填写占位符，在单次通过文件 [英] Fill placeholders in file in single pass

问题描述

我有一个占位符字符串的框架文本文件：

I have a skeleton text file with placeholder strings:

blah blah blah
blah $PLACEHOLDER_1$
blah
$PLACEHOLDER_2$

和等。具体的占位符的形式没关系 - 我可能会改变他们什么最舒服的具体实施

and so on. Specific "form" of placeholders does not matter -- I may change them to whatever most comfortable for specific implementation.

我有一个bash脚本，我知道占位符值，我需要生成一个新的文件，用值替换的占位符。

I have a bash script where I know values for placeholders, and I need to generate a new file, with placeholders replaced with values.

#! /bin/sh
PLACEHOLDER_1 = 'string 1'
PLACEHOLDER_2 = 'multiline 
string 
2'
# TODO: Generate file output.txt from file output.template 
#       using placeholders above.

我可以多遍与SED做到这一点，但它是不好玩。我做的不可以想使用Perl。我想用文本实用程序，只来砸自己。

I may do this in multiple passes with sed, but it is not fun. I do not want to use Perl. I want to use textutils and bash itself only.

什么是做我想做的一次通过最好的方式？

What is the best way to do what I want in a single pass?

推荐答案

您仍可以使用SED做一次通过替换。你只需要在一个命令来指定所有的替代品。

You can still use sed to do the replace in a single pass. You just need to specify all the replacements in one command.

如：

sed -i 's/PLACEHOLDER_1/string 1/g;s/PLACEHOLDER_2/string 2/g' <file>

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