bash 如何用文本文件中的命令输出替换值？
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How to replace a value with the output of a command in a text file?
提问by Benjamin
I have a file that contains:
我有一个包含以下内容的文件：
<?php return 0;
I want to replace in bash, the value 0by the current timestamp.
我想用 bash 替换0当前时间戳的值。
I know I can get the current timestamp with:
我知道我可以通过以下方式获取当前时间戳：
date +%s
And I can replace strings with sed:
我可以用以下内容替换字符串sed：
sed 's/old/new/g' input.txt > output.txt
But how to combine the two to achieve what I want? Solutions not involving sedand dateare welcome as well, as long as they only use shell tools.
但是如何将两者结合起来实现我想要的呢？解决方案不涉及sed，并date欢迎为好，只要他们只使用shell工具。
回答by fedorqui 'SO stop harming'
In general, do use this syntax:
通常，请使用以下语法：
sed "s/<expression>/$(command)/" file
This will look for <expression>and replace it with the output of command.
这将查找<expression>并替换为command.
For your specific problem, you can use the following:
对于您的具体问题，您可以使用以下方法：
sed "s/0/$(date +%s)/g" input.txt > output.txt
This replaces any 0present in the file with the output of the command date +%s. Note you need to use double quotes to make the command in $()be interpreted. Otherwise, you would get a literal $(date +%s).
这将0用命令的输出替换文件中的任何内容date +%s。请注意，您需要使用双引号来$()解释命令。否则，你会得到一个文字$(date +%s).
If you want the file to be updated automatically, add -ito the sed command: sed -i "s/.... This is called in-place editing.
如果您希望文件自动更新，请添加-i到 sed 命令：sed -i "s/.... 这称为就地编辑。
Test
测试
Given a file with this content:
给定一个包含此内容的文件：
<?php return 0;
Let's see what it returns:
让我们看看它返回什么：
$ sed "s/0/$(date +%s)/g" file
<?php return 1372175125;
回答by Walter A
When the replacement string has newlines and spaces, you can use something else.
We will try to insert the output of ls -lin the middle of some template file.
当替换字符串有换行符和空格时，您可以使用其他内容。我们将尝试ls -l在一些模板文件的中间插入输出。
awk 'NR==FNR {a[NR]=sed '/^Insert after this$/r'<(ls -l) template.file
;next}
{print}
/Insert index here/ {for (i=1; i <= length(a); i++) { print a[i] }}'
<(ls -l) template.file
or
或者
##代码##