NSRegularExpression模式

Question

我坚持定义这种模式，它将产生我所寻找的结果。任何帮助都将不胜感激。

NSError *regexError = nil;
NSRegularExpression *parsingRegex = [NSRegularExpression regularExpressionWithPattern:@"(vector)\\((.*?)(?:,\\s*(.*?))*\\)"
                                            options:0
                                            error:&regexError];

NSString *mystring = @"vector(0.1, 0.0, 0.0, 1.0)";

NSTextCheckingResult *parse = [parsingRegex firstMatchInString:mystring
                                        options:0
                                        range:NSMakeRange(0, [string length])];


NSLog(@"string0 =%@\n"[mystring substringWithRange:[parse rangeAtIndex:0]]);
NSLog(@"string1 =%@\n"[mystring substringWithRange:[parse rangeAtIndex:1]]);
NSLog(@"string2 =%@\n"[mystring substringWithRange:[parse rangeAtIndex:2]]);
NSLog(@"string3 =%@\n"[mystring substringWithRange:[parse rangeAtIndex:3]]);
NSLog(@"string4 =%@\n"[mystring substringWithRange:[parse rangeAtIndex:4]]);
NSLog(@"string4 =%@\n"[mystring substringWithRange:[parse rangeAtIndex:5]]);

我希望得到以下输出：

string0 = vector(0.1, 0.0, 0.0, 1.0)
string1 = vector
string2 = 0.1
string3 = 0.0
string4 = 0.0
string5 = 1.0

我得到以下输出：

string0 = vector(0.1, 0.0, 0.0, 1.0)
string1 = vector
string2 = 0.1
string3 = 1.0
*** Terminating app due to uncaught exception 'NSInvalidArgumentException', reason: '-[NSExtendedRegularExpressionCheckingResult rangeAtIndex:]: index 4 out of range'

谢谢

所以这是可行的：

模式是：

NSRegularExpression *parsingRegex = [NSRegularExpression 
                regularExpressionWithPattern:@"(vector)\\((.*?)(?:,\\s*(.*?))(?:,\\s*(.*?))(?:,\\s*(.*?))*\\)"
                options:0
                error:&regexError];

但仅当字符串为

NSString *mystring = @"vector(0.1, 0.0, 0.0, 1.0)";

如果字符串为：

NSString *mystring = @"value(0.1)";

我预料到了：

NSRegularExpression *parsingRegex = [NSRegularExpression 
                regularExpressionWithPattern:@"(vector|value)\\((.*?)(?:,\\s*(.*?))(?:,\\s*(.*?))(?:,\\s*(.*?))*\\)"
                options:0
                error:&regexError];


NSLog(@"string0 =%@\n"[mystring substringWithRange:[parse rangeAtIndex:0]]);
NSLog(@"string1 =%@\n"[mystring substringWithRange:[parse rangeAtIndex:1]]);
NSLog(@"string2 =%@\n"[mystring substringWithRange:[parse rangeAtIndex:2]]);

将会工作并返回

string0 = value(0.1)
string1 = value
string2 = 0.1

但事实并非如此。

有什么想法吗？

Answer 1

(vector)\((.*?)(?:,\s*(.*?))*\)有3个捕获组。它永远不会有更多。

(vector)\(([^,\s]+)\s*,\s*([^,\s]+)\s*,\s*([^,\s]+)\s*,\s*([^,\s]+)\s*,\s*([^)\s]+)\s*\)更接近你想要的。