linux脚本查找文件名中的特定单词

Question

I need help writing a script to do the following stated below in part a.

The following code will output all of the words found in $filename, each word on a separate line.

for word in “cat $filename”
do
    echo $word
done 

a. Write a new script which receives two parameters. The first is a file's name ($1 instead of $filename) and the second is a word you want to search for ($2). Inside the for loop, instead of echo $word, use an if statement to compare $2 to $word. If they are equal, add one to a variable called COUNT. Before the for loop, initialize COUNT to 0 and after the for loop, output a message that tells the user how many times $2 appeared in $1. That is, output $COUNT, $2 and $1 in an echo statement but make sure you have some literal words in here so that the output actually makes sense to the user. HINTS: to compare two strings, use the notation [ $string1 == $string2 ]. To add one to a variable, use the notation X=$((X+1)). If every instruction is on a separate line, you do not need any semicolons. Test your script on /etc/fstab with the word defaults (7 occurrences should be found)

This is what I got so far, but it does not work right. It says it finds 0 occurrences of the word "defaults" in /etc/fstab. I am sure my code is wrong but can't figure out the problem. Help is appreciated.

count=0
echo “what word do you want to search for?: “
read two
for word in “cat $1”
do
    if [ “$two” == “$word” ]; then
        count=$((count+1))
    fi
done
echo $two appeared $count times in $1

Answer 1

You need to use command substitution, you were looping over this string: cat first_parameter .

for word in $(cat "$1")

Better way to do this using grep , paraphrasing How do I count the number of occurrences of a word in a text file with the command line?

grep -o "\<$two\>" "$1" | wc -l