I want to find out in which of my 1200 *.gz files a certain IP address occurs on a certain date (17.07.2025). I could not find the IP address in my current logs: grep 'IP address' *.logs My attempt to see if the IP even exists in the older, zipped log files returns: 'The argument list is too long'. zgrep 'IP address' * A combination of find and grep was also unsuccessful. I am grateful for any assistance.

I know how to do it with grep, but the command doesn't work with **zgrep**

grep -E 'Pattern1.*Patter2' fileName

I'm using **zgrep** to match patterns inside a .json.gz file. Because the files are too big, I want to zgrep BOTH pattern1 **AND** pattern2, the order doesn't matter. Possible to achieve?

I am preparing a script that will try to detect two texts on the same line.  This pattern is present in my application log files, which then further used to identify which file contains the pattern. Prepared a script like below:

#!/bin/bash

#Takes command line argument#
PROC_CNTL_ID="$1"

#Extracts the file name#
FILE_NAME="$(zgrep -e 'ERROR.*"$PROC_CNTL_ID"' /opt/applications/cds2/logs/intraday/log/*.gz | cut -d '/' -f 8 | cut -d ':' -f 1 | awk "NR==1")"

#copies to pn02398 user specific directory#
cp -r /opt/applications/cds2/logs/intraday/log/"$FILE_NAME" /home/pn02398/logs/

Now my problem is that the script runs with exit status of 0 but still the copy command is not working.  If I execute zgrep command by giving a specific value for the command line argument, it also gives the correct filename. What's wrong with the command substitution?

I have the following lines result from zcat command

1 - URL Template: https://www.test.com 

2 - Response:

I want to add the line 2 result from call to first line 1 like this

URL Template: https://www.test.com 

Can I do that?

I want to search a directory's contents for a certain string on Ubuntu, and output all files that contain it. Problem: if zgrep tries to read a damaged file, the process stops: zgrep -n 'test' /var/log/tomcat/archive/* Result: gzip: /var/log/tomcat/archive/my.log.gz: unexpected end of file Question: how can I tell zgrep to ignore files that give errors?

I need to search "335664324" in the file A33256.xml.gz. I tried the following code. sftp> zgrep -c "335664324" A33256.xml.gz A33256.xml.gz Invalid command. I'm in the same directory as the file. How can I overcome this issue to display the output?

I have a big size file like 5GB with .gz. Inside that file, we have few XML files that contains values that I want to search and extract just in case if those values are there. For example I want to extract the tags that contains the name NOOSS and also the subcontent of this tags like `, , , from the the .gz` file

1
    
        NOOSSCONTROLExample
        ACTIVE
        FIVE_MIN
        MEASUREMENTJOB
        HIGH
        FIVE_MIN
        Sla
        CHANGED_ONLY
        
            mr_2
            
                Anything
            
            PER_SECOND
        
        
            mr_1
            
                ManagedElement=1,SystemFunctions=1,Pm=1,PmGroup=OSProcessingLogicalUnit,MeasurementType=CPULoad.Total
            
            PER_SECOND

I was using cat *gz 1 zgrep -a "PmJobId" but the output only show the `` value and not the rest of the information or tags. Please your help, I'm noobie on this. Im using CentOS - RedHat Linux. Thanks

I have a flat database based on 65536 files, each one containing one word by line starting with two hexadecimal characters. They look like this:

afword
46word2
Feword3
...

I make tens of thousands of requests a day on this, so I'm looking for a better way to find a line starting with two hexadecimal characters. The files were sorted before being gzipped. As of now, I do: LC=ALL zgrep --text '^af' file Is there any other faster way to do this in perl or bash or any command line?

I have a **9.8GB** gzip file A.gz and other file i have is **79MB** B.txt which has some text in each line. I want to grep B's text in A.gz and write to a new file. Initially, I used this command zgrep -f B.txt A.gz > C.xml But this command was hanged and it created an empty C.xml for very long time. Then after googling I came to know that because B.txt is huge it hangs when it keeps the text in buffer. So I splitted the text file in 20000 text each split -l 20000 -a 4 B.txt B I created files like Baaaa, Baaab.... and then iterated over each file cd B for f in B*; do zgrep -f "$f" ../A.gz >> C.xml done It is very slow and still running. Any better approach for this? Will gunzipping the gz file will improve performance? **Update** I tried with -F zgrep -F -f "$f" ../A.gz >> C.xml this is little faster but still want other option I have xml like source1 123 ABC source2 123 XYZ Here id is same 123 but category is different ABC and XYZ (Input is limited set of categories e.g ABC,DEF,GHI,JKLM,NOP) Initially I have category as ABC and So based on the category ABC I find out its id i.e. 123 like this i keep on writing all id's that belong to these input categories to a new file i.e. B.txt(list of id's) like below zgrep -E 'ABC|DEF|GHI|JKLM|NOP' A.gz | sed -n 's:.*\(.*\).*:\1:p' | uniq > B.txt later on I iterate over this id's and fetch all xmls, this way I get both xml tags with category ABC and XYZ which belongs to id 123

I am trying to match pattern for a particular column in the thousands of gzipped files on a Linux machine, and based on the match I want to print the file name how to do that. Below options are not working for me , any suggestions please. Thanks zgrep 12345 *| awk -F"^" '{if($8==12345) print}' find . -type f |xargs zcat | awk -F"^" '{if($8==12345) print}'

I have many gzipped files and I am trying to find some string pattern from the thousands of file but I am getting extra .gz while trying to zgrep it. Please refer the below, could you please advise. Thanks. $ zcat abc.gz 1234^911^is^good this^is^test $ ls *.gz| xargs zgrep 911 gzip: abc.gz.gz: No such file or directory Edit 1: @waltinator, thank you. In the echo it is showing the correct stuff. $ ls *.gz| xargs echo zgrep 911 zgrep 911 abc.gz $ ls *.gz| xargs zgrep 911 gzip: abc.gz.gz: No such file or directory Edit2: find is giving the correct output $ find . -name *.gz | xargs zgrep 911 1234^911^is^good

I have a lot of .gz files in a folder /a/b/c1.gz /a/b/c2.gz /a/b/c3.gz and so on. Some of the files have a single pipe delimiter, some have two, three and four and so on, in such a way: xyz|abc xyz|abc|wty xyz|abc|wty|asd and so on. How do I find all the files that have two pipe delimiters overall, three delimiters overall etc ?

Production logs file are huge in size and are stored in compressed format. I use the following to find specific log line zgrep searchString LargeLogFiles*.gz Typically this takes huge amount of time, when this runs on multiple files at once. In my case, I am aware that the **search string I'm trying to find (if present) will lie in 50% to 70% of the part of the file for sure**. Is there any way I can leverage this information for faster search? **Possible solution** zcat the file and filter the part using other unix commands. This is a possbility, but interested to know if there is a good/efficient way to achieve the above said problem.