bash 逐行遍历文件时跳过空行
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原文地址: http://stackoverflow.com/questions/22080937/
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Skip blank lines when iterating through file line by line
提问by MAXGEN
I am iterating through a file line by line and put each word into a array and that works. But it also picks up blank lines and puts it as an item in the array, how can I skip the blank lines?
我正在逐行遍历一个文件，并将每个单词放入一个数组中，这样就可以了。但它也会拾取空行并将其作为数组中的一个项目，如何跳过空行？
example file
示例文件
Line 1
line 2
line 3
line 4
line 5
line 6
My code
我的代码
while read line ; do
myarray[$index]="$line"
index=$(($index+1))
done < $inputfile
Possible psuedo code
可能的伪代码
while read line ; do
if (line != space);then
myarray[$index]="$line"
fi
index=$(($index+1))
done < $inputfile
采纳答案by SzG
Remove the blank lines first with sed.
先用 删除空行sed。
for word in `sed '/^$/d' $inputfile`; do
myarray[$index]="$word"
index=$(($index+1))
done
回答by Bohdan
Be more elegant:
更优雅：
echo "\na\nb\n\nc" | grep -v "^$"
cat $file | grep -v "^$" | next transformations...
回答by isedev
Implement the same test as in your pseudo-code:
实现与伪代码相同的测试：
while read line; do
if [ ! -z "$line" ]; then
myarray[$index]="$line"
index=$(($index+1))
fi
done < $inputfile
The -ztest means true if empty. !negates (i.e. true if not empty).
该-z检测手段true if empty。!否定（即如果不为空则为真）。
You can also use expressions like [ "x$line" = x ]or test "x$line" = xto test if the line is empty.
您还可以使用像[ "x$line" = x ]或test "x$line" = x这样的表达式来测试该行是否为空。
However, any line which contains whitespace will notbe considered empty. If that is a problem, you can use sedto remove such lines from the input (including empty lines), before they are passed to the whileloop, as in:
但是，任何包含空格的行都不会被视为空行。如果这是一个问题，您可以使用sed从输入中删除这些行（包括空行），然后再将它们传递给while循环，如下所示：
sed '/^[ \t]*$/d' $inputfile | while read line; do
myarray[$index]="$line"
index=$(($index+1))
done
回答by Kostas Andrianos
cat -b -s file |grep -v '^$'
cat -b -s file |grep -v '^$'
I know it's solved but, I needed to output numbered lines while ignoring empty lines, so I thought of putting it right here in case someone needs it. :)
我知道它已经解决了，但是，我需要在忽略空行的同时输出编号的行，所以我想把它放在这里以防有人需要它。:)
回答by NFTX
Use grep to remove the blank lines:
使用 grep 删除空行：
for word in $(cat ${inputfile} | grep -v "^$"); do
myarray[$index]="${word}"
index=$(($index+1))
done
回答by Galik
This version is very fast compared to solutions that invoke external commands like sedand grep. Also it skips lines that contain only spaces, the lines don't need to be empty to be skipped.
与调用sed和等外部命令的解决方案相比，此版本非常快grep。此外，它会跳过仅包含空格的行，这些行不需要为空即可被跳过。
#!/bin/bash
myarray=()
while read line
do
if [[ "$line" =~ [^[:space:]] ]]; then
myarray+=("${line}")
fi
done < test.txt
for((i = 0; i < ${#myarray[@]}; ++i))
do
echo ${myarray[$i]}
done
