xargs wget 使用参数从 URL 中提取文件名

Question

我想进行并行下载，但问题是 wget 输出的文件名不正确。

url.txt

http://example.com/file1.zip?arg=tereef&arg2=okook
http://example.com/file2.zip?arg=tereef&arg2=okook

命令

xargs -P 4 -n 1 wget <url.txt

输出文件名

file1.zip?arg=tereef&arg2=okook
file2.zip?arg=tereef&arg2=okook

预期输出

file1.zip
file2.zip

我是 bash 的新手，请建议我如何输出正确的文件名，请不要建议 for 循环或 & 因为它会阻塞。

谢谢

Answer 1

处理您的输入以生成所需的命令，然后运行通过 xargs。

perl -ne - 遍历输入文件的行并执行内联程序

-e : Execute perl one-liner

-n : Loop over all input lines, assigning each to $_ in turn.

xargs -P 4 -n 1 -i -t wget "{}"

-P 4 : Max of 4 Processes at a time

-n 1 : Consume one input line at a time

-i : Use the replace string "{}"

-t : Print the command before executing it

perl -ne '
    chomp(my ($url) = $_);                         # Remove trailing newline
    my ($name) = $url =~ m|example.com/(.+)\?|;    # Grab the filename
    print "$url -O $name\n";                       # Print all of the wget params
' url.txt | xargs -P 4 -n 1 -i -t wget "{}"

输出

wget http://example.com/file1.zip?arg=tereef&arg2=okook -O file1.zip
wget http://example.com/file2.zip?arg=tereef&arg2=okook -O file2.zip
--2016-07-21 22:24:44--  http://example.com/file2.zip?arg=tereef&arg2=okook%20-O%20file2.zip
--2016-07-21 22:24:44--  http://example.com/file1.zip?arg=tereef&arg2=okook%20-O%20file1.zip
Resolving example.com (example.com)... Resolving example.com (example.com)...     93.184.216.34, 2606:2800:220:1:248:1893:25c8:1946
93.184.216.34, Connecting to example.com (example.com)|93.184.216.34|:80... 2606:2800:220:1:248:1893:25c8:1946
Connecting to example.com (example.com)|93.184.216.34|:80... connected.
connected.
HTTP request sent, awaiting response... HTTP request sent, awaiting   response... 404 Not Found
2016-07-21 22:24:44 ERROR 404: Not Found.

404 Not Found
2016-07-21 22:24:44 ERROR 404: Not Found.

Answer 2

您可以使用必须导出的 bash 函数才能在当前 shell

之外看到

function mywget()
{
    wget -O ${1%%\?*} "''"
}
export -f mywget
xargs -P 4 -n 1 -I {} bash -c "mywget '{}'" < url.txt

Answer 3

GNU Parallel 看起来像这样：

parallel -P 4 wget -O '{= s/\?.*//;s:.*/:: =}' {} <url.txt

xargs wget 使用参数从 URL 中提取文件名

xargs wget extract filename from URL with Parameter

bash

wget

xargs