Linux AWK 去除空行
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AWK remove blank lines
提问by general exception
The /./is removing blank lines for the first condition { print "a"$0 }only, how would I ensure the script removes blank lines for every condition ?
该/./被删除空行的第一个条件{ print "a"$0 }而已，我怎么会确保每个条件的脚本删除空行？
awk -F, '/./ { print "a"awk -F, '
/./ {
print "a"/^$/ {next}
;
if (NR!=1) { print "b"awk 'NF > 0' filename
}
print "c"awk NF file
}
END { print "d"awk '!/^$/' file
}
' MyFile
} NR!=1 { print "b"##代码## } { print "c"##代码## } END { print "d"##代码## }' MyFile
采纳答案by Birei
Put following conditions inside the first one, and check them with ifstatements, like this:
将以下条件放在第一个条件中，并使用if语句检查它们，如下所示：
回答by glenn Hymanman
if you want to ignore all blank lines, put this at the beginning of the script
如果要忽略所有空行，请将其放在脚本的开头
##代码##回答by Aaykay
Awk command to remove blank lines from a file:
从文件中删除空行的 awk 命令：
##代码##回答by oliv
A shorter form of the already proposed answer could be the following:
已经提出的答案的较短形式可能如下：
##代码##Any awkscript follows the syntax condition {statement}. If the statement block is not present, awkwill print the whole record (line) in case the conditionis not zero.
任何awk脚本都遵循语法condition {statement}。如果语句块不存在，awk则在condition不为零的情况下打印整个记录（行）。
NFvariable in awkholds the number of fields in the line. So when the line is non empty, NFholds a positive value which trigger the default awkaction (print the whole line). In case of empty line, NFis zero and the condition is not met, so awkdoes nothing.
NF变量 inawk保存行中的字段数。因此，当该行非空时，NF持有一个正值，触发默认awk操作（打印整行）。在空行的情况下，NF为零且不满足条件，因此awk什么也不做。
Note that you don't even need quote because this 2 letters awk script doesn't contain any space or character that could be interpreted by the shell.
请注意，您甚至不需要引号，因为这个 2 个字母的 awk 脚本不包含任何可以被 shell 解释的空格或字符。
or
或者
##代码##^$is the regex for an empty line. The 2 /is needed to let awkunderstand the string is a regex. !is the standard negation.
^$是空行的正则表达式。/需要2才能awk理解字符串是正则表达式。!是标准否定。
