3

我尝试使用以下代码将 NSData 转换为十六进制字符串值,但都是错误的?

这里的“结果”是我的 NSdata

我需要一个十六进制字符串作为输出,但我没有得到那个

NSUInteger dataLength = [result length];
  NSLog(@"%d",dataLength);
    NSMutableString *string = [NSMutableString stringWithCapacity:dataLength*2];
   const unsigned char *dataBytes = [result bytes];
    for (NSInteger idx = 0; idx < dataLength; ++idx) {        [string appendFormat:@"%02x", dataBytes[idx]];
 }    
    NSLog(@"%@",string);

请问我该如何转换

更新:结果数据包含加密字符串。我想将其转换为十六进制值

4

5 回答 5

14

这是我这样做的一个类别。这只是为您提供十六进制数据的所有字符。没有换行或任何东西。

@implementation NSData (NSData_Conversion)

#pragma mark - String Conversion
- (NSString *)hexadecimalString
{
    /* Returns hexadecimal string of NSData. Empty string if data is empty.   */

    const unsigned char *dataBuffer = (const unsigned char *)[self bytes];

    if (!dataBuffer)
    {
        return [NSString string];
    }

    NSUInteger          dataLength  = [self length];
    NSMutableString     *hexString  = [NSMutableString stringWithCapacity:(dataLength * 2)];

    for (int i = 0; i < dataLength; ++i)
    {
        [hexString appendFormat:@"%02x", (unsigned int)dataBuffer[i]];
    }

    return [NSString stringWithString:hexString];
}
于 2013-05-13T16:01:19.223 回答
3

最简单的解决方案是遍历 NSData 的字节并从中构造 NSString。使用 [yourData bytes] 访问字节,并将字符串构建为 NSMutableString。

这是一个使用 NSData 类别实现它的示例

interface NSData(Hex)
-(NSString*)hexRepresentationWithSpaces_AS:(BOOL)spaces;
@end

@implementation NSData(Hex)
-(NSString*)hexRepresentationWithSpaces_AS:(BOOL)spaces
{
    const unsigned char* bytes = (const unsigned char*)[self bytes];
    NSUInteger nbBytes = [self length];
    //If spaces is true, insert a space every this many input bytes (twice this many output characters).
    static const NSUInteger spaceEveryThisManyBytes = 4UL;
    //If spaces is true, insert a line-break instead of a space every this many spaces.
    static const NSUInteger lineBreakEveryThisManySpaces = 4UL;
    const NSUInteger lineBreakEveryThisManyBytes = spaceEveryThisManyBytes * lineBreakEveryThisManySpaces;
    NSUInteger strLen = 2*nbBytes + (spaces ? nbBytes/spaceEveryThisManyBytes : 0);

    NSMutableString* hex = [[NSMutableString alloc] initWithCapacity:strLen];
    for(NSUInteger i=0; i<nbBytes; ) {
        [hex appendFormat:@"%02X", bytes[i]];
        //We need to increment here so that the every-n-bytes computations are right.
        ++i;

        if (spaces) {
            if (i % lineBreakEveryThisManyBytes == 0) [hex appendString:@"\n"];
            else if (i % spaceEveryThisManyBytes == 0) [hex appendString:@" "];
        }
    }
    return [hex autorelease];
}
@end





  NSData* data = ...
NSString* hex = [data hexRepresentationWithSpaces_AS:YES];
于 2013-05-13T11:50:38.383 回答
2

这是整数运算的转换:

-(NSString*)hex:(NSData*)data{
     NSMutableData *result = [NSMutableData dataWithLength:2*data.length];
     unsigned const char* src = data.bytes;
     unsigned char* dst = result.mutableBytes;
     unsigned char t0, t1;

     for (int i = 0; i < data.length; i ++ ) {
          t0 = src[i] >> 4;
          t1 = src[i] & 0x0F;

          dst[i*2] = 48 + t0 + (t0 / 10) * 39;
          dst[i*2+1] = 48 + t1 + (t1 / 10) * 39;
     }

     return [[NSString alloc] initWithData:result encoding:NSASCIIStringEncoding];
}
于 2017-01-31T20:40:55.470 回答
0

这个答案是在 Xcode 10 Playground 中运行的 Swift 4.2 中。

import Foundation

let d = Data(bytes: [0x3f, 0x09, 0x0a, 0xa3, 0x01])
let s:String = d.compactMap({ String(format: "%02x ", $0) }).joined()
print(s)

这是输出:

3f 09 0a a3 01 
于 2018-11-14T17:25:25.613 回答
0

斯威夫特 2 版本

private let kHexChars = Array("0123456789abcdef".utf8) as [UInt8];

extension NSData {

    public func hexString() -> String {
        guard length > 0 else {
            return ""
        }

        let buffer = UnsafeBufferPointer<UInt8>(start: UnsafePointer(bytes), count: length)
        var output = [UInt8](count: length*2 + 1, repeatedValue: 0)
        var i: Int = 0
        for b in buffer {
            let h = Int((b & 0xf0) >> 4)
            let l = Int(b & 0x0f)
            output[i++] = kHexChars[h]
            output[i++] = kHexChars[l]
        }

        return String.fromCString(UnsafePointer(output))!
    }
}
于 2015-08-13T17:06:32.713 回答