bash 如何测试shell脚本中的行是否为空？
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How can I test if line is empty in shell script?
提问by planetp
I have a shell script like this:
我有一个这样的shell脚本：
cat file | while read line
do
# run some commands using $line
done
Now I need to check if the line contains any non-whitespace character ([\n\t ]), and if not, skip it. How can I do this?
现在我需要检查该行是否包含任何非空白字符 ([\n\t ])，如果没有，请跳过它。我怎样才能做到这一点？
回答by Arkku
Since readreads whitespace-delimited fields by default, a line containing only whitespace should result in the empty string being assigned to the variable, so you should be able to skip empty lines with just:
由于read默认情况下读取以空格分隔的字段，因此仅包含空格的行应该会导致将空字符串分配给变量，因此您应该能够跳过空行，只需：
[ -z "$line" ] && continue
回答by NSC
try this
尝试这个
while read line;
do
if [ "$line" != "" ]; then
# Do something here
fi
done < $SOURCE_FILE
回答by Ignacio Vazquez-Abrams
bash:
重击：
if [[ ! $line =~ [^[:space:]] ]] ; then
continue
fi
And use done < fileinstead of cat file | while, unless you know why you'd use the latter.
并使用done < file代替cat file | while，除非您知道为什么要使用后者。
回答by Ignacio Vazquez-Abrams
if ! grep -q '[^[:space:]]' ; then
continue
fi
回答by ghostdog74
cati useless in this case if you are using while read loop. I am not sure if you meant you want to skip lines that is empty or if you want to skip lines that also contain at least a white space.
cat如果您使用 while 读取循环，我在这种情况下没用。我不确定您的意思是要跳过空行还是要跳过至少还包含空格的行。
i=0
while read -r line
do
((i++)) # or $(echo $i+1|bc) with sh
case "$line" in
"") echo "blank line at line: $i ";;
*" "*) echo "line with blanks at $i";;
*[[:blank:]]*) echo "line with blanks at $i";;
esac
done <"file"
回答by Rounak
blank=`tail -1 <file-location>`
if [ -z "$blank" ]
then
echo "end of the line is the blank line"
else
echo "their is something in last line"
fi
