Question

我正在尝试编写一个bash命令来查找字符串是否包含一个接一个地重复相同的字母数字字符。

例如：

input: aaa, abc, aba, abc111;

output: aaa, abc111.

“aba”不被视为一个接一个地重复相同的字母数字字符。 “11”，“111”，“aa”，“aaaaaaaaa”等被认为是“重复”。不要担心输入。假设已被定义为其他地方的字符串。

说：INPUT =“aa”

Answer 1

使用tr可以提供帮助。（使用-s选项挤压定义的集合中的重复。）

示例：

$ a="abcdef"
$ [ $(echo ${a} | tr -s '[:alnum:]') == "${a}" ] && echo "No repeating characters" || echo "Repeating characters present"
No repeating characters
$ a="abc1111"
$ [ $(echo ${a} | tr -s '[:alnum:]') == "${a}" ] && echo "No repeating characters" || echo "Repeating characters present"
Repeating characters present
$ a="abc123"
$ [ $(echo ${a} | tr -s '[:alnum:]') == "${a}" ] && echo "No repeating characters" || echo "Repeating characters present"
No repeating characters
$ a="aaa123"
$ [ $(echo ${a} | tr -s '[:alnum:]') == "${a}" ] && echo "No repeating characters" || echo "Repeating characters present"
Repeating characters present
$ a="abacdc"
$ [ $(echo ${a} | tr -s '[:alnum:]') == "${a}" ] && echo "No repeating characters" || echo "Repeating characters present"
No repeating characters

Answer 2

实际上正则表达式与您的目标匹配：

'([0-9a-zA-Z])\1'

如果你想打印这些单词，试试这个：

grep -oE '\b\w*([0-9a-zA-Z])\1\w*\b'

用你的例子测试：

kent$  echo "aaa, abc, aba, abc111"|grep -oE '\b\w*([0-9a-zA-Z])\1\w*\b'
aaa
abc111

Answer 3

如果您在数组中有一组单词，则此代码将起作用：

#!/bin/bash

WORDS=(aaa abc aba abc111)

for W in "${WORDS[@]}"; do
    P=''
    for (( I = 0; I < ${#W}; ++I )); do
        C=${W:I:1}
        [[ $C == "$P" ]] && {
            echo "$W"
            break
        }
        P=$C
    done
done

输出：

aaa
abc111

使用bash查找相同字符（字母数字）的重复

3 个答案: