Question

我有一个包含以下信息的文本文件

info.txt

1,susan,12345678,partTimeStaff,1
2,john,54243214,fullTimeStaff,3
3,mary,53214567,contractStaff,3
4,gerald,14546752,partTimeStaff,0
5,joe,14234567,fullTimeStaff,2
6,joshua,11234567,contractStaff,2

我正在努力获得员工的名字和工作经验的数量，并打印出类似的东西（“staffName总年工作经验：WorkExperience”）

这就是我目前所拥有的

printOutName=$(cat "info.txt" | cut -d, -f2,4,5 | sort -r -t, -k2 | grep "partTimeStaff" | cut -d, -f1)
printOutYOfExp=$(cat "info.txt" | cut -d, -f2,4,5 | sort -r -t, -k2 | grep "partTimeStaff" | cut -d, -f3)

echo "part Time Staff"
echo -e "$printOutName total years of work experience: $printOutYOfExp"

我注意到下面显示的输出存在问题输出

part Time Staff
gerald 
susan  total years of work experience : 0
1

预期产出

part Time Staff
gerald total years of work experience : 0
susan  total years of work experience : 1

提前致谢

Answer 1

使用awk执行此任务，因为它有助于处理分隔数据。

$ awk -F, '/partTimeStaff/{print $2" total years of work experience : "$5}' info.txt
susan total years of work experience : 1
gerald total years of work experience : 0

Answer 2

您可以逐行解析文件，将每个字段放在数组字段中，选择所需的行，然后输出格式化字段，如：

echo "part Time Staff"
while IFS=, read -r -a array; do
    [[ "${array[3]}" == partTimeStaff ]] || continue
    printf "%s total years of work experience: %s\n" "${array[1]}" "${array[4]}"
done < info.txt

并且全部是100％bash，没有外部工具，也没有子shell！

Answer 3

使用perl：

$ perl -F, -lane 'print "$F[1] total years of work experience: $F[4]" if $F[3] eq "partTimeStaff"' info.txt
susan total years of work experience: 1
gerald total years of work experience: 0

Answer 4

只是为了好玩：

( IFS=$'\n,'; printf '%.0s%s total years of work experience:%.0s%.0s %s\n' $(grep 'partTimeStaff' info.txt ) )

Answer 5

这是一个解决方案，循环查找每个类别名称并为每个类别名称生成所需的输出。我把它放在一个函数中，所以你可以直接调用它：

编辑2：缩短时间！

countstaff(){
    TYPE=$(cut -d, -f4 info.txt | sort -u )
    for T in $TYPE; do 
        echo "--- $T ---"
        printf "%s total years of work experience: %s\n" $(grep ${T} info.txt | cut -d, -f2,5 | tr ',' ' ')
    done
    }

输出：

--- contractStaff ---
mary total years of work experience: 3
joshua total years of work experience: 2
--- fullTimeStaff ---
john total years of work experience: 3
joe total years of work experience: 2
--- partTimeStaff ---
susan total years of work experience: 1
gerald total years of work experience: 0

编辑：为了回应同伴压力<coff @gniourf_gniourf coff/>我稍微简化了一下这个功能，虽然文件格式的灵活性稍差（sed必须匹配更准确地输入字段）。输出如上（在子标题上有缩进）：

countstaff(){
    FILEN="info.txt"  # can change this to "$1" to change file at run time
    TYPE=$(cut -d, -f4 "$FILEN" | sort | uniq)
    for T in $TYPE; do 
        echo "--- $T ---"
        printf "  %s - worked days: %s\n" $(grep "$T" "$FILEN" | sed -E 's/^[0-9]+,([A-Za-z]+).*,([0-9]+)$/\1 \2/g')
    done
}

剪切grep并排序并打印出文本文件中的值

5 个答案: