计算grep结果在bash脚本中不起作用

Question

My question is not easy to ask, I try explain the problem with the following example:

/home/luther/tipical_surnames.txt

Smith
Johnson
Williams
Jones
Brown
#Davis
Miller
Wilson
#Moore
Taylor
Anderson

/home/luther/employers.txt

2000    Johnson     A lot-of details / BJC3000,6000, i550                0
2101    Smith       A lot-of details / BJC3000,6000, i550                0
2102    Smith       A lot-of details / BJC3000,6000, i550                0
2103    Jones       A lot-of details / BJC3000,6000, i550                0
2104    Johnson     A lot-of details / BJC3000,6000, i550                0
2100    Smith       A lot-of details / BJC3000,6000, i550                0

I have a list with the favorite surnames and another with the name of employers. Let's check how many people have the most popular surname in the company, using console:

grep -v "#" /home/luther/tipical_surnames.txt | sed -n 1'p' | cut -f 1
Smith
grep Smith /home/luther/employers.txt | wc -l
230

Work perfect. Now lets check the first 5 most popular surnames using a simple bash script:

#!/bin/bash
counter=1
while [ $counter -le 5 ]
 do
  surname=`grep -v "#" /home/luther/tipical_surnames.txt | sed -n "$counter"'p' | cut -f 1`
  qty=`grep "$surname" /home/luther/employers.txt | wc -l`
  echo $surname
  echo $qty
  counter=$(( $counter + 1 ))
 done

And the result the follows:

Smith
0
Johnson
0
Williams
0
Jones
0
Brown
0

Whats wrong?

Update: Like I wrote I tested the script on other computer and everything is works fine. After I try the follow:

root@problematic:/var/www# cat testfile.bash
#!/bin/bash
for (( c=1; c<=5; c++ ))
{
echo $c
}

root@problematic:/var/www# bash testfile.bash
testfile.bash: line 2: syntax error near unexpected token `$'\r''
'estfile.bash: line 2: `for (( c=1; c<=5; c++ ))
root@problematic:/var/www# echo $BASH_VERSION
4.2.37(1)-release
root@problematic:/var/www#

Of course on other computer this simply script work as expected, without error.

Answer 1

This is obviously untested since you haven't posted sample input but this is the kind of approach you should use:

awk '
NR==FNR { if (!/#/) cnt[$1]=0; next }
{ cnt[$WHATEVER]++ }
END {
    PROCINFO["sorted_in"] = "@val_num_desc"
    for (name in cnt) {
        print name, cnt
        if (++c == 5) {
            break
        }
    }
}
' /home/luther/tipical_surnames.txt /home/luther/employers.txt

Replace "WHATEVER" with the field number where employee surnames are stored in employers.txt.

The above uses GNU awk for sorted_in, with other awks I'd just remove the PROCINFO line and the count from the output loop and pipe the output to sort then head, eg:

awk '
NR==FNR { if (!/#/) cnt[$1]=0; next }
{ cnt[$WHATEVER]++ }
END {
    for (name in cnt) {
        print name, cnt
    }
}
' /home/luther/tipical_surnames.txt /home/luther/employers.txt | sort -k2,1nr | head -5

or whatever the right sort options are.

Answer 2

I'm actually not quite sure. I tested your script, by copying it and pasting it, with imagined data ( /usr/share/dict/words ) and it seems to work as expected. I wonder if there is a difference between the script you posted and the script you're running?

While at it, I took the liberty of making it run a bit smoother. Notice how, in the loop, you read the entirety of the surnames file in each iteration? Also, grep + wc -l may be replaced by grep -c . I'm also adding -F to the first invocation of grep since the pattern ( # ) is fixed strings. The grep into the employee file uses \\<$name\\> to make sure we only get the Johns and no Johnssons when $name is John .

#!/bin/bash

employees_in="/usr/share/dict/words"
names_in="/usr/share/dict/words"

grep -v -F "#" "$names_in" | head -n 5 | cut -f 1 |
while read -r name; do
    count="$( grep -c "\<$names\> " "$employees_in" )"
    printf "name: %-10s\tcount: %d\n" "$name" "$count"
done

Testing it:

$ bash script.sh
name: A             count: 1
name: a             count: 1
name: aa            count: 1
name: aal           count: 1
name: aalii         count: 1

Note: I get only ones in the count because the dictionary (not surprisingly) contains only unique words.