在 bash 上垂直翻转多个字符串

Question

我正在尝试垂直翻转任何句子的内容。因此，任何字符串的每个字符都将在同一行中垂直打印。例如：

Sample Text: This is an Example

Output expected: T i a E
                 h s n x
                 i     a
                 s     m
                       p
                       l 
                       e

按照以下方向，我正在努力实现这一目标，但目前还做不到。

    echo "Input provided by user is $@"
    for i in $(seq 1 $#); do

        echo ${!i} | sed 's/./ &/g' | xargs |tr ' ' '\n' 

        done


Current output:

T
h
i
s
i
s
a
n
E
x
a
m
p
l
e

此外，这也无济于事

echo Print text vertically | fold -c -w1
T
h
i
s

i
s

a
n

E
x
a
m
p
l
e

更多无效的替代方案：

#echo "Input provided by user is $@"

for i in $(seq 1 $#); do

content[i]=$(echo ${!i}|fold -c -w1)

   #echo ${content[i]}


    done

    echo ${content[@]}

Answer 1

max 变量保存所有单词中的最大长度。对于您的文本，它将是：length('Example')，即 7（所有单词的最大长度）

使用 awk 脚本文件：

$ awk -f script.awk <<< "This is an Example"
TiaE
hsnx
i  a
s  m
   p
   l
   e

这是脚本：

{
        max=0
        for(i=1;i<=NF;i++)
                max=length($i)>max?length($i):max;

        for(j=1;j<=max;j++)
        {
                for(i=1;i<=NF;i++)
                {
                        temp=substr($i, j, 1);
                        printf temp==""?" ":temp
                }
                printf "\n"
        }
}

Answer 2

#!/bin/bash
function abc(){
maxIteration=0;
for i in $(seq 1 $#); do
j=$(echo ${!i})
if [ $maxIteration -lt ${#j} ]
then
maxIteration=${#j};
fi
done
COUNTER=0;
while [ $COUNTER -lt $maxIteration ]; do
             for i in $(seq 1 $#); do
                j=$(echo ${!i})
                if [ ${#j} -gt $COUNTER ]
                then
                        echo ${j:$COUNTER:1} | tr '\n' ' ';
                else
                        echo " " | tr '\n' ' ';
                fi
             done
echo -e "\n"

                let COUNTER=COUNTER+1
         done
}

abc $@| grep .

Answer 3

我以前创建过一些类似的脚本。简短但完整的 POC：

#!/bin/bash
count=0
max=0
#first determine the longest string so we can later pad shorter strings with spaces
for i in $(echo "" | xargs -d: -i  echo {})
do
    size=$(echo $i | wc -c)
    if [[ $size > $max ]]
    then
    max=$size
    fi 
done

files=""
#then echo the strings vertically inside the tmp files
for i in $(echo "" | xargs -d: -i  echo {})
do
    res=$(echo $i | sed 's/./ &/g'  | xargs |tr ' ' '\n' > /tmp/$count.out)
    #and add spaces at the end
    add_space=$((max-$(echo $i | wc -c)))
    for space in $(seq 0 $add_space)
    do                   
    echo " " >> /tmp/$count.out
    done
    files=$files" $count.out"
    count=$((count+1))
done
#and finally print them side by side
pr -t -J -m -w 70 -S" " $files

我在 /tmp 下创建 tmp 文件，垂直回显字符串，然后使用 pr 打印出来。

% ./s.sh "This is an Example"
T i a E
h s n x
i     a
s     m
      p
      l
      e

Answer 4

因为 Perl 很有趣：

perl -a script.pl <<< 'This is an Example'
T i a E 
h s n x 
i     a 
s     m 
      p 
      l 
      e

和脚本：

@F = map { [/./g] } @F;
while (grep @{$_}, @F) {
  printf "%s ", shift @{$_} || ' ' for @F;
  print "\n"
}

备选脚本：

perl -pe '$r.=$/while/\S/&&s/(\S)(\S*)|\B/$r.=(||$").$";/ge}{$_=$r' \
  <<< 'This is an Example'
T i a E   
h s n x   
i     a   
s     m   
      p   
      l   
      e

在 bash 上垂直翻转多个字符串

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