如何将/ dev/input/event*中的Linux密钥代码转换为Perl中的ASCII？

Question

我正在写一个从臭名昭着的数据中读取数据的Perl脚本,/dev/input/event*我没有找到将内核生成的密钥代码转换为ASCII的方法.

我说的这个表中的Linux的关键代码在这里和我似乎无法找到的东西,会帮我翻译他们没有硬编码的数组到脚本.我错过了什么吗？

我想跳过数组部分,因为它似乎不是一个好习惯,所以任何想法？:)

Answer 1

不幸的是,我没有在Perl中编程,但这是一个用C编写的简单示例.或许它可能会帮助你.

/*
 * Based on keytable.c by Mauro Carvalho Chehab
 *
 * This program is free software; you can redistribute it and/or modify
 * it under the terms of the GNU General Public License as published by
 * the Free Software Foundation, version 2 of the License.
 *
 * This program is distributed in the hope that it will be useful,
 * but WITHOUT ANY WARRANTY; without even the implied warranty of
 * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
 * GNU General Public License for more details.
 */

#include <stdlib.h>
#include <stdio.h>
#include <fcntl.h>

#include <linux/input.h>

#include <string.h>
#include <linux/input.h>
#include <sys/ioctl.h>

#define KEY_RELEASE 0
#define KEY_PRESS 1
#define KEY_KEEPING_PRESSED 2

#include "parse.h"

void prtcode(int codes) {
    struct parse_key *p;

    for (p = keynames; p->name != NULL; p++) {
        if (p->value == (unsigned) codes) {
            printf("scancode %s (0x%02x)\n", p->name, codes);
            return;
        }
    }

    if (isprint(codes)) {
        printf("scancode '%c' (0x%02x)\n", codes, codes);
    } else {
        printf("scancode 0x%02x\n", codes);
    }
}

int main (int argc, char *argv[]) {
    int i, fd;
    struct input_event ev[64];

    if (argc != 2) {
        fprintf(stderr, "usage: %s event-device (/dev/input/eventX)\n", argv[0]);
        return 1;
    }

    if ((fd = open(argv[1], O_RDONLY)) < 0) {
        perror("Couldn't open input device");
        return 1;
    }

    while (1) {
        size_t rb = read(fd, ev, sizeof(ev));

        if (rb < (int) sizeof(struct input_event)) {
            perror("short read");
            return 1;
        }

        for (i = 0; i < (int) (rb / sizeof(struct input_event)); i++) {
            if (EV_KEY == ev[i].type) {
                if ((ev[i].value == KEY_PRESS) || (ev[i].value == KEY_KEEPING_PRESSED)) {
                    prtcode(ev[i].code);
                    printf("type %d code %d value %d\n", ev[i].type, ev[i].code, ev[i].value);
                    printf("\n");
                }
            }
        }
    }

    return 0;
}

要生成parse.h,请将其放入Makefile:

parse.h: /usr/include/linux/input.h
    @echo generating parse.h
    @echo -en "struct parse_key {\n\tchar *name;\n\tunsigned int value;\n} " >parse.h
    @echo -en "keynames[] = {\n" >>parse.h

    @more /usr/include/linux/input.h |perl -n \
    -e 'if (m/^\#define\s+(KEY_[^\s]+)\s+(0x[\d\w]+|[\d]+)/) ' \
    -e '{ printf "\t{\"%s\", %s},\n",$$1,$$2; }' \
    -e 'if (m/^\#define\s+(BTN_[^\s]+)\s+(0x[\d\w]+|[\d]+)/) ' \
    -e '{ printf "\t{\"%s\", %s},\n",$$1,$$2; }' \
    >> parse.h
    @echo -en "\t{ NULL, 0}\n};\n" >>parse.h

然后,像这样使用它:

./keytable /dev/input/by-path/platform-i8042-serio-0-event-kbd

Answer 2

这基本上是一个地图问题.您必须使用密钥代码并查找其ASCII等效代码.你认为"阵列部分"不是一个好习惯吗？

我没有在CPAN上看到这个模块,但这意味着你有机会成为第一个上传它的人.:)