bash 带有 wc -l 的 Shell 脚本，如果语句不起作用
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Shell script with wc -l, if statement ain't working
提问by user1204032
The problem is with this code:
问题在于这段代码：
words=`wc -l /home/tmp/logged.log | awk '{print }'`;
if [ $words == 26 ]
then
echo $words
echo Good
else
echo Not so good
fi
it always returns the else statement. Even tho the result is 26. I also tried
它总是返回 else 语句。即使结果是 26。我也试过
words=`wc -l < /home/jonathan/tmp/logged.log`;
采纳答案by triclosan
try to use [ $words -eq 26 ]instead of [ $words == 26 ]
尝试使用[ $words -eq 26 ]而不是[ $words == 26 ]
or [ 26 == 26 ]to check that statement works properly
或[ 26 == 26 ]检查该语句是否正常工作
回答by Shiplu Mokaddim
Because ==is not valid. Use =
因为==无效。用=
if [ $words = 26 ]
By the way you can use cutinstead of awk.
顺便说一句，您可以使用cut代替awk.
wc -l /home/tmp/logged.log | cut -f1 -d" "
回答by rkosegi
Try this:
尝试这个：
words=`wc -l /home/tmp/logged.log | awk '{print }'`;
if test $words -eq 26; then
echo $words
echo Good
else
echo Not so good
fi
