finding contents of one file into another file in

问题:

I'm using following shell script to find contents of one file into another:

#!/bin/ksh
file="/home/nimish/contents.txt"

while read -r line; do
    grep $line /home/nimish/another_file.csv
done < "$file"

I'm executing the script but it is not displaying the contents from the csv file. My contents.txt file contains number such as "08915673" or "123223" which are present in the csv file as well. Is there anything wrong am I doing?

回答1:

grep itself is able to do so. Simply use the flag -f:

grep -f <patterns> <file>

<patterns> is a file containing one pattern in each line; and <file> is the file in which you want to search things.

Note that, to force grep to consider each line a pattern, even if the contents of each line look like a regular expression, you should use the flag -F, --fixed-strings.

grep -F -f <patterns> <file>

If your file is a CSV, as you said, you may do:

grep -f <(tr ',' '\n' < data.csv) <file>

---

As an example, consider the file "a.txt", with the following lines:

alpha
0891234
beta

Now, the file "b.txt", with the lines:

Alpha
0808080
0891234
bEtA

The output of the following command is:

grep -f "a.txt" "b.txt"
0891234

You don't need at all to for-loop here; grep itself offers this feature.

---

Now using your file names:

#!/bin/bash
patterns="/home/nimish/contents.txt"
search="/home/nimish/another_file.csv"
grep -f <(tr ',' '\n' < "${patterns}") "${search}"

You may change ',' to the separator you have in your file.

回答2:

Another solution:

• use awk, create your own hash(e.g. ahash), all control by yourself.
• replace $0 to $i, you can match any fields you want.

---

awk -F"," '
{  
   if (nowfile==""){ nowfile = FILENAME;  }

   if(FILENAME == nowfile)
   {
     hash[$0]=$0;
   }
   else
   {
       if($0 ~ hash[$0])
       {  
           print $0
       }
   }
} '  xx yy

回答3:

I don't think you really need a script to perform what you're trying to do.

One command is enough. In my case, in needed an identification number in column 11 in a csv file (with ";" as separator)

grep -f <(awk -F";" '{print $11}' FILE_TO_EXTRACT_PATTERNS_FROM.csv) TARGET_FILE.csv 

I hope this helps.