这是表格...
Group Name Designation
2 (John) Front End Developer
12 (Jim) Back End Developer
8 (Jill) Full Stack Developer
21 (Jack) Front End Developer
2 (James) Front End Developer
12 (Jane) Full Stack Developer
我想提取属于同一组的人名。这里 John 和 James 属于第 2 组。我应该使用哪些 bash 命令或脚本(组合)来显示以下输出
John
James
我使用了不同类型的 grep 组合。但似乎不起作用。
答案1
你可以sed
这样使用:
sed -n '/^2 /s/.*(\([^)]\+\)).*/\1/p' file.txt
或者awk
像这样:
awk -F "[()]" '/^2 / {print $2}' file.txt
第一个解决方案在打印之前用括号内的字符串替换该行。第二个解决方案使用括号作为字段分隔符,然后仅打印字段二(括号内的字符串)。
答案2
我已经编写了一个 bash 脚本来执行此操作,但要使用各种命令作为输入。
脚本如下:
#!/bin/bash
# Flags - to avoid potential conflicts when the labels are numbers
# 1st
# -n = only number for column
# -t = only text for column
# -a = any for column
# 2nd
# -n = only number for row
# -t = only text for row
# -a = any for row
nc=1; tc=1; nr=1; tr=1
if [ ${1:0:1} == "-" ]; then
if [ ${1:1:1} == "n" ]; then
tc=0
elif [ ${1:1:1} == "t" ]; then
nc=1
fi
if [ ${1:2:1} == "n" ]; then
tr=0
elif [ ${1:2:1} == "t" ]; then
nr=1
fi
shift
fi
command="$1"
# Number or text value
column="$2"
row="$3"
ltrim="$4"
rtrim="$5"
columnNo=-1
rowNo=-1
exec < /dev/null
while read -r -a columns; do
(( rowNo++ ))
if (( columnNo == -1 )); then
total="${#columns[@]}"
columnNo=$(for (( i=0; i<$total; i++ ))
{
if [[ "${columns[$i]}" == "$column" && "$tc" == 1 ]] || [[ "$column" == "$i" && "$nc" == 1 ]]; then
echo "$i"
break
elif (( i >= total - 1 )); then
echo -1
break
fi
})
else
if [[ "${columns[0]}" == "$row" && "$tr" == 1 ]] || [[ "$rowNo" -eq "$row" && "$nr" == 1 ]]; then
str="${columns[columnNo]}"
str=${str#"$ltrim"}
str=${str%"$rtrim"}
echo "$str"
exit 0
fi
fi
done < <($command 2> /dev/null)
echo "Error! Could not be found."
exit 1
它接受一个可选的标志加上 3 到 5 个参数、命令、列(数字或文本值)和行(数字或文本值)以及可选的ltrim
值rtrim
。
当输出包含多个表,或者输出包含不属于表的额外文本时,它会起作用。
./extract-table-entry.sh "cat table.txt" "Name" 1
# Output = (John)
./extract-table-entry.sh "cat table.txt" "Name" 5
# Output = (James)
要删除括号,我们可以简单地指定ltrim
参数和rtrim
参数:
例如:
./extract-table-entry.sh "cat table.txt" "Name" 5 "(" ")"
# Output = James