以编程方式构建目录排除列表以查找命令(重击) [英] Build Directory Exclusion List Programmatically for Find Command (Bash)

问题描述

我有一个目录数组要从我的find命令的结果中排除，例如EXCLUDE=("foo" "bar").

我可以像这样从交互式终端运行它:

find . -name 'hvr.yml' -not -path "foo/*" -not -path "bar/*"

所以我试图像这样建立论点:

getServersToCheck() {

    # Build command arguments to exclude given dirs
    local exclude_command="-not -path"
    local excount=${#EXCLUDE[@]}
    for ((i=0;i<excount;i++)); do
        EXCLUDE[i]="${exclude_command} ${EXCLUDE[i]}/*"
    done

    find . -name 'hvr.yml' "${EXCLUDE[@]}"
}

但这会导致find抛出未知的谓词错误:'-not -path foo/*'

有没有办法做到这一点?当我回显该命令时，它看起来是正确的，但是必须有一些bash语法规则导致该命令无法按我期望的那样工作.

更新:

我在路径周围添加了\"以排除，因为我读到globation仅发生在带引号的字符串中. xtrace显示以下内容:

find . -name hvr.yml -not -path '"foo/*"' -not -path '"bar/*"'

单引号可能是问题

删除\"并使用xtrace运行表明，该globbing正在for循环中应用，导致:

find . -name hvr.yml -not -path "foo/fileinfoo" "foo/somethingelseinfoo" -not -path "bar/*" "bar/fileinbar" "bar/otherfilesinbar"

所以find抱怨把随机路径作为参数给出.

是否可以扩展EXCLUDE数组并将/*添加到命令中每个元素的末尾?

解决方案

找到了我尝试使用grep实现的替代解决方案:

EXCLUDE=("abc" "def")

getPaths() {
    local exclude_count=${#EXCLUDE[@]}
    if [ $exclude_count -eq 0 ]; then
        find . -name $FILENAME | tr '\n' ' '
        return $?
    fi

    # Concat excluded servers as grep regex
    local regex="(${EXCLUDE[0]}"
    for ((i=1;i<exclude_count;i++)); do
        regex="${regex}|${EXCLUDE[i]}"
    done
    regex="${regex})"

    find . -name $FILENAME | grep -Ev "${regex}" | tr '\n' ' '
    return $?
}

• 如果exclude为空，则运行常规的find命令.
• 否则，它会为grep建立一个正则表达式，以过滤掉最终看起来像(abc | def)的示例.

I have an array of directories to exlude from the result of my find command, something like EXCLUDE=("foo" "bar").

I can run this from the interactive terminal like so:

find . -name 'hvr.yml' -not -path "foo/*" -not -path "bar/*"

And so i tried to build the argument up like this:

getServersToCheck() {

    # Build command arguments to exclude given dirs
    local exclude_command="-not -path"
    local excount=${#EXCLUDE[@]}
    for ((i=0;i<excount;i++)); do
        EXCLUDE[i]="${exclude_command} ${EXCLUDE[i]}/*"
    done

    find . -name 'hvr.yml' "${EXCLUDE[@]}"
}

But this results in find throwing an unknown predicate error: '-not -path foo/*'

Is there way of achieving this? When I echo the command it looks correct, but there must be some bash syntax rules that are causing this not to work how I expect.

UPDATE:

I added \" around the path to exclude as I read that globbing only occurs with quoted strings. xtrace shows this following:

find . -name hvr.yml -not -path '"foo/*"' -not -path '"bar/*"'

The single quotes possibly being the problem

Removing the \" and running with xtrace shows that the globbing is being applied in the for loop, resulting in:

find . -name hvr.yml -not -path "foo/fileinfoo" "foo/somethingelseinfoo" -not -path "bar/*" "bar/fileinbar" "bar/otherfilesinbar"

And so find is complaining about random paths being given as arguments.

Is there way to expand the EXCLUDE array and add /* to the end of each element in the command?

解决方案

Found an alternative solution to what I was trying to achieve by using grep:

EXCLUDE=("abc" "def")

getPaths() {
    local exclude_count=${#EXCLUDE[@]}
    if [ $exclude_count -eq 0 ]; then
        find . -name $FILENAME | tr '\n' ' '
        return $?
    fi

    # Concat excluded servers as grep regex
    local regex="(${EXCLUDE[0]}"
    for ((i=1;i<exclude_count;i++)); do
        regex="${regex}|${EXCLUDE[i]}"
    done
    regex="${regex})"

    find . -name $FILENAME | grep -Ev "${regex}" | tr '\n' ' '
    return $?
}

• If exclude is empty, this runs a normal find command.
• Otherwise, it builds up a regex for grep to filter out that ends up looking like (abc|def) for this example.

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