Linux 在bash追加换行符中连接两个字符串变量
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Concatenating two string variables in bash appending newline
提问by Aman Deep Gautam
I have a variable final_listwhich is appended by a variable urlin a loop as:
我有一个变量final_list，它由url循环中的变量附加为：
while read url; do
final_list="$final_list"$'\n'"$url"
done < file.txt
To my surprise the \nis appended as an space, so the result is:
令我惊讶的\n是，它被附加为一个空格，所以结果是：
url1 url2 url3
while I wanted:
虽然我想要：
url1
url2
url3
What is wrong?
怎么了？
采纳答案by anubhava
New lines are very much there in the variable "$final_list". echoit like this with double quotes:
变量中有很多新行"$final_list"。echo它像这样带双引号：
echo "$final_list"
url1
url2
url3
OR better use printf:
或更好地使用printf：
printf "%s\n" "$final_list"
url1
url2
url3
回答by Emo Mosley
It may depend on how you're trying to display the final result. Try outputting the resulting variable within double-quotes:
这可能取决于您尝试显示最终结果的方式。尝试在双引号内输出结果变量：
echo "$final_list"
