In a file looking like below I would to find if all lines with "PRIO" has any value after that and if there are some values missing I would like to write it as output.
I've tried to do this with grep, but it only matches if there's even one occurrence of looking word.
cat PATH_TO_FILE | grep 'PRIO' &> /dev/null
if [ $? == 0 ]; then
echo "matched"
else
echo "not found"
fi
File structure looks simillar to this one below
name1
sdgk
PRIO 3
name2
PRIO
dsl dfhhhdf
name3
fnslkf hsdhfd
jlkg;jslk sgdgdsg
kfasdjmgkdlsgl sdggsehg
PRIO 1
name4
sdgds
dsdsgdg
PRIO 2
sdgg
With awk this is very simple by checking the number of fields.
awk '/PRIO/{ str=(NF>1)?"matched":"not found"; print str }' <file>
This does :
/PRIO/ : if a line contains the word PRIO perform action {...}
{...} : if the number for fields is bigger then 1 `(NF>1)1, it matched otherwise it did not.
If you want to ensure that PRIO is the first word, then use $1=="PRIO", and if you want to print the line number then use
awk '($1=="PRIO"){ str=(NF>1)?"matched":"not found"; print NR,str }' <file>
not sure what you want to test but
$ grep -q 'PRIO\s*$' file
checks whether there are any PRIO without any value after. For your sample input this will succeed, which you can use this as an error condition.
if grep -q 'PRIO\s*$' file
then echo "found missing value instance"
else echo "all instances have values"
fi
Related
2001-06-30T11:33:33,543 DEBUG (Bss-Thread-948:[]) SUNCA#44#77#CALMED#OK#58#NARDE#4356#68654768961#BHR#TST#DEV
2001-06-30T11:33:33,543 DEBUG (Bss-Thread-948:[]) SUNCA#44#77#CALMED#OK#58#NARDE#89034#1234567#BHR#TST#DEV
2001-06-30T11:33:33,543 DEBUG (Bss-Thread-948:[]) SUNCA#44#77#OK#58#BHREDD#234586#4254567#BHR#TST#DEV
2001-06-30T11:33:33,543 DEBUG (Bss-Thread-948:[]) SUNCA#44#77#OK#58#NARDE#89034#1034567#BHR#TST#DEV
I have log file mentioned above. I would like to print lines only if value between patterns # and #BHR is greater than 1100000.
I can see in my log file lines with values 68654768961, 1234567, 4254567, 1034567. As per the requirement the output should conatin only first 3 lines.
I am looking for regex to get desired output.
One questions, this #58#BHR should be ignore in third line ? If yes, I will get value between patterns # and #BHR#.
Normally, it should be solved this question by writing scripting according the business logical. But you could try this one line command by awk.
awk '{if (0 == system("[ $(echo \"" $0 "\"" " | grep -oP \"" "(?<=#)\\d+(?=#BHR#)\" || echo 0) -gt 1100000 ]")) {print $0}}' log_file
Mainly, it use system() to scratch the value by grep:
# if can't get the pattern value by grep, the value will assign 0
echo $one_line | grep -oP "(?<=#)\d+(?=#BHR#)" || echo 0`
and compare the value to 1100000 by [ "$value" -gt 1100000 ] in awk.
FYI, so if the value greater than 1100000 it will return 0.
system(cmd): executes cmd and returns its exit status
I have a problem. This is my script:
#!/bin/bash
file_name="eq3_luteina_horyzontalna"
file_name2="wiazanie_PO4"
tmp=$(mktemp) || exit 1
for index in {1..405000}
do
if ! [ -s "${file_name}_$index.ndx" ];then
echo "0" >> ${file_name2}_POP42.txt
else
awk '{if($2==/POP42/) print "5"; else print "0"}' ${file_name}_$index.ndx >> ${file_name2}_POP42.txt
fi
done
The problem is here
awk '{if($2==/POP42/) print "5"; else print "0"}' ${file_name}_$index.ndx
I want to only check if POP42 is in the file in the second column and print 5
but I have data like that
162 POP87
1851 POP42
so it will print into my output file ${file_name2}_POP42.txt, something like that:
0
5
but I want to have
5
Another situation
3075 POP42
2911 POP42
It will print to output
5
5
but I want only
5
How can I manage my problem?
awk '$2=="POP42"{s=5; exit} END{print s+0}' file
By the way - $2==/POP42/ doesn't do what you think it does, i.e. look for lines with $2 equal to (or even containing) POP42. It's actually shorthand for $2==($0 ~ /POP42/ ? 1 : 0) courtesy of the regexp delimiters /.../ you used and what THAT does is see if a string matching the regexp POP42 occurs anywhere on the current line and, if it does, then test to see if $2 has the value 1, otherwise test to see if $2 has the value 0. It's important to know the difference between string (") and regexp (/) delimiters and string (e.g. ==) and regexp (e.g. ~) comparison operators when using awk.
I have a file which is created after some manual configuration.
I need to check this file automatically with a shell script.
The file looks like this:
eth0;eth0;1c:98:ec:2a:1a:4c
eth1;eth1;1c:98:ec:2a:1a:4d
eth2;eth2;1c:98:ec:2a:1a:4e
eth3;eth3;1c:98:ec:2a:1a:4f
eth4;eth4;48:df:37:58:da:44
eth5;eth5;48:df:37:58:da:45
eth6;eth6;48:df:37:58:da:46
eth7;eth7;48:df:37:58:da:47
I want to compare it to a pattern like this:
eth0;eth0;*
eth1;eth1;*
eth2;eth2;*
eth3;eth3;*
eth4;eth4;*
eth5;eth5;*
eth6;eth6;*
eth7;eth7;*
If I would only have to check this pattern I could run this loop:
c=0
while [ $c -le 7 ]
do
if [ "$(grep "eth"${c}";eth"${c}";*" current_mapping)" ];
then
echo "eth$c ok"
fi
(( c++ ))
done
There are 6 or more different patterns possible. A pattern could also look like this for example (depending and specific configuration requests):
eth4;eth0;*
eth5;eth1;*
eth6;eth2;*
eth7;eth3;*
eth0;eth4;*
eth1;eth5;*
eth2;eth6;*
eth3;eth7;*
So I don't think I can run a standard grep per line command in a loop. The eth numbers are not consistently the same.
Is it possible somehow to compare the whole file to pattern like it would be possible with grep for a single line?
Assuming file is your data file and patt is your file that contains above pattern. You can use this grep -f in conjunction with sed in a process substitution that replaces * with .* and ? with . to make it a workable regex.
grep -f <(sed 's/\*/.*/g; s/?/./g' patt) file
eth0;eth0;1c:98:ec:2a:1a:4c
eth1;eth1;1c:98:ec:2a:1a:4d
eth2;eth2;1c:98:ec:2a:1a:4e
eth3;eth3;1c:98:ec:2a:1a:4f
eth4;eth4;48:df:37:58:da:44
eth5;eth5;48:df:37:58:da:45
eth6;eth6;48:df:37:58:da:46
eth7;eth7;48:df:37:58:da:47
I wrote this loop now and it does the job (current_mapping being the file with the content in the first code block of the question). I would have to create arrays with different patterns and use a case for every pattern. I was just wondering if there is something like grep for multiple lines, that could the same without writing this loop.
array=("eth0;eth0;*" "eth1;eth1;*" "eth2;eth2;*" "eth3;eth3;*" "eth4;eth4;*" "eth5;eth5;*" "eth6;eth6;*" "eth7;eth7;*")
c=1
while [ $c -le 8 ]
do
if [ ! "$(sed -n "${c}"p current_mapping | grep "${array[$c-1]}")" ];
then
echo "somethings wrong"
fi
(( c++ ))
done
Try any:
grep -P '(eth[0-9]);\1'
grep -E '(eth[0-9]);\1'
sed -n '/\(eth[0-9]\);\1/p'
awk -F';' '$1 == $2'
There are commands only. Apply them to a pipe or file.
Updated the answer after the question was edited.
As we can see the task requirements are as follows:
a file (a set of lines) formatted like ethN;ethM;MAC
examine each line for equality ethN and ethM
if they are equal, output a string ethN ok
If I understand the task correctly we can achieve this using the following code without loops:
awk -F';' '$1 == $2 { print $1, "ok" }'
The following code does not work as I would expect:
(the original purpose of the script is to make a relation between items of two files where the identifiers are not sorted in the same order, but my question raises rather a curiosity about basic shell functionalities)
#!/bin/sh
process_line() {
id="$1"
entry=$(grep $id index.txt) # the "grep" line
if [ "$entry" = "" ]; then
echo 00000 $id
else
echo $entry | awk '{print $2, $1;}'
fi
}
cat << EOF > index.txt
xyz 33333
abc 11111
def 22222
EOF
cat << EOF | while read line ; do process_line "$line"; done
abc
def
xyz
EOF
The output is:
11111 abc
22222 def
00000
But I would expect:
11111 abc
22222 def
00000
33333 xyz
(the last line is missing in the actual output)
My investigations show that the "grep" line is the one that leads to the early interruption of the while loop. However I cannot see the causal relationship.
That's because in the third iteration with the empty line, you call process_line with an empty id. This leads to grep index.txt, i.e. no file name. This grep reads from stdin and that'll consume all your input you pipe into the while loop.
To see this in action, add set -x at the top of your script.
You can get the desired behaviour if you replace the empty id with a string guaranteed to be not found, such as
entry=$(grep "${id:-NoSuchString}" index.txt)
Changing the "process_line" function to the following might help...
process_line() {
id=$1
if [ "$id" = "" ]
then
echo "00000"
else
entry=$(grep "${id}" index.txt)
echo "$entry" | awk '{ print $2, $1 }'
fi
}
Explanation:
if the "id" passed in is empty then just output the default
move the grep to the else clause so it only executes when "id" has a value
solves the problem with the missing quotes around id in the grep statement
another thing to consider is the case where "id" is not-empty but not found in the index.txt file. This could result in a blank output. Adding an if statement after the grep call to handle this case may be a good idea depending on what the overall intention is.
Hope that helps
I need to be able to output a record if a condition is true.
Suppose this is the existing file,
Record_ID,Name,Last Name,Phone Number
I am trying to output record if the last name matches. I collect user input to get last name and then perform the following operation.
read last_name
cat contact_records.txt | awk -F, '{if($3=='$last_name')print "match"; else print "no match";}'
This script outputs no match for every record within contact_records.txt
Your script has two problems:
First, $last_name is not considered quoted in the context of 'awk'. For example, if "John" is to be queried, you are comparing $3 with the variable John rather than string "John". This can be fixed by adding two double-quotes as below:
read last_name
cat contact_records.txt | awk -F, '{if($3=="'$last_name'")print "match"; else print "no match";}'
Second, it actually scans the whole contact_records.txt and prints match/no match for each line of comparison. For example, contact_records.txt has 100 lines, with "John" in it. Then, querying if John is in it by this script yields 1 "match"'s and 99 "no match"'s. This might not be what you want. Here's a fix:
read last_name
if [ `cat contact_records.txt | cut -d, -f 3 | grep -c "$last_name"` -eq 0 ]; then
echo "no match"
else
echo "match"
fi