删除awk中的换行符
本文介绍了删除awk中的换行符的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我想删除一个日期函数换行字符,并将它包含空格。我节省了使用这个变量:
I am trying to remove the new line character for a date function and have it include spaces. I am saving the variables using this:
current_date=$(date "+%m/%d/ AT %y%H:%M:%S" )
我可以看到,这是正确的格式,我需要做的回声$ CURRENT_DATE。
然而,当我需要使用这个变量它不采取行动,我想它的方式。
I can see that this is the right format I need by doing a echo $current_date.However, when I need to use this variable it does not act the way I would like it.
awk '(++n==47) {print "1\nstring \nblah '$current_date' blah 2; n=0} (/blah/) {n=0} {print}' input file > output file
我需要的日期停留在文本的当前行,没有换行继续,除非指定。
I need the date to stay in the current line of text and continue with no newline unless specified.
在此先感谢。
推荐答案
,而不是试图将变量插入到你正在做的命令字符串,你可以通过它来awk的是这样的:
Rather than attempting to insert the variable into the command string as you are doing, you can pass it to awk like this:
awk -v date="$(date "+%m/%d/ AT %y%H:%M:%S")" '# your awk one-liner here' input_file
然后换行符,您可以使用变量日期作为脚本中一个awk变量:
You can then use the variable date as an awk variable within the script:
print "1\nstring \nblah " date " blah 2";
顺便说一句,它看起来像你原来的打印语句被打破了换行符,因为有双引号,从它的结束丢失。
As an aside, it looks like your original print statement was broken, as there were double quotes missing from the end of it.
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