Question

public class HexASCIITest {
public static void main(String[] args) throws DecoderException, UnsupportedEncodingException {

    String test = "src=\"test/__test/path/path2/AA_5F00_20140915_5F00_15_5F00_11_5F00_55_5F00_image_5F005F00_name.jpg\"";

    Pattern patternImages = Pattern.compile("src=\"[^\"]*?/__test/[^/]*?/[^/]*?/([^\"/]*?)\"");
    Matcher matcherImages = patternImages.matcher(test);

    while(matcherImages.find()) {
        String imageName = matcherImages.group(1);

        Pattern pattern = Pattern.compile("_((?:[01234567890ABCDEF]{4}){1,})_");            
        Matcher matcher = pattern.matcher(imageName);

        while(matcher.find()) {     
            byte[] bytes = Hex.decodeHex(matcher.group(1).toCharArray());       
            String imagePath = new String(bytes, "latin1");             
            imagePath = imagePath.replaceAll("\0", "");
            imageName = imageName.replaceFirst("_((?:[01234567890ABCDEF]{4}){1,})_", imagePath.trim());
        }           
        System.out.println(imageName);
    }   
}   
}

大家好，这是我的一个程序，实际上应该将十六进制代码转换为ASCII，但似乎我遇到了逻辑问题，有人可以帮我吗？

初始图像名称为：AA_5F00_20140915_5F00_15_5F00_11_5F00_55_5F00_image_5F005F00_name.jpg 完成所有替换后：AA_15_11_55__image_5F005F00_name.jpg

当20140915日期消失并且 5F005F00 仍然存在时，它不应该如何工作。谢谢你的帮助！

Answer 1

找到它正则表达式应该是 - 模式模式= Pattern.compile（＆＃34;（（[0123456789ABCDEF] {4}）{1，} ）＆＃34;）;

然后 - ＆gt; byte [] bytes = Hex.decodeHex（matcher.group（2）.toCharArray（））;

最后更换 imageName = imageName.replaceFirst（matcher.group（1），imagePath.trim（））;

Java Matcher过度替换组

1 个答案: