使用正则表达式在NSString中查找/替换子字符串 [英] Use regular expression to find/replace substring in NSString

问题描述

我想使用正则表达式来查找正则表达式模式的每个实例。我的字符串中的& *; 并将其删除，因此返回值是没有任何匹配项的原始字符串。也想使用相同的函数来匹配单词之间的多个空格，而是使用单个空格。找不到这样的功能。

I would like to use regular expression to find every instances of a regular expression pattern I.e. &*; in my string and remove that from so the return value is the original string without any of the matches. Also would like to use the same function to match multiple spaces between words and have a single space instead. Could not find such a function.

示例输入字符串

NSString *str = @"123 &1245; Ross Test  12";

返回值应为

123 Ross Test 12

如果符合此模式的任何内容& * 或多个空格并替换为 @;

If anything matching this pattern "&* or multiple white spaces and replaces it with @"";

推荐答案

NSString *string = @"123 &1245; Ross Test 12";
NSError *error = nil;
NSRegularExpression *regex = [NSRegularExpression regularExpressionWithPattern:@"&[^;]*;" options:NSRegularExpressionCaseInsensitive error:&error];
NSString *modifiedString = [regex stringByReplacingMatchesInString:string options:0 range:NSMakeRange(0, [string length]) withTemplate:@""];
NSLog(@"%@", modifiedString);

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