linux - 创建 aspell 样式文件编辑 bash 脚本

Question

我正在创建一个在 Linux 上运行的命令行 lint 工具。

我的输出目前是这样的:

./ex4/task6.7/SumOfCubedDigits.java
> Line 15 has inconsistent indenting
> Line 16 has inconsistent indenting
./ex2/task3.2/YearsBeforeRetirement.java
> Line 0 has a curly brace on the end
./ex2/task3.4/YearsBeforeRetirement.java
> Line 0 has a curly brace on the end
./ex2/task3.7/ThreeWeights.java
> Line 18 has inconsistent indenting
> Line 29 has inconsistent indenting
./ex2/task3.7/fourWeightsCoffeeTime/FourWeights.java
> Line 9 has inconsistent indenting
> Line 11 has inconsistent indenting
./ex2/task2.9/Limerick.java
> Line 0 has a curly brace on the end

通过将输出传送到 awk '/.\/ex/{print;}' 我可以只提取文件名:

./ex4/task6.7/SumOfCubedDigits.java
./ex2/task3.2/YearsBeforeRetirement.java
./ex2/task3.4/YearsBeforeRetirement.java
./ex2/task3.7/ThreeWeights.java
./ex2/task3.7/fourWeightsCoffeeTime/FourWeights.java
./ex2/task2.9/Limerick.java

我想依次打开这些文件中的每一个并编辑它们，也许在我打开它们时向用户发送一条消息，告知每个文件中的错误。类似于 aspell 的作用。

这可能吗？

Answer 1

与 dialog和 awk :

演示:

代码:

file=/path/to/file.txt

trap '\rm -f /tmp/out_file' 0 1 2 3 15

if dialog \
    --clear \
    --title "Pick up one of these files" \
    --menu "Files/errors" 80 300 100 $(
        awk '/>/{
            $1=""
            gsub(/ +/, "_", $0)
            arr[k]=arr[k] $0
            next
        }
        {k=$0}
        END{for (a in arr) printf "%s ", a " " arr[a]}
    ' "$file") 2>/tmp/out_file
then
    $EDITOR "$(</tmp/out_file)"
fi