Question

看起来并不那么简单。至少对于我来说。

我需要在printf文本中有一个变量。 它类似于：

FOO="User data"
+++++++++++++++++++ $FOO +++++++++++++++++++++

输出

+++++++++++++++++++ User Data +++++++++++++++++++++

但是

FOO="Fooooooo barrrr"
+++++++++++++++++++ $FOO +++++++++++++++++++++

应输出

++++++++++++++++ Fooooooo barrrr ++++++++++++++++++

和

FOO="Foooooooooooooooooooo barrrrr"
+++++++++++++++++++ $FOO +++++++++++++++++++++

应该是

+++++++++ Foooooooooooooooooooo barrrrr +++++++++++

正如您所看到的，我需要一个变量位于n长度线的中间，由+标记包围。如何使用printf和其他默认可用命令实现？

（Debian 8）

Answer 1

foo="stackoverflow.com"

输出：

+++++++++++++++++++ User data ++++++++++++++++++++

使用System.register("foo", [], function(exports_1) { "use strict"; var App; return { setters:[], execute: function() { App = (function () { function App() { this.bar = 'Hello world.'; console.log(this.bar); } return App; })(); exports_1("App", App); ; } } }); System.register("bar", ["foo"], function(exports_1) { "use strict"; var App; return { setters:[], execute: function() { App = (function () { function App() { this.bar = 'Mony a mickle maks a muckle.'; console.log(this.bar); } return App; })(); exports_1("App", App); ; } } });：

+++++++++++++++ stackoverflow.com ++++++++++++++++

Answer 2

#!/usr/bin/env bash
linelen=100
char="+"
text=$1
len=$(echo -n $text | wc -m)
fillerlen=$((($linelen - $len - 2) / 2))
filler=$(printf "$char%.0s" $(seq 1 $fillerlen))

echo $filler $text $filler

Answer 3

在printf的格式字符串中，您可以使用%${p}s指定字符串的“精度”，其中$p是精度。您可以通过打印所需的次数（扩展到空格）然后将空格转换为“+”来利用它：

$ p=10
$ printf "%${p}s\n" | tr ' ' +
++++++++++

这个函数取你的线的长度和你想放在它中心的字符串，然后用加号打印它：

pad () {
    len=$1
    string=$2

    # ${#string} expands to the length of $string
    n_pad=$(( (len - ${#string} - 2) / 2 ))

    printf "%${n_pad}s" | tr ' ' +
    printf ' %s ' "$string"
    printf "%${n_pad}s\n" | tr ' ' +
}

像这样工作：

$ pad 50 Test
++++++++++++++++++++++ Test ++++++++++++++++++++++
$ pad 50 "A longer string to be padded"
++++++++++ A longer string to be padded ++++++++++

注意你必须引用由多个单词组成的字符串，或者只使用第一个单词。

如果您的线的长度不能被2整除，则填充将向下舍入，但始终是对称的。

Answer 4

试试这个：

#!/bin/bash
n=50; # You can change the value of n as you please.
var="fooo baar";
size=${#var}
n=$(( n - size ))
n=$(( n / 2 ))
s=$(printf "%-${n}s" "*")
echo "${s// /*} "$var" ${s// /*}" #white-spaces included here.

printf bash - 在标记包围的现有行的中间打印文本

4 个答案: