使用curl循环访问URL的Shell脚本 [英] Shell script using curl to loop through urls

问题描述

我一直在尝试创建一个简单的脚本，该脚本将从.txt文件中获取查询列表，附加主url变量，然后抓取内容并将其输出到文本文件.

I've been trying to create a simple script that will take a list of queries from a .txt file, append the main url variable, then scrape the content and output it to a text file.

这是我到目前为止所拥有的:

Here's what I have so far:

#!/bin/bash

url="example.com/?q="
for i in $(cat query.txt); do
    content=$(curl -o $url $i)
    echo $url $i
    echo $content >> output.txt
done

列表:

images
news
stuff
other

错误日志:

curl: (6) Could not resolve host: other; nodename nor servname provided, or not known
example.com/?q= other

如果我直接从命令行使用此命令，则会将一些输出输出到文件中:

If I use this command straight from the command line I get some output into the file:

curl -L http://example.com/?q=other >> output.txt

最终，我希望输出为:

fetched:    http://example.com/?q=other
content:    the output of the page

followed by the next query in the list.

推荐答案

使用更多引号！

• http://mywiki.wooledge.org/Quotes
• http://mywiki.wooledge.org/Arguments
• http://wiki.bash-hackers.org/syntax/words

• http://mywiki.wooledge.org/Quotes
• http://mywiki.wooledge.org/Arguments
• http://wiki.bash-hackers.org/syntax/words

尝试以下方法:

url="example.com/?q="
for i in $(cat query.txt); do
    content="$(curl -s "$url/$i")"
    echo "$content" >> output.txt
done

这篇关于使用curl循环访问URL的Shell脚本的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！