我有一个像这样的文本文件:
+------------------+------------+----------+
| col_name | data_type | comment |
+------------------+------------+----------+
| _id | bigint | |
| starttime | string | |
+------------------+------------+----------+
我如何使用bash获得这样的结果
(_id bigint, starttime string )
所以只是列名和类型
#remove first 3 lines
sed -e '1,3d' < columnnames.txt >clean.txt
#remove first character from each line
sed 's/^.//' < clean.txt >clean.txt
#remove last character from each line
sed 's/.$//' < clean.txt >clean.txt
# remove certain characters
sed 's/[+-|]//g' < clean.txt >clean.txt
# remove last line
sed '$ d' < clean.txt >clean.txt
所以这就是我到目前为止的情况,如果有更好的实现方法,请告诉我!
类似的,仅使用awk
:
awk -F ' *[|]' 'BEGIN {printf("(")} NR>3 && NF>1 {printf("%s%s%s", NR>4 ? "," : "", $2, $3)} END {printf(" )\n")}' columnnames.txt