Question

我想检查一行是否以标点符号（。，？！）结尾，如果没有，则添加句号。

A button is missing!
A button on their back allowed them to hug
A cat is carried by the scruff of its neck by a teenage boy?
A cattery is where cats are commercially housed

输出：

A button is missing!
A button on their back allowed them to hug.
A cat is carried by the scruff of its neck by a teenage boy?
A cattery is where cats are commercially housed.

我真的很难过，我试着点击字符串的最后一个字符 - 有什么帮助吗？

Answer 1

使用sed：

sed -i.bak 's/[^[:punct:]]$/&./' file
A button is missing!
A button on their back allowed them to hug.
A cat is carried by the scruff of its neck by a teenage boy?
A cattery is where cats are commercially housed.

Answer 2

使用awk：

awk '!/[[:punct:]]$/ && NF{$NF=$NF"."}1' file

<强>解释

我们使用!/[[:punct:]]$/正则表达式查找不以标点符号结尾的行。
使用&& NF可以让我们只抓取非空行。
当我们找到这样的行时，我们会使用awk内置变量$NF来保存最后一个字段的值。
我们通过.

$NF=$NF"."

1用于打印该行，因为它会生成true。

$ cat file
A button is missing!
A button on their back allowed them to hug
A cat is carried by the scruff of its neck by a teenage boy?
A cattery is where cats are commercially housed

$ awk '!/[[:punct:]]$/ && NF{$NF=$NF"."}1' file
A button is missing!
A button on their back allowed them to hug.
A cat is carried by the scruff of its neck by a teenage boy?
A cattery is where cats are commercially housed.

Answer 3

纯粹的bash解决方案，假设您的消息位于文件msg中：

while read
do
    r=${REPLY: -1}
    if [[ $r != "." ]] && [[ $r != "," ]] && [[ $r != "!" ]] && [[ $r != "?" ]]      
    then
        REPLY="$REPLY." 
    fi
    echo "$REPLY" 
done < msg

r被分配了$REPLY的最后一个字符。

编辑：在较新的bash上，您可以使用[[:punct:]]字符类并使用+=连接字符串。此版本也不会将.添加到空行：

while read
do
    if [[ ! -z "$REPLY" ]] && [[ $REPLY != *[[:punct:]] ]]
    then 
        REPLY+="."
    fi
    echo "$REPLY"
done < msg

Answer 4

这是使用正则表达式模式匹配的另一个仅使用bash的解决方案，请参阅注释以获取详细信息。如果您的数据在文件中，则必须使用readarray或类似技术将数据导入数组。我只是将文本粘贴到一个数组中，以便在本演示中使用：

#!/usr/bin/env bash

text=(
    "A button is missing!"
    "A button on their back allowed them to hug."
    "A cat is carried by the scruff of its neck by a teenage boy?"
    "A cattery is where cats are commercially housed,"
    "A cattery is where cats are commercially housed" # Demo sentence missing a punctuation
)

# Regex pattern
regex="[!?,.]$"

# Counter used to call the index of the array inside the for loop
counter=0

# Iterate over each sentence inside the array
for sentence in "${text[@]}"; do

    # If a sentence does not match the regex pattern
    # append a dot.
    if [[ ! "$sentence" =~ $regex ]]; then
        text[$counter]="${sentence}."
    fi

    # Increment the counter to get the next item of the array
    (( counter++ ))
done

# We are done print the results
printf "%s\n" "${text[@]}"

检查行是否以标点符号结尾;如果没有，请添加期间

4 个答案: