我四处寻找名字叫这个,但我找不到任何.... 它是在朋友发送给我的.java文件中(他以为我会直接解码它),不知道我也是这个菜鸟.... 这是我想要在不编译Java文件的情况下解码的字符串数据。
String[] descriptorData = {
"\n0com/google/javascript/jscomp/function_" +
"info.proto\022\006jscomp\"\277\002\n\026FunctionInformati" +
"onMap\0223\n\005entry\030\001 \003(\n2$.jscomp.FunctionIn" +
"formationMap.Entry\0225\n\006module\030e \003(\n2%.jsc" +
"omp.FunctionInformationMap.Module\032\207\001\n\005En" +
"try\022\n\n\002id\030\002 \002(\005\022\023\n\013source_name\030\003 \002(\t\022\023\n\013" +
"line_number\030\004 \002(\005\022\023\n\013module_name\030\005 \002(\t\022\014" +
"\n\004size\030\006 \002(\005\022\014\n\004name\030\007 \002(\t\022\027\n\017compiled_s" +
"ource\030\010 \002(\t\032/\n\006Module\022\014\n\004name\030f \002(\t\022\027\n\017c" +
"ompiled_source\030g \002(\tB \n\034com.google.javas","cript.jscompP\001"
答案 0 :(得分:0)
What you see there is if I'm not mistaken a piece of auto generated code that describes
message FunctionInformationMap {
repeated group Entry = 1 {
required int32 id = 2;
required string source_name = 3;
required int32 line_number = 4;
required string module_name = 5;
required int32 size = 6;
required string name = 7;
required string compiled_source = 8;
}
}
and the generated code here can be found for example here
Ps: I just googled "com/google/javascript/jscomp/function_info.proto" but you can actually reverse the process. Hints here for example https://www.sysdream.com/reverse-engineering-protobuf-apps
答案 1 :(得分:0)
我只是通过使用System.Out.PrintIn
将字符串数据打印到TXT文件来解决这个问题......
那是...... 感谢...