使用printf格式化输出:截断或填充 [英] Formatting output with printf: truncating or padding

问题描述

我想产生以下输出:

> Avril Stewart  99  54
> Sally Kinghorn 170 60
> John Young     195 120
> Yutte Schim... 250 40

如您所见，少于14个字符的名称用空格填充.长度超过15个字符的名称将被截断:"Yutte Schimmelpenninck"将截断为"Yutte Schim ...".

As you can see, names shorter than 14 characters are padded with spaces. Names longer than 15 characters are truncated: 'Yutte Schimmelpenninck' truncates to 'Yutte Schim...'.

这是我尝试实现的目标(变量$name，$height和$weight是从文件中提取的，并且循环在每个文件数据上运行printf命令):

Here is what I have tried to achieve this (the variables $name, $height, and $weight are extracted from files, and a loop runs the printf command on each file data):

printf '%-14s -%3s -%3s\n' "$name" "$height" "$weight"
> Avril Stewart  99  54
> Sally Kinghorn 170 60
> John Young     195 120
> Yutte Schimmelpenninck 250 40

printf单线是理想的解决方案.

A printf one-liner is the desired solution.

什么代码将产生第一块输出?

What code will produce the first block of output?

推荐答案

在printf命令之前，您要检查名称是否长于14个字符，如果是，则截断该名称并将其后三个字符替换为点.这个命令可以做到:

Before your printf command, you want to check if a name is longer than 14 characters, and if yes, truncate it and replace the last three characters with dots. This command does that:

(( ${#name} > 14 )) && name="${name:0:11}..."

它将name替换为前11个字符，并附加....

It replaces name with its first eleven characters and appends ....

您还必须修复printf格式字符串:而不是

You also have to fix the printf format string: instead of

'%-14s -%3s -%3s\n'

必须是

'%-14s %-3s -%-3s\n'

或者您得到诸如

Avril Stewart  - 99 - 54

不过，也许这只是一个错字，因为示例输出中没有连字符.

Maybe that was just a typo, though, as your example output didn't have the hyphens.

总共:

$ name='Avril Stewart'; weight=99; height=54
$ (( ${#name} > 14 )) && name="${name:0:11}..."
$ printf '%-14s %-3s %-3s\n' "$name" $weight $height
Avril Stewart  99  54
$ name='Sally Kinghorn'; weight=170; height=60
$ (( ${#name} > 14 )) && name="${name:0:11}..."
$ printf '%-14s %-3s %-3s\n' "$name" $weight $height
Sally Kinghorn 170 60
$ name='Yutte Schimmelpeninck'; weight=250; height=40
$ (( ${#name} > 14 )) && name="${name:0:11}..."
$ printf '%-14s %-3s %-3s\n' "$name" $weight $height
Yutte Schim... 250 40

因此，如果您从文件中读取此内容(例如，逗号分隔)，则会遇到如下所示的循环:

So if you read this from a file, for example comma separated, you'd end up with a loop like this:

while IFS=, read -r name weight height; do
    (( ${#name} > 14 )) && name="${name:0:11}..."
    printf '%-14s %-3s %-3s\n' "$name" $weight $height
done < inputFile

导致

Avril Stewart  99  54
Sally Kinghorn 170 60
John Young     195 120
Yutte Schim... 250 40

---

我认为不可能一口气.我尝试了三元运算符，并尝试了类似的方法

---

I don't think it's possible in a one-liner. I experimented with the ternary operator and tried something like

printf '%s\n' $(( ${#name} > 14 ? "${name:0:11}..." : "$name" ))

但是这里的问题是它仅适用于整数，并且字符串在算术上下文中扩展为零.

but the problem here is that it only works for integers, and strings expand to zero in arithmetic context.

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