获取和替换字符串中特定单词的最佳方法是什么? 例如,我有
NSString * currentString = @"one {two}, thing {thing} good";
现在我需要找到每个{currentWord}
并为其应用功能
[self replaceWord:currentWord]
然后将currentWord替换为函数
的结果-(NSString*)replaceWord:(NSString*)currentWord;
答案 0 :(得分:3)
以下示例说明如何使用NSRegularExpression
和enumerateMatchesInString
完成任务。我刚刚使用uppercaseString
作为替换单词的函数,但您也可以使用replaceWord
方法:
编辑:如果替换的单词是,我的答案的第一个版本无法正常工作 比原始单词更短或更长(感谢Fabian Kreiser注意到这一点!)。 现在它应该在所有情况下都能正常工作。
NSString *currentString = @"one {two}, thing {thing} good";
// Regular expression to find "word characters" enclosed by {...}:
NSRegularExpression *regex;
regex = [NSRegularExpression regularExpressionWithPattern:@"\\{(\\w+)\\}"
options:0
error:NULL];
NSMutableString *modifiedString = [currentString mutableCopy];
__block int offset = 0;
[regex enumerateMatchesInString:currentString
options:0
range:NSMakeRange(0, [currentString length])
usingBlock:^(NSTextCheckingResult *result, NSMatchingFlags flags, BOOL *stop) {
// range = location of the regex capture group "(\\w+)" in currentString:
NSRange range = [result rangeAtIndex:1];
// Adjust location for modifiedString:
range.location += offset;
// Get old word:
NSString *oldWord = [modifiedString substringWithRange:range];
// Compute new word:
// In your case, that would be
// NSString *newWord = [self replaceWord:oldWord];
NSString *newWord = [NSString stringWithFormat:@"--- %@ ---", [oldWord uppercaseString] ];
// Replace new word in modifiedString:
[modifiedString replaceCharactersInRange:range withString:newWord];
// Update offset:
offset += [newWord length] - [oldWord length];
}
];
NSLog(@"%@", modifiedString);
输出:
one {--- TWO ---}, thing {--- THING ---} good