i am new to unix shell scripting
I need to write a shell script that should increment a variable value (numeric) and when i run the script next time the variable should take the incremented value and it should increment once agian
any help in this would be really thankful
thanks
Try the following code:
vale=`expr 0000000000 + 1`
ed -s $0 <<EOT
1s/ ........../`printf ' %010d' $vale`/
w
q
EOT
echo $vale
You need for the value of the variable to persist somehow. One way is to save it to a file.
outputfile="~/variable.txt"
value=`cat $outputFile`
newValue=`expr $value + 1`
echo $newValue > $outputFile
Related
I have a file named param1.txt which contains certain variables. I have another file as source1.txt which contains place holders. I want to replace the place holders with the values of the variables that I get from the parameter file.
I have basically hard coded the script where the variable names in the parameter.txt file is known before hand. I want to know a dynamic solution to the problem where the variable names will not be known beforehand. In other words, is there any way to find out the variable names in a file using the source command in UNIX?
Here is my script and the files.
Script:
#!/bin/bash
source /root/parameters/param1.txt
sed "s/{DB_NAME}/$DB_NAME/gI;
s/{PLANT_NAME}/$PLANT_NAME/gI" \
/root/sources/source1.txt >
/root/parameters/Output.txt`
param1.txt:
PLANT_NAME=abc
DB_NAME=gef
source1.txt:
kdashkdhkasdkj {PLANT_NAME}
jhdbjhasdjdhas kashdkahdk asdkhakdshk
hfkahfkajdfk ljsadjalsdj {PLANT_NAME}
{DB_NAME}
I cannot comment since I don't have enough points.
But is it correct that this is what you're looking for:
How to reference a file for variables using Bash?
Your problem statement isn't very clear to me. Perhaps you can simplify your problem and desired state.
Don't understand why you try to source param1.txt.
You can try with this awk :
awk '
NR == FNR {
a[$1] = $2
next
}
{
for ( i = 1 ; i <= NF ; i++ ) {
b = $i
gsub ( "^{|}$" , "" , b )
if ( b in a )
sub ( "{" b "}" , a[b] , $i )
}
} 1' FS='=' param1.txt FS=" " source1.txt
A basic Unix question.
I have a script which counts the number of records in a delta file.
awk '{
n++
} END {
if(n >= 1000) print "${completeFile}"; else print "${deltaFile}";
}' <${deltaFile} >${fileToUse}
Then, depending on the IF condition, I want to process the appropriate file:
cut -c2-11 < ${fileToUse}
But how do I use the contents of the file as the filename itself?
And if there are any tweaks to be made, feel free.
Thanks in advance
Cheers
Simon
To use as a filename the contents of a file which is itself identified by a variable (as asked)
cut -c2-11 <"$( cat $filetouse )"
// or in zsh just
cut -c2-11 <"$( < $filetouse )"
unless the filename in the file ends with one or more newline character(s), which people rarely do because it's quite awkward and inconvenient, then something like:
read -rdX var <$filetouse; cut -c2-11 < "${var%?}"
// where X is a character that doesn't occur in the filename
// maybe something like $'\x1f'
Tweaks: your awk prints the variable reference ${completeFile} or ${deltaFile} (because they're within the single-quoted awk script) not the value of either variable. If you actually want the value, as I'd expect from your description, you should pass the shell vars to awk vars like this
awk -vf="$completeFile" -vd="$deltaFile" '{n++} END{if(n>=1000)print f; else print d}' <"$deltaFile"`
# the " around $var can be omitted if the value contains no whitespace and no glob chars
# people _often_ but not always choose filenames that satisfy this
# and they must not contain backslash in any case
or export the shell vars as env vars (if they aren't already) and access them like
awk '{n++} END{if(n>=1000) print ENVIRON["completeFile"]; else print ENVIRON["deltaFile"]}' <"$deltaFile"
Also you don't need your own counter, awk already counts input records
awk -vf=... -vd=... 'END{if(NR>=1000)print f;else print d}' <...
or more briefly
awk -vf=... -vd=... 'END{print (NR>=1000?f:d)}' <...
or using a file argument instead of redirection so the name is available to the script
awk -vf="$completeFile" 'END{print (NR>=1000?f:FILENAME)}' "$deltaFile" # no <
and barring trailing newlines as above you don't need an intermediate file at all, just
cut -c2-11 <"$( awk -vf="$completeFile" -'END{print (NR>=1000?f:FILENAME)}' "$deltaFile")"
Or you don't really need awk, wc can do the counting and any POSIX or classic shell can do the comparison
if [ $(wc -l <"$deltaFile") -ge 1000 ]; then c="$completeFile"; else c="$deltaFile"; fi
cut -c2-11 <"$c"
i'm confused about the $symbol for unix.
according to the definition, it states that it is the value stored by the variable following it. i'm not following the definition - could you please give me an example of how it is being used?
thanks
You define a variable like this:
greeting=hello
export name=luc
and use like this:
echo $greeting $name
If you use export that means the variable will be visible to subshells.
EDIT: If you want to assign a string containing spaces, you have to quote it either using double quotes (") or single quotes ('). Variables inside double quotes will be expanded whereas in single quotes they won't:
axel#loro:~$ name=luc
axel#loro:~$ echo "hello $name"
hello luc
axel#loro:~$ echo 'hello $name'
hello $name
In case of shell sctipts. When you assign a value to a variable you does not need to use $ simbol. Only if you want to acces the value of that variable.
Examples:
VARIABLE=100000;
echo "$VARIABLE";
othervariable=$VARIABLE+10;
echo $othervariable;
The other thing: if you use assignment , does not leave spaces before and after the = simbol.
Here is a good bash tutorial:
http://linuxconfig.org/Bash_scripting_Tutorial
mynameis.sh:
#!/bin/sh
finger | grep "`whoami` " | tail -n 1 | awk '{FS="\t";print $2,$3;}'
finger: prints all logged in user example result:
login Name Tty Idle Login Time Office Office Phone
xuser Forname Nickname tty7 3:18 Mar 9 07:23 (:0)
...
grep: filter lines what containing the given string (in this example we need to filter xuser if our loginname is xuser)
http://www.gnu.org/software/grep/manual/grep.html
whoami: prints my loginname
http://linux.about.com/library/cmd/blcmdl1_whoami.htm
tail -n 1 : shows only the last line of results
http://unixhelp.ed.ac.uk/CGI/man-cgi?tail
the awk script: prints the second and third column of the result: Forname, Nickname
http://www.staff.science.uu.nl/~oostr102/docs/nawk/nawk_toc.html
This question already has answers here:
How can I repeat a character in Bash?
(36 answers)
Closed 8 years ago.
I need to generate a string of dots (.characters) as a variable.
I.e., in my Bash script, for input 15 I need to generate this string of length 15: ...............
I need to do so variably. I tried using this as a base (from Unix.com):
for i in {1..100};do printf "%s" "#";done;printf "\n"
But how do I get the 100 to be a variable?
You can get as many NULL bytes as you want from /dev/zero. You can then turn these into other characters. The following prints 16 lowercase a's
head -c 16 < /dev/zero | tr '\0' '\141'
len=100 ch='#'
printf '%*s' "$len" | tr ' ' "$ch"
Easiest and shortest way without a loop
VAR=15
Prints as many dots as VAR says (change the first dot to any other character if you like):
printf '.%.0s' {1..$VAR}
Saves the dotted line in a variable to be used later:
line=`printf '.%.0s' {1..$VAR}`
echo "Sign here $line"
-Blatantly stolen from dogbane's answer https://stackoverflow.com/a/5349842/3319298
Edit: Since I have now switched to fish shell, here is a function defined in config.fish that does this with convenience in that shell:
function line -a char -a length
printf '%*s\n' $length "" | tr ' ' $char
end
Usage: line = 8 produces ========, line \" 8 produces """""""".
On most systems, you could get away with a simple
N=100
myvar=`perl -e "print '.' x $N;"`
I demonstrated a way to accomplish this task with a single command in another question, assuming it's a fixed number of characters to be produced.
I added an addendum to the end about producing a variable number of repeated characters, which is what you asked for, so my previous answer is relevant here:
https://stackoverflow.com/a/17030976/2284005
I provided a full explanation of how it works there. Here I'll just add the code to accomplish what you're asking for:
n=20 # This the number of characters you want to produce
variable=$(printf "%0.s." $(seq 1 $n)) # Fill $variable with $n periods
echo $variable # Output content of $variable to terminal
Outputs:
....................
You can use C-style for loops in Bash:
num=100
string=$(for ((i=1; i<=$num; i++));do printf "%s" "#";done;printf "\n")
Or without a loop, using printf without using any externals such as sed or tr:
num=100
printf -v string "%*s" $num ' ' '' $'\n'
string=${string// /#}
The solution without loops:
N=100
myvar=`seq 1 $N | sed 's/.*/./' | tr -d '\n'`
num=100
myvar=$(jot -b . -s '' $num)
echo $myvar
When I have to create a string that contains $x repetitions of a known character with $x below a constant value, I use this idiom:
base='....................'
# 0 <= $x <= ${#base}
x=5
expr "x$base" : "x\(.\{$x\}\)" # Will output '\n' too
Output:
.....
Hi I want to delete a line from a file which matches particular pattern
the code I am using is
BEGIN {
FS = "!";
stopDate = "date +%Y%m%d%H%M%S";
deletedLineCtr = 0; #diagnostics counter, unused at this time
}
{
if( $7 < stopDate )
{
deletedLineCtr++;
}
else
print $0
}
The code says that the file has lines "!" separated and 7th field is a date yyyymmddhhmmss format. The script deletes a line whose date is less than the system date. But this doesn't work. Can any one tell me the reason?
Is the awk(1) assignment due Tuesday? Really, awk?? :-)
Ok, I wasn't sure exactly what you were after so I made some guesses. This awk program gets the current time of day and then removes every line in the file less than that. I left one debug print in.
BEGIN {
FS = "!"
stopDate = strftime("%Y%m%d%H%M%S")
print "now: ", stopDate
}
{ if ($7 >= stopDate) print $0 }
$ cat t2.data
!!!!!!20080914233848
!!!!!!20090914233848
!!!!!!20100914233848
$ awk -f t2.awk < t2.data
now: 20090914234342
!!!!!!20100914233848
$
call date first to pass the formatted date as a parameter:
awk -F'!' -v stopdate=$( date +%Y%m%d%H%M%S ) '
$7 < stopdate { deletedLineCtr++; next }
{print}
END {do something with deletedLineCrt...}
'
You would probably need to run the date command - maybe with backticks - to get the date into stopDate. If you printed stopDate with the code as written, it would contain "date +...", not a string of digits. That is the root cause of your problem.
Unfortunately...
I cannot find any evidence that backticks work in any version of awk (old awk, new awk, GNU awk). So, you either need to migrate the code to Perl (Perl was originally designed as an 'awk-killer' - and still includes a2p to convert awk scripts to Perl), or you need to reconsider how the date is set.
Seeing #DigitalRoss's answer, the strftime() function in gawk provides you with the formatting you want (check 'info gawk' as I did).
With that fixed, you should be getting the right lines deleted.