我正在编写一个从臭名昭著的数据中读取数据的 Perl 脚本,但/dev/input/event*
我没有找到将内核生成的关键代码转换为 ASCII 的方法。
我在这里谈论的是这个表中的 linux 键代码,我似乎找不到可以帮助我翻译它们的东西,而无需将数组硬编码到脚本中。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,所以有什么想法吗?:)
不幸的是,我没有用 Perl 编程,但这里有一个用 C 编写的简单示例。也许它可能对您有所帮助。
/*
* 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,请将其放入您的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
这基本上是一个地图问题。您必须获取一个键码并查找其 ASCII 等价物。您认为“数组部分”不是一个好做法吗?
我在 CPAN 上没有看到这个模块,但这意味着你有机会成为第一个上传它的人。:)
示例 1 只为您返回已经来自 linux 内核的相同关键代码值。例如,您获得 KEY_A 0x1e 用于按“a”键。你想要的是(我想要的)是 ascii 转换,所以如果按下“a”,我想看到小写的 0x61 和大写的 0x41。
为了从条形码阅读器读取条形码,我错过了一个简单的应用程序,可以将纯击键输入字符串。到目前为止,进行完整的键盘翻译要容易得多,因为条形码通常包含大部分数字和一些正常的 ascii 字符。所以,也许,这个简单的 python3 脚本也可以帮助其他人开始。它需要python3-evdev
作为库。当然,您可能必须适应InputDevice
. 这适用于曼哈顿读者。
from evdev import InputDevice, categorize, ecodes
dev = InputDevice('/dev/input/by-id/usb-040b_6543-if01-event-kbd')
print(dev)
shiftPressed = False
ctrlPressed = False
string = ""
for event in dev.read_loop():
if event.type == ecodes.EV_KEY:
keyEvent = categorize(event)
# handle release of special keys
if keyEvent.keystate == 0:
if keyEvent.keycode=="KEY_LEFTSHIFT":
shiftPressed = False
continue
if keyEvent.keycode=="KEY_LEFTCTRL":
ctrlPressed = False
continue
# handle key presses
if keyEvent.keystate == 1:
if keyEvent.keycode=="KEY_LEFTSHIFT":
shiftPressed = True
continue
if keyEvent.keycode=="KEY_LEFTCTRL":
ctrlPressed = True
continue
if ctrlPressed:
continue
key = keyEvent.keycode[4:]
if key == "ENTER":
print(string)
string = ""
continue
dict2 = {"Z" : "Y", "Y": "Z"}
if key in dict2:
key = dict2[key]
if not (shiftPressed):
key = key.lower()
else:
dict = {"0" : "=",
"1" : "!",
"2" : "\"",
"3" : "§",
"4" : "$",
"5" : "%",
"6" : "&",
"7" : "/",
"8" : "(",
"9" : ")"}
if key in dict:
key = dict[key]
string+=key