Using grep
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Comments
Some confuusion while searching using grep
I hav a dir and on that file present like this
-rw-r--r-- 1 vs users 1390254 Jun 2 01:00 abcdefg.1006010100.dat
-rw-r--r-- 1 vs users 1388800 Jun 3 01:00 abcdefg.1006020100.dat
-rw-r--r-- 1 vs users 1388555 Jun 4 01:00 abcdefg.1006030100.dat
-rw-r--r-- 1 vs users 1392184 Jun 5 01:00 abcdefg.1006040100.dat
-rw-r--r-- 1 vs users 1391747 Jun 6 01:00 abcdefg.1006050100.dat
-rw-r--r-- 1 vs users 1392099 Jun 7 01:00 abcdefg.1006060100.dat
-rw-r--r-- 1 vs users 1389362 Jun 8 01:00 abcdefg.1006070100.dat
-rw-r--r-- 1 vs users 1392676 Jun 9 01:00 abcdefg.1006080100.dat
-rw-r--r-- 1 vs users 1436696 Jun 10 01:00 abcdefg.1006090100.dat
-rw-r--r-- 1 vs users 1060539 Jun 10 18:39 abcdefg.1006100100.dat
please check output while i am using grep command in below manner ..
vs@vodalksmvs2 /var/opt/nokia/smvs/tmp > ls -ltr | grep abc*
vs@vodalksmvs2 /var/opt/nokia/smvs/tmp > ls -ltr | grep "abc*"| tail
-rw-r--r-- 1 vs users 1390254 Jun 2 01:00 abcdefg.1006010100.dat
-rw-r--r-- 1 vs users 1388800 Jun 3 01:00 abcdefg.1006020100.dat
-rw-r--r-- 1 vs users 1388555 Jun 4 01:00 abcdefg.1006030100.dat
-rw-r--r-- 1 vs users 1392184 Jun 5 01:00 abcdefg.1006040100.dat
-rw-r--r-- 1 vs users 1391747 Jun 6 01:00 abcdefg.1006050100.dat
-rw-r--r-- 1 vs users 1392099 Jun 7 01:00 abcdefg.1006060100.dat
-rw-r--r-- 1 vs users 1389362 Jun 8 01:00 abcdefg.1006070100.dat
-rw-r--r-- 1 vs users 1392676 Jun 9 01:00 abcdefg.1006080100.dat
-rw-r--r-- 1 vs users 1436696 Jun 10 01:00 abcdefg.1006090100.dat
-rw-r--r-- 1 vs users 1059029 Jun 10 18:37 abcdefg.1006100100.dat
vs@vodalksmvs2 /var/opt/nokia/smvs/tmp > ls -ltr | grep abc| tail
-rw-r--r-- 1 vs users 1390254 Jun 2 01:00 abcdefg.1006010100.dat
-rw-r--r-- 1 vs users 1388800 Jun 3 01:00 abcdefg.1006020100.dat
-rw-r--r-- 1 vs users 1388555 Jun 4 01:00 abcdefg.1006030100.dat
-rw-r--r-- 1 vs users 1392184 Jun 5 01:00 abcdefg.1006040100.dat
-rw-r--r-- 1 vs users 1391747 Jun 6 01:00 abcdefg.1006050100.dat
-rw-r--r-- 1 vs users 1392099 Jun 7 01:00 abcdefg.1006060100.dat
-rw-r--r-- 1 vs users 1389362 Jun 8 01:00 abcdefg.1006070100.dat
-rw-r--r-- 1 vs users 1392676 Jun 9 01:00 abcdefg.1006080100.dat
-rw-r--r-- 1 vs users 1436696 Jun 10 01:00 abcdefg.1006090100.dat
-rw-r--r-- 1 vs users 1060539 Jun 10 18:39 abcdefg.1006100100.dat
vs@vodalksmvs2 /var/opt/nokia/smvs/tmp > ls -ltr | grep abc.*| tail -2
-rw-r--r-- 1 vs users 1436696 Jun 10 01:00 abcdefg.1006090100.dat
-rw-r--r-- 1 vs users 1060539 Jun 10 18:39 abcdefg.1006100100.dat
vs@vodalksmvs2 /var/opt/nokia/smvs/tmp > ls -ltr | grep *abc| tail -2
vs@vodalksmvs2 /var/opt/nokia/smvs/tmp > ls -ltr | grep *.abc| tail -2
vs@vodalksmvs2 /var/opt/nokia/smvs/tmp >
I want to know in what manner grep runs and giving output for above commands.
grep (-A|-B|-C)
I use linux as my development environment. Grep with -A, -B and -C option provides me with context for what I'm searching. These options print out lines either above or below the target line.
For example I search for a function call and get grep to display the lines above and/or below the said line.
grep -p
I come from the AIX world. We have many scripts that use grep -p to get the paragraph containing a search string. I cannot find anything comparable to grep -p in Red Hat Linux. Any ideas?
grep -p
Never mind, I found this code. (Shoulda done more googling first! :-) )
#!/bin/sh
# usage: pargrep
inFile="$1"
searchString="$2"
awk '
BEGIN {
FS="\n"
RS=""
}
/'"$searchString"'/ { print }
' ${inFile}
non-standard args - could be made less confusing
That's good as far as it goes, but you should reverse $1 and $2. NORMAL grep is:
grep
That pargrep is
grep
At least you should minimize the differences. Also, if you go with the "standard" way, a very small change to that script could be made to search across multiple files, just like grep would.
[sigh] stupid html. Ok, that
[sigh] stupid html.
Ok, that should have read:
...
grep pattern filename
that pargrep is:
grep filename pattern
...
Question
Dear Sir,
How can i search in my lunix server for a word in the sources code for any file that is located in the vhost. for example:
I need to search for the word "iframe" in all the "vhosts" folder on my server. this word is in the "Sources code for all my files for my websites"
Can u help me in this please ?
i will too much greatfull for you
regards
samer
The fastest method is...
The following, replacing "*.php" by your source files' extension:
$ find . -name "*.php" -print0 | xargs -0 grep iframeSearch
There are a number of ways of doing that, try:
$ find /path/to/vhosts -type f -exec grep --with-filename iframe {} \;or try:
Mitch Frazier is an Associate Editor for Linux Journal.