如何将文本字符串转换为十六进制字符串？ [英] How to convert a text string to hex string?

问题描述

可以说我有一个文本字符串：
向我的小朋友问好

函数应该返回十六进制值：

5361792048656c6c6f20746f204d79204c6974746c6520467269656e64

解决方案

无法用swift编写，下面的代码可能是你正在寻找的：

  NSString * str = @向我的小朋友问好; 
 
 NSString * hexString = [NSString stringWithFormat：@％@，
 [NSDataDataWithBytes：[str cStringUsingEncoding：NSUTF8StringEncoding] 
 length：strlen（[str cStringUsingEncoding：NSUTF8StringEncoding] ）]]; 
 
 for（NSArray arrayWithObjects：@<，@> @ ]; 
 
 NSLog（@hexStr：％@，hexString）; 
  

上面的代码给出了确切的字符串：

5361792048656c6c6f20746f204d79204c6974746c6520467269656e64

希望它有助于：）

Lets say I have a text string : "Say Hello to My Little Friend"

A function should return hex value as:

5361792048656c6c6f20746f204d79204c6974746c6520467269656e64

解决方案

Couldn't able to write in swift, but in Objective-C below code is may be what you are looking for:

    NSString * str = @"Say Hello to My Little Friend";

    NSString * hexString = [NSString stringWithFormat:@"%@",
                         [NSData dataWithBytes:[str cStringUsingEncoding:NSUTF8StringEncoding]
                                        length:strlen([str cStringUsingEncoding:NSUTF8StringEncoding])]];

    for(NSString * toRemove in [NSArray arrayWithObjects:@"<", @">", @" ", nil])
        hexString = [hexString stringByReplacingOccurrencesOfString:toRemove withString:@""];

    NSLog(@"hexStr:%@", hexString);

Above code gives exact string as you given:

5361792048656c6c6f20746f204d79204c6974746c6520467269656e64

Hope it will help:)

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