bash读取循环仅读取输入变量的第一行 [英] bash read loop only reading first line of input variable

问题描述

我有一个读取循环，该循环正在读取变量，但未达到我期望的方式。我想读取变量的每一行并处理每一行。这是我的循环：

I have a read loop that is reading a variable but not behaving the way I expect. I want to read every line of my variable and process each one. Here is my loop:

while read -r line
do
   echo $line | sed 's/<\/td>/<\/td>$/g' | cut -d'$' -f2,3,4 >> file.txt
done <<< "$TABLE"

我希望它能够处理文件的每一行，但它只会执行第一行一。如果我的中间只是回声$ line >> file.txt，它将按预期工作。这里发生了什么？如何获得所需的行为？

I expect it to process every line of the file but instead it just does the first one. If my the middle is simply echo $line >> file.txt it works as expected. What's going on here? How do I get the behavior I want?

推荐答案

似乎您的行由 \r分隔而不是 \n 。

It seems your lines are delimited by \r instead of \n.

使用此while循环使用 read -d $'\r'来迭代输入：

Use this while loop to iterate the input with use of read -d $'\r':

while read -rd $'\r' line; do
    echo "$line" | sed 's~</td>~</td>$~g' | cut -d'$' -f2,3,4 >> file.txt
done <<< "$TABLE"

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