linux - 从终端创建多行文件时如何缩进？

Question

所以我需要一个命令来使输出看起来像这样:

    [core]
            repositoryformatversion = 0
            filemode = true
            bare = false
            logallrefupdates = true
    [remote "origin"]
            url = ssh://git@github.com/user/address.git
            fetch = +refs/heads/*:refs/remotes/origin/*
    [branch "master"]
            remote = origin
            merge = refs/heads/master

我试过以下，

printf "[core]\n    repositoryformatversion = 0\n   filemode = true\n   bare = false\n  logallrefupdates = true\n   logallrefupdates = true\n[remote "origin"]\n    url = ssh://git@github.com/user/address.git\n   fetch = +refs/heads/*:refs/remotes/origin/*\n[branch "master"]\n    remote = origin\n   merge = refs/heads/master" > config

但是它创建了一个包含这些内容的文件

[core]
repositoryformatversion = 0
filemode = true
bare = false
logallrefupdates = true
logallrefupdates = true
[remote origin]
url = ssh://git@github.com/user/address.git
fetch = +refs/heads/*:refs/remotes/origin/*
[branch master]
remote = origin
merge = refs/heads/master

我怎样才能让它以最初规定的格式输出？

Answer 1

根本不要为此使用 printf。只需将此处文档与 cat 一起使用:

cat <<'EOF' > config
[core]
        repositoryformatversion = 0
        filemode = true
        bare = false
        logallrefupdates = true
[remote "origin"]
        url = ssh://git@github.com/user/address.git
        fetch = +refs/heads/*:refs/remotes/origin/*
[branch "master"]
        remote = origin
        merge = refs/heads/master
EOF

或者，用内容定义一个变量

config='
[core]
        repositoryformatversion = 0
        filemode = true
        bare = false
        logallrefupdates = true
[remote "origin"]
        url = ssh://git@github.com/user/address.git
        fetch = +refs/heads/*:refs/remotes/origin/*
[branch "master"]
        remote = origin
        merge = refs/heads/master
'

printf '%s\n' "$config" > config