Occurrence of a string in all the file names within a folder in Bash - bash

I am trying to make a script which allow me to select files with 2 or more occurrences of a string in their name.
Example:
test.txt // 1 occurrence only, not ok
test-test.txt // 2 occurrences OK
test.test.txt // 2 occurrences OK
I want the script to return me only the files 2 and 3. I tried like that but this didn't work:
rows=$(ls | grep "test")
for file in $rows
do
if [[ $(wc -w $file) == 2 ]]; then
echo "the file $file matches"
fi
done

grep and wc are overkill. A simple glob will suffice:
*test*test*
You can use this like so:
ls *test*test*
or
for file in *test*test*; do
echo "$file"
done

You can use :
result=$(grep -o "test" yourfile | wc -l)
-wc is a word count
In shell script if $result>1 do stuff...

Related

Counting number of lines in file and saving it in a bash file

I am trying to loop through all the files in a folder and add the file name of those files with 10 lines to a txt file but I don't know how to write the if statement.
As of right now, what I have is:
for FILE in *.txt do if wc $FILE == 10; then "$FILE" >> saved_names.txt fi done
I am getting stuck in how to format the statement that will evaluate to a boolean for the if statement.
I have already tried the if statement as:
if [ wc $FILE != 10 ]
if "wc $FILE" != 10
if "wc $FILE != 10"
as well as other ways but I don't seem to get it right. I know I am new to Bash but I can't seem to find a solution to this question.
There are a few problems in your code.
To count the number of lines in the file you should run "wc -l" command. However, that command will result in the number of lines and the name of the file (so for example - 10 a.txt - you can test it by running the command on a file in your terminal). To receive only the number of lines you need to pass the file's name to the standard input of that command
"==" is used in bash to compare strings. To compare integers as in that case, you should use "-eq" (take a look here https://tldp.org/LDP/abs/html/comparison-ops.html)
In terms of brackets: To get the wc command result you need to run it in a terminal and switch the command in the code to the result. To do that, you need correct brackets - $(wc -l). To receive a result of the comparison as a bool, you need to use square brackets with spaces [ 1 -eq 1 ].
To save the name of the file in another file using >> you need to first put the name to the standard output (as >> redirect the standard output to the chosen place). To do that you can just use the echo command.
The code should look like this:
#!/bin/bash
for FILE in *.txt
do
if [ "$(wc -l < "$FILE")" -eq 10 ]
then
echo "$FILE" >> saved_names.txt
fi
done
Try:
for file in *.txt; do
if [[ $(wc -l < "$file") -eq 10 ]]; then
printf '%s\n' "$file"
fi
done > saved_names.txt
Change > to >> if you want to append the filenames.
Related docs:
Command Substitution
Conditional Constructs
Extract the actual number of lines from a file with wc -l $FILE | cut -f1 -d' ' and use -eq operator:
for FILE in *.txt; do if [ "$(wc -l $FILE | cut -f1 -d' ')" -eq 10 ]; then "$FILE" >> saved_names.txt; fi; done

Counting grep results

I'm new to bash so I'm finding trouble doing something very basic.
Through playing with various scripts I found out that the following script prints the lines that contain the "word"
for file in*; do
cat $file | grep "word"
done
doing the following:
for file in*; do
cat $file | grep "word" | wc -l
done
had a result of printing in every iteration how many times did the "word" appeared on file.
How can I implement a counter for all those appearances and in the
end just echo the counter?
I used a counter that way but it appeared 0.
let x+=cat $filename | grep "word"
You can pipe the entire loop to wc -l.
for file in *; do
cat $file | grep "word"
done | wc -l
This is a useless use of cat. How about:
for file in *; do
grep "word" "$file"
done | wc -l
Actually, the entire loop is unnecessary if you pass all the file names to grep at once.
grep "word" * | wc -l
Note that if word shows up more than once on the same line these solutions will only count the entire line once. If you want to count same-line occurrences separately you can use -o to print each match on a separate line:
grep -o "word" * | wc -l
The oneliner in John's answer is the way to go. Just to satisfy your curiosity:
sum=0
for f in *; do
x="$(grep 'word' "$f" | wc -l)"
echo "x: $x"
(( sum += x ))
done
echo "sum: $sum"
If the line containing the grep and wc does not yield a number you are SOL. That is why you should stick to the other solution or do a pure bash implementation with things like read, 'case and *word*)' or if [[ "$line" =~ "$re_containing_word" ]]; then ...

How to match a folder name and use it in an if condition using grep in bash?

for d in */ ; do
cd $d
NUM = $(echo ${PWD##*/} | grep -q "*abc*");
if [[ "$NUM" -ne "0" ]]; then
pwd
fi
cd ..
done
Here I'm trying to match a folder name to some substring 'abc' in the name of the folder and check if the output of the grep is not 0. But it gives me an error which reads that NUM: command not found
An error was addressed in comments.
NUM = $(echo ${PWD##*/} | grep -q "*abc*"); should be NUM=$(echo ${PWD##*/} | grep -q "*abc*");.
To clarify, the core problem would be to be able to match current directory name to a pattern.
You can probably simply the code to just
if grep -q "*abc*" <<< "${PWD##*/}" 2>/dev/null; then
echo "$PWD"
# Your rest of the code goes here
fi
You can use the exit code of the grep directly in a if-conditional without using a temporary variable here ($NUM here). The condition will pass if grep was able to find a match. The here-string <<<, will pass the input to grep similar to echo with a pipeline. The part 2>/dev/null is to just suppress any errors (stderr - file descriptor 2) if grep throws!
As an additional requirement asked by OP, to negate the conditional check just do
if ! grep -q "*abc*" <<< "${PWD##*/}" 2>/dev/null; then

ksh: shell script to search for a string in all files present in a directory at a regular interval

I have a directory (output) in unix (SUN). There are two types of files created with timestamp prefix to the file name. These file are created on a regular interval of 10 minutes.
e. g:
1. 20140129_170343_fail.csv (some lines are there)
2. 20140129_170343_success.csv (some lines are there)
Now I have to search for a particular string in all the files present in the output directory and if the string is found in fail and success files, I have to count the number of lines present in those files and save the output to the cnt_succ and cnt_fail variables. If the string is not found I will search again in the same directory after a sleep timer of 20 seconds.
here is my code
#!/usr/bin/ksh
for i in 1 2
do
grep -l 0140127_123933_part_hg_log_status.csv /osp/local/var/log/tool2/final_logs/* >log_t.txt; ### log_t.txt will contain all the matching file list
while read line ### reading the log_t.txt
do
echo "$line has following count"
CNT=`wc -l $line|tr -s " "|cut -d" " -f2`
CNT=`expr $CNT - 1`
echo $CNT
done <log_t.txt
if [ $CNT > 0 ]
then
exit
fi
echo "waiitng"
sleep 20
done
The problem I'm facing is, I'm not able to get the _success and _fail in file in line and and check their count
I'm not sure about ksh, but while ... do; ... done is notorious for running off with whatever variables you're using in bash. ksh might be similar.
If I've understand your question right, SunOS has grep, uniq and sort AFAIK, so a possible alternative might be...
First of all:
$ cat fail.txt
W34523TERG
ADFLKJ
W34523TERG
WER
ASDTQ34T
DBVSER6
W34523TERG
ASDTQ34T
DBVSER6
$ cat success.txt
abcde
defgh
234523452
vxczvzxc
jkl
vxczvzxc
asdf
234523452
vxczvzxc
dlkjhgl
jkl
wer
234523452
vxczvzxc
And now:
egrep "W34523TERG|ASDTQ34T" fail.txt | sort | uniq -c
2 ASDTQ34T
3 W34523TERG
egrep "234523452|vxczvzxc|jkl" success.txt | sort | uniq -c
3 234523452
2 jkl
4 vxczvzxc
Depending on the input data, you may want to see what options sort has on your system. Examining uniq's options may prove useful too (it can do more than just count duplicates).
Think you want something like this (will work in both bash and ksh)
#!/bin/ksh
while read -r file; do
lines=$(wc -l < "$file")
((sum+=$lines))
done < <(grep -Rl --include="[1|2]*_fail.csv" "somestring")
echo "$sum"
Note this will match files starting with 1 or 2 and ending in _fail.csv, not exactly clear if that's what you want or not.
e.g. Let's say I have two files, one starting with 1 (containing 4 lines) and one starting with 2 (containing 3 lines), both ending in `_fail.csv somewhere under my current working directory
> abovescript
7
Important to understand grep options here
-R, --dereference-recursive
Read all files under each directory, recursively. Follow all
symbolic links, unlike -r.
and
-l, --files-with-matches
Suppress normal output; instead print the name of each input
file from which output would normally have been printed. The
scanning will stop on the first match. (-l is specified by
POSIX.)
Finaly I'm able to find the solution. Here is the complete code:
#!/usr/bin/ksh
file_name="0140127_123933.csv"
for i in 1 2
do
grep -l $file_name /osp/local/var/log/tool2/final_logs/* >log_t.txt;
while read line
do
if [ $(echo "$line" |awk '/success/') ] ## will check the success file
then
CNT_SUCC=`wc -l $line|tr -s " "|cut -d" " -f2`
CNT_SUCC=`expr $CNT_SUCC - 1`
fi
if [ $(echo "$line" |awk '/fail/') ] ## will check the fail file
then
CNT_FAIL=`wc -l $line|tr -s " "|cut -d" " -f2`
CNT_FAIL=`expr $CNT_FAIL - 1`
fi
done <log_t.txt
if [ $CNT_SUCC > 0 ] && [ $CNT_FAIL > 0 ]
then
echo " Fail count = $CNT_FAIL"
echo " Success count = $CNT_SUCC"
exit
fi
echo "waitng for next search..."
sleep 10
done
Thanks everyone for your help.
I don't think I'm getting it right, but You can't diffrinciate the files?
maybe try:
#...
CNT=`expr $CNT - 1`
if [ $(echo $line | grep -o "fail") ]
then
#do something with fail count
else
#do something with success count
fi

Count mutiple occurences of a word on the same line using grep

Here I made a small script that take input from user searching some pattern from a file and displays required no of lines from that file where the pattern is found. Although this code is searching the pattern line wise due to standard grep practice. I mean if the pattern occurs twice on the same line, i want the output to print twice. Hope I make some sense.
#!/bin/sh
cat /dev/null>copy.txt
echo "Please enter the sentence you want to search:"
read "inputVar"
echo "Please enter the name of the file in which you want to search:"
read "inputFileName"
echo "Please enter the number of lines you want to copy:"
read "inputLineNumber"
[[-z "$inputLineNumber"]] || inputLineNumber=20
cat /dev/null > copy.txt
for N in `grep -n $inputVar $inputFileName | cut -d ":" -f1`
do
LIMIT=`expr $N + $inputLineNumber`
sed -n $N,${LIMIT}p $inputFileName >> copy.txt
echo "-----------------------" >> copy.txt
done
cat copy.txt
As I understood, the task is to count number of pattern occurrences in line. It can be done like so:
count=$((`echo "$line" | sed -e "s|$pattern|\n|g" | wc -l` - 1))
Suppose you have one file to read. Then, code will be following:
#!/bin/bash
file=$1
pattern="an."
#reading file line by line
cat -n $file | while read input
do
#storing line to $tmp
tmp=`echo $input | grep "$pattern"`
#counting occurrences count
count=$((`echo "$tmp" | sed -e "s|$pattern|\n|g" | wc -l` - 1))
#printing $tmp line $count times
for i in `seq 1 $count`
do
echo $tmp
done
done
I checked this for pattern "an." and input:
I pass here an example of many 'an' letters
an
ananas
an-an-as
Output is:
$ ./test.sh input
1 I pass here an example of many 'an' letters
1 I pass here an example of many 'an' letters
1 I pass here an example of many 'an' letters
3 ananas
4 an-an-as
4 an-an-as
Adapt this to your needs.
How about using awk?
Assume the pattern you are searching for is in variable $pattern and the file you are checking is $file
The
count=`awk 'BEGIN{n=0}{n+=split($0,a,"'$pattern'")-1}END {print n}' $file`
or for a line
count=`echo $line | awk '{n=split($0,a,"'$pattern'")-1;print n}`

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