计算NSString的行数?
可能重复:Objective c 有没有办法计算NSString的行数?,objective-c,cocoa,Objective C,Cocoa,可能重复: 有没有办法计算NSString的行数 NSString * myString = @"line 1 \n line 2 \n"; 直线=3 谢谢请尝试以下代码 NSString * myString = @"line 1 \n line 2 \n"; NSArray *list = [myString componentsSeparatedByString:@"\n"]; NSLog(@"No of lines : %d",[list count]); 请尝试以下代码 NSSt
有没有办法计算NSString的行数
NSString * myString = @"line 1 \n line 2 \n";
直线=3
谢谢请尝试以下代码
NSString * myString = @"line 1 \n line 2 \n";
NSArray *list = [myString componentsSeparatedByString:@"\n"];
NSLog(@"No of lines : %d",[list count]);
请尝试以下代码
NSString * myString = @"line 1 \n line 2 \n";
NSArray *list = [myString componentsSeparatedByString:@"\n"];
NSLog(@"No of lines : %d",[list count]);
试试这个
NSString * myString = @"line 1 \n line 2 \n";
int count = [[myString componentsSeparatedByString:@"\n"] count];
NSLog(@"%d", count);
试试这个
NSString * myString = @"line 1 \n line 2 \n";
int count = [[myString componentsSeparatedByString:@"\n"] count];
NSLog(@"%d", count);
小心
由字符串分隔的组件不够智能,无法检测mac/windows/unix行结束之间的连接。在\n
上分离将适用于windows/unix行结尾,但不适用于经典的mac文件(有些流行的mac编辑器默认情况下仍然使用这些文件)。您确实应该检查\r\n
和\r
如果是加法,组件通过字符串分离:
速度慢且内存不足。如果您关心性能,则应重复搜索换行符并计算结果数:
NSString * myString = @"line 1 \n line 2 \n";
int lineCount = 1;
NSUInteger characterLocation = 0;
NSCharacterSet *newlineCharacterSet = [NSCharacterSet newlineCharacterSet];
while (characterLocation < myString.length) {
characterLocation = [myString rangeOfCharacterFromSet:newlineCharacterSet options:NSLiteralSearch range:NSMakeRange(characterLocation, (myString.length - characterLocation))].location;
if (characterLocation == NSNotFound) {
break;
}
// if we are at a \r character and the next character is a \n, skip the next character
if (myString.length >= characterLocation &&
[myString characterAtIndex:characterLocation] == '\r' &&
[myString characterAtIndex:characterLocation + 1] == '\n') {
characterLocation++;
}
lineCount++;
characterLocation++;
}
NSLog(@"%i", lineCount);
NSString*myString=@“第1行\n第2行\n”;
int lineCount=1;
NSU整数字符位置=0;
NSCharacterSet*newlineCharacterSet=[NSCharacterSet newlineCharacterSet];
while(characterLocation=字符位置&&
[myString characterAtIndex:characterLocation]=='\r'&&
[myString characterAtIndex:characterLocation+1]=='\n'){
characterLocation++;
}
lineCount++;
characterLocation++;
}
NSLog(@“%i”,行数);
小心由字符串分隔的组件
不够智能,无法在mac/windows/unix行结束之间进行检测。在\n
上分离将适用于windows/unix行结尾,但不适用于经典的mac文件(有些流行的mac编辑器默认情况下仍然使用这些文件)。您确实应该检查\r\n
和\r
如果是加法,组件通过字符串分离:
速度慢且内存不足。如果您关心性能,则应重复搜索换行符并计算结果数:
NSString * myString = @"line 1 \n line 2 \n";
int lineCount = 1;
NSUInteger characterLocation = 0;
NSCharacterSet *newlineCharacterSet = [NSCharacterSet newlineCharacterSet];
while (characterLocation < myString.length) {
characterLocation = [myString rangeOfCharacterFromSet:newlineCharacterSet options:NSLiteralSearch range:NSMakeRange(characterLocation, (myString.length - characterLocation))].location;
if (characterLocation == NSNotFound) {
break;
}
// if we are at a \r character and the next character is a \n, skip the next character
if (myString.length >= characterLocation &&
[myString characterAtIndex:characterLocation] == '\r' &&
[myString characterAtIndex:characterLocation + 1] == '\n') {
characterLocation++;
}
lineCount++;
characterLocation++;
}
NSLog(@"%i", lineCount);
NSString*myString=@“第1行\n第2行\n”;
int lineCount=1;
NSU整数字符位置=0;
NSCharacterSet*newlineCharacterSet=[NSCharacterSet newlineCharacterSet];
while(characterLocation=字符位置&&
[myString characterAtIndex:characterLocation]=='\r'&&
[myString characterAtIndex:characterLocation+1]=='\n'){
characterLocation++;
}
lineCount++;
characterLocation++;
}
NSLog(@“%i”,行数);
后一种解决方案也将正确处理CRLF序列,而Components SeparatedByCharactersInSet:
将为每个这样的序列返回一个空的组件。谢谢@PeterHosey,我认为与您所说的完全相反的是正确的,但我做了一些测试,您是对的。我的代码在windows新行中被破坏。我已编辑了我的答案以删除单行示例,并更新了长/快示例以检查\r\n
。后一种解决方案也将正确处理CRLF序列,然而,componentsSeparatedByCharactersInSet:
将为每个这样的序列返回一个空组件。谢谢@PeterHosey,我认为与你所说的完全相反的是真的,但我做了一些测试,你是对的。我的代码在windows新行中被破坏。我已编辑了我的答案以删除单行示例,并更新了长/快示例以检查\r\n
。您应该参考文本布局编程指南@Patrick请在SO上发布问题之前搜索SO和google。你们应该参考文本布局编程指南@Patrick请先搜索SO和google,然后再在SO上发布问题。