Question

我有以下bash stript打印出project.json文件路径：

projectDirectories=./src/*/
projectJsonFiles=${projectDirectories%%/}/project.json
for projectFilePath in $projectJsonFiles; do echo $projectFilePath; done

给了我这个输出：

./src/foo-aspnet-configuration/project.json
./src/foo.Common/project.json
./src/foo.bar.MongoDB/project.json
./src/foo.bar/project.json
./src/foo.bar.Queries/project.json
./src/foo.Graph/project.json
./src/foo-http-frontend/project.json
./src/foo.Http.Model/project.json
./src/foo.Infrastructure/project.json
./src/foo.Search/project.json

我还想做的是列出./workers/*/路径下的project.json文件，但我不知道如何组合它。我的最终目标是能够将两个文件整合在一起，例如：

projectDirectories=./src/*/,./workers/*/
projectJsonFiles=${projectDirectories%%/}/project.json
for projectFilePath in $projectJsonFiles; do echo $projectFilePath; done

会给我：

./src/foo-aspnet-configuration/project.json
./src/foo.Common/project.json
./src/foo.bar.MongoDB/project.json
./src/foo.bar/project.json
./src/foo.bar.Queries/project.json
./src/foo.Graph/project.json
./src/foo-http-frontend/project.json
./src/foo.Http.Model/project.json
./src/foo.Infrastructure/project.json
./src/foo.Search/project.json
./workers/foo.Sync.Common/project.json
./workers/foo-sync-foobar/project.json
./workers/foo-sync-bar/project.json

但显然，projectDirectories=./src/*/,./workers/*/不起作用。知道如何在bash中完成这项工作吗？

Answer 1

使用GNU bash 4：

printf -v projectDirectories "%s" ./src/*/ ./workers/*/
echo "$projectDirectories"

Answer 2

您应该使用数组而不是依赖于不包含空格的文件路径：

dirs=(./src/*/ ./workers/*/)
for dir in "${dirs[@]}"; do
  file=${dir}project.json; do
    echo "$file"
  done
done

在bash中连接两个文件globbings

2 个答案: