替换 NSString 中的特定单词 [英] Replace specific words in NSString

问题描述

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}

并为其应用函数

 [self replaceWord:currentWord]

然后用函数的结果替换 currentWord

-(NSString*)replaceWord:(NSString*)currentWord;

解决方案

以下示例展示了如何使用 NSRegularExpression 和 enumerateMatchesInString 来完成任务.我刚刚使用了 uppercaseString 作为替换单词的函数，但您也可以使用 replaceWord 方法:

如果替换的单词是比原词更短或更长(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";//正则表达式查找 {...} 包围的单词字符":NSRegularExpression *regex;regex = [NSRegularExpression regularExpressionWithPattern:@"\\{(\\w+)\\}"选项:0错误:NULL];NSMutableString *modifiedString = [currentString mutableCopy];__block int offset = 0;[正则表达式 enumerateMatchesInString:currentString选项:0范围:NSMakeRange(0，[当前字符串长度])usingBlock:^(NSTextCheckingResult *result, NSmatchingFlags 标志, BOOL *stop) {//range = currentString 中正则表达式捕获组(\\w+)"的位置:NSRange range = [result rangeAtIndex:1];//调整已修改字符串的位置:range.location += 偏移量；//获取旧词:NSString *oldWord = [modifiedString substringWithRange:range];//计算新词://在你的情况下，那就是//NSString *newWord = [self replaceWord:oldWord];NSString *newWord = [NSString stringWithFormat:@"--- %@ ---", [oldWord uppercaseString] ];//替换 modifiedString 中的新单词:[修改字符串替换字符输入范围:范围与字符串:新词];//更新偏移量:偏移量 += [新词长度] - [旧词长度];}];NSLog(@"%@", modifiedString);

输出:

<前>一 {--- 二 ---}，东西 {--- 东西 ---} 好

what is the best way to get and replace specific words in string ? for example I have

NSString * currentString = @"one {two}, thing {thing} good";

now I need find each {currentWord}

and apply function for it

 [self replaceWord:currentWord]

then replace currentWord with result from function

-(NSString*)replaceWord:(NSString*)currentWord;

解决方案

The following example shows how you can use NSRegularExpression and enumerateMatchesInString to accomplish the task. I have just used uppercaseString as function that replaces a word, but you can use your replaceWord method as well:

EDIT: The first version of my answer did not work correctly if the replaced words are shorter or longer as the original words (thanks to Fabian Kreiser for noting that!) . Now it should work correctly in all cases.

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);

Output:

one {--- TWO ---}, thing {--- THING ---} good

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