Linux 在用户选择的文件中查找单词的 Shell 脚本
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Shell script that finds a word within a file selected by a user
提问by user2957449
I am training to do shell scripting as a hobby, I was stumbled on a task I was given by my tutor.
我正在接受作为业余爱好的 shell 脚本训练，我偶然发现了我的导师给我的任务。
The task is to make a shell script that you can enter a file name you want to search in and then it will respond if it exists or not; then if it exists you have another option to find a certain word in the file that exist in there the certain word must be shown.
任务是制作一个shell脚本，您可以输入您要搜索的文件名，然后它会响应是否存在；然后，如果它存在，您还有另一个选项可以在文件中找到某个单词，该单词必须显示在那里。
Here is what I've done so far. My tutor only gave me a hint it has something to do with grep??
这是我到目前为止所做的。我的导师只给了我一个暗示它与grep有关？？
#!/bin/bash
echo "search the word you want to find"
read strfile
echo "Enter the file you wish to search in"
grep $strfile
"strword" strfile
Here's the start of my improved work.
这是我改进工作的开始。
#!/bin/bash
printf "Enter a filename:
"
read str
if [[ -f "$str" ]]; then
echo "The file '$str' exists."
else
echo "The file '$str' does not exists"
It seems the file isnt then asking for the word I want to find after searching for a file name.
在搜索文件名后，该文件似乎并没有询问我想找到的单词。
What am I doing wrong?
我究竟做错了什么？
!/bin/bash
!/bin/bash
read -p "enter a filename:" filename
read -p "输入文件名：" 文件名
if [[ -f $ filename ]] ;
如果 [[ -f $ 文件名 ]] ;
echo " the filename exists " then
回声“文件名存在”然后
read -p "enter the word you want to find. : word
read -p "输入要查找的单词。: word
[grep -c $word $filename
[grep -c $word $文件名
else echo "the file $str doesn't exist." fi
否则回显“文件 $str 不存在。” 菲
回答by damienfrancois
One solution:
一种解决方案：
#!/bin/bash
read -p "Enter a filename: " filename
if [[ -f $filename ]] ; then
echo "The file $filename exists."
read -p "Enter the word you want to find: " word
grep "$word" "$filename"
else
echo "The file $filename does not exist."
fi
Quite a few variants possible.
相当多的变体是可能的。
回答by Darth Hunterix
You can do the word counting part by:
您可以通过以下方式进行字数统计：
exits=$(grep -c $word $file)
if [[ $exists -gt 0 ]]; then
echo "Word found"
fi
That's what you're missing, the rest of your script is ok.
这就是你所缺少的，你的脚本的其余部分没问题。
"grep -c" counts lines which contain $word, so a file:
"grep -c" 计算包含 $word 的行，因此是一个文件：
word word other word
word
nothing
will produce value "2". Putting grep in $() let's you store the result in a variable. I think the rest is self-explanatory, especially, that you already have it in you post :)
将产生值“2”。将 grep 放入 $() 可以让您将结果存储在变量中。我认为其余的都是不言自明的，尤其是你已经在你的帖子中:)
回答by Ranjithkumar T
Try,
尝试，
# cat find.sh
#!/bin/bash
echo -e "Enter the file name:"
read fi
echo -e "Enter the full path:"
read pa
se=$(find "$pa" -type f -name "$fi")
co=$(cat $se | wc -l)
if [ $co -eq 0 ]
then
echo "File not found on current path"
else
echo "Total file found: $co"
echo "File(s) List:"
echo "$se"
echo -e "Enter the word which you want to search:"
read wa
sea=$(grep -rHn "$wa" $se)
if [ $? -ne 0 ]
then
echo "Word not found"
else
echo "File:Line:Word"
echo "$sea"
fi
fi
output:
输出：
# ./find.sh
Enter the file name:
best
Enter the full path:
.
Total file(s) found: 1
File(s) List:
./best
Enter the word which you want to search:
root
File:Line:Word
./best:1:root
# ./find.sh
Enter the file name:
besst
Enter the full path:
.
File not found on current path
