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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