awk - 用于读取每行中具有不同字段数的输入文件的 awk 脚本

Question

输入文件在每行中具有不同数量的字段。

有没有办法用 awk 获得预期的输出？

awk -F',' '{print "echo "$1; for (i = 2; i <= NF; i++) print "command1 "$i " command2"}' test.txt

测试.txt

"abc",4,21,22,25
"standard",1 
"test",4,5,10,11,12

输出：

echo "abc"
command1 4 command2
command1 21 command2
command1 22 command2
command1 25 command2
echo "standard"
command1 1 command2
echo "test"
command1 4 command2
command1 5 command2
command1 10 command2
command1 11 command2
command1 12 command2

预期输出：

echo "abc" command1 4 command2
echo "abc" command1 21 command2
echo "abc" command1 22 command2
echo "abc" command1 25 command2
echo "standard" command1 1 command2
echo "test" command1 4 command2
echo "test" command1 5 command2
echo "test" command1 10 command2
echo "test" command1 11 command2
echo "test" command1 12 command2

标签： awk

Answer 1

请试试这个：

awk -F, '{for (i=2;i<=NF;i++) print "echo", $1, "command1", $i, "command2"}'

例如：

$ cat file
"abc",4,21,22,25
"standard",1
"test",4,5,10,11,12

$ awk -F, '{for (i=2;i<=NF;i++) print "echo", $1, "command1", $i, "command2"}' file
echo "abc" command1 4 command2
echo "abc" command1 21 command2
echo "abc" command1 22 command2
echo "abc" command1 25 command2
echo "standard" command1 1  command2
echo "test" command1 4 command2
echo "test" command1 5 command2
echo "test" command1 10 command2
echo "test" command1 11 command2
echo "test" command1 12 command2

默认OFS是一个空格，所以我只是使用逗号来分隔需要打印的不同内容。

为了好玩，GNUsed解决方案：

sed -r '/,/!d;/,/{s/([^,]*),([^,]*)/"echo" \1 "command1" \2 "command2"\n\1/; P; D;}'

awk另一种类似于RavinderSingh13's answer 的noloop方法，但简洁：

awk -F, '{gsub(/,/,"\necho " $1 " command1 ");sub(/[^\n]*\n/,"");gsub(/\n|$/," command2\n");printf $0}' file