bash GAWK 脚本 - 在 BEGIN 部分打印文件名
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GAWK Script - Print filename in BEGIN section
提问by jonseager
I am writing a gawk script that begins
我正在写一个开始的 gawk 脚本
#!/bin/gawk -f
BEGIN { print FILENAME }
I am calling the file via ./script file1.htmlbut the script just returns nothing. Any ideas?
我通过调用文件，./script file1.html但脚本只返回任何内容。有任何想法吗？
回答by kurumi
you can use ARGV[1] instead of FILENAME if you really want to use it in BEGIN block
如果你真的想在 BEGIN 块中使用它，你可以使用 ARGV[1] 而不是 FILENAME
awk 'BEGIN{print ARGV[1]}' file
回答by Hai Vu
You can print the file name when encounter line 1:
您可以在遇到第 1 行时打印文件名：
FNR == 1
If you want to be less cryptic, easier to understand:
如果你想不那么神秘，更容易理解：
FNR == 1 {print}
UPDATE
更新
My first two solutions were incorrect. Thank you Dennis for pointing it out. His way is correct:
我的前两个解决方案是不正确的。谢谢丹尼斯指出。他的方法是正确的：
FNR == 1 {print FILENAME}
回答by JUST MY correct OPINION
Straight from the man page (slightly reformatted):
直接从手册页（稍微重新格式化）：
FILENAME: The name of the current input file. If no files are specified on the command line, the value of FILENAME is “-”. However, FILENAME is undefined inside the BEGIN block (unless set by getline).
FILENAME：当前输入文件的名称。如果命令行没有指定文件，FILENAME 的值为“-”。但是， FILENAME 在 BEGIN 块内未定义（除非由 getline 设置）。
回答by Warren Wright
Building on Hai Vu's answer I suggest that if you only want the filename printed once per file it needs to be wrapped in a conditional.
以 Hai Vu 的回答为基础，我建议如果您只想为每个文件打印一次文件名，则需要将其包装在条件中。
if(FNR == 1) { print FILENAME };
