用变量替换命令？

Question

#!/bin/bash
```
download()
{
    local url=
    echo -n "    "
    wget -nc --progress=dot $url 2>&1 | grep --line-buffered "%" | \
    sed -u -e "s,\.,,g" | awk '{printf("\b\b\b\b%4s", )}'
    echo -ne "\b\b\b\b"
    echo " DONE"
}
```
file="adaway.org.txt"
echo -n "Downloading $file:"
download "https://raw.githubusercontent.com/EnergizedProtection/block/master/assets/active/filter/$file"

这仍然很粗糙，但可以工作。我只是想做一些变量来缩短底部，然后让我的 bash 从 cat 文件中读取。

Answer 1

您的问题不清楚，但听起来您可能正在尝试这样做：

$ cat a.txt
http://google.com/foo
http://yahoo.com/stuff/bar

$ cat tst.sh
#!/bin/env bash

input=""
while IFS= read -r line; do
    path="${line%/*}"
    file="${line##*/}"
    printf '\nline="%s"\n' "$line"
    printf 'path="%s"\n' "$path"
    printf 'file="%s"\n' "$file"
    echo download "${path}/${file}"
done < "$input"

$ ./tst.sh a.txt

line="http://google.com/foo"
path="http://google.com"
file="foo"
download http://google.com/foo

line="http://yahoo.com/stuff/bar"
path="http://yahoo.com/stuff"
file="bar"
download http://yahoo.com/stuff/bar

如果不是，请编辑您的问题以澄清。

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