bash 脚本上的回车 \r
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Carriage return \r on bash script
提问by James Wise
I have this bash script on my CentOS 5.3.
我的 CentOS 5.3 上有这个 bash 脚本。
#!/bin/bash
RESULT=$(cat /tmp/logfile.log | grep -i "Backlog File Count" | awk '{print }')
if [ "${RESULT}" -lt "5" ];
then
echo "${RESULT} is less than 5"
else
echo "${RESULT} is greater than 5"
fi
/tmp/logfile.log:
/tmp/logfile.log：
Backlog File Names (first 1 files)
This bash script should supposedly to get the value "1" on the log file and print the output. However, when I run the script, this is the error message:
这个 bash 脚本应该应该在日志文件中获取值“1”并打印输出。但是，当我运行脚本时，这是错误消息：
: integer expression expected
So when I set the debug mode, I found the "RESULT" variable output:
所以当我设置调试模式时，我发现了“RESULT”变量输出：
+ RESULT=$'1\r' ..... + '[' $'1\r' -lt 5 ']'
I've noticed that "\r" output is attached on the value.
我注意到 "\r" 输出附加在值上。
I would appreciate if any one could lead me why there is "\r" on that output and how to get rid of that error. I tried on CentOS 6.3 and there is no issue.
如果有人能引导我为什么该输出上有“\ r”以及如何摆脱该错误，我将不胜感激。我在 CentOS 6.3 上试过，没有问题。
# rpm -qa bash bash-3.2-32.el5_9.1
Thank you. James
谢谢你。詹姆士
回答by devnull
Your input file contains CR+LF line endings. Strip the CR before reading the variable. Say:
您的输入文件包含 CR+LF 行尾。在读取变量之前剥离 CR。说：
RESULT=$(tr -d '\r' < /tmp/logfile.log | grep -i "Backlog File Count" | awk '{print }')
Alternatively, you could remove the carriage returns from the input file by using dos2unixor any other utility.
或者，您可以使用dos2unix或任何其他实用程序从输入文件中删除回车符。
Moreover, saying:
此外，说：
cat file | grep foobar
is equivalent to saying:
相当于说：
grep foobar file
and avoids the Useless Use of Cat.
