Question

我有这段代码：

NSString *formatted = original;
formatted = [original stringByReplacingOccurrencesOfString:@"+" withString:@"%2B"];
formatted = [original stringByReplacingOccurrencesOfString:@" " withString:@"+"];

正如您可能看到的那样，我正在尝试执行以下操作：

'+'替换为'％2B'
''替换为'+'

当然，这不起作用，因为第二行会覆盖第一行，所以现在只有''被'+'替换，第一行代码无效。

有解决方法吗？

Answer 1

NSString *formatted = original;
formatted = [formatted stringByReplacingOccurrencesOfString:@"+" withString:@"%2B"];
formatted = [formatted stringByReplacingOccurrencesOfString:@" " withString:@"+"];

Answer 2

简单的解决方案：

NSString *original = @"some string+here";

NSString *formatted = [original stringByReplacingOccurrencesOfString:@"+" withString:@"%2B"];
formatted = [formatted stringByReplacingOccurrencesOfString:@" " withString:@"+"];

NSLog(@"formatted: %@", formatted);

输出：

格式化：some + string％2Bhere

Answer 3

最好的方法是使用Mutable string对象而不是不可变对象，因为每次更换字符串时都必须在内存中创建一个新对象，这会导致编码错误。

   NSString *original = @"some string here";

    /**
     *  Create a Mutable string object as you will always be changing its value depending on the request
     */
    NSMutableString *newString = [NSMutableString stringWithString:original];

    /**
     *  replace all occurrences within the current object
     */
    [newString replaceOccurrencesOfString:@"+"
                               withString:@"%2B"
                                  options:NSCaseInsensitiveSearch range:NSMakeRange(0, newString.length)];

    [newString replaceOccurrencesOfString:@" "
                               withString:@"+"
                                  options:NSCaseInsensitiveSearch range:NSMakeRange(0, newString.length)];

    NSLog(@"Result: %@", newString);

防止'stringByReplacingOccurrencesOfString'覆盖

3 个答案: