problem with while loop - bash

here is the script I wrote but it seems it has problem with while ! while suppose to compare the content of K with SDATE and while they are not equal go to the loop !
for d in \
$(sed -nre 's/.*\[(..)\/(...)\/(....):(..:..:..) .*/\1 \2 \3 \4/p' thttpd.log | date +%s -f-);
do echo $d >s1; done
time=$(expr 60 \* 60 \* 24 \* 5)
EDATE=`tail -1 s1`
SDATE=$[$EDATE - $time]
time=$(expr 60 \* 60 \* 24 \* 5)
EDATE=`tail -1 s1`
SDATE=$[$EDATE - $time]
k=`tail -1 s1`
echo $k
echo $SDATE
while [$k -ne $SDATE](k and SDATE contain numbers)
do
k=`tail -1 s1`
sed '1d' < s1 > tempfile
mv s1 s1.old
mv tempfile s1
echo $K| awk '{print strftime("%d/%m/%Y:%T",$1)}'|tee -a ass
done

The problem is that you do not have spaces around [ or ]. Which causes BASH to parse the line incorrectly.
With the following line, BASH will attempt to run the program [$K, probably not what you are intending.
while [$k -ne $SDATE]
What you need to have is the following:
while [ $k -ne $SDATE ]

Try this:
while [[ $k != $SDATE ]]

Ah, shell programming is so touchy...
k=0
while [ $k != 4 ]; do echo $k; k=`expr $k + 1`; done
Works and prints 0, 1, 2, 3 on separate lines as expected and this essentially looks like what you have except for spaces, newlines, and sources of values. Maybe it is a string versus value problem but I did not think shell variables had types.
I would try stripping what you have back until it works then add back what you want it to do.

Related

Absolute value of a number

I want to take the absolute of a number by the following code in bash:
#!/bin/bash
echo "Enter the first file name: "
read first
echo "Enter the second file name: "
read second
s1=$(stat --format=%s "$first")
s2=$(stat -c '%s' "$second")
res= expr $s2 - $s1
if [ "$res" -lt 0 ]
then
res=$res \* -1
fi
echo $res
Now the problem I am facing is in the if statement, no matter what I changes it always goes in the if, I tried to put [[ ]] around the statement but nothing.
Here is the error:
./p6.sh: line 13: [: : integer expression expected
You might just take ${var#-}.
${var#Pattern} Remove from $var the shortest part of $Pattern that matches the front end of $var. tdlp
Example:
s2=5; s1=4
s3=$((s1-s2))
echo $s3
-1
echo ${s3#-}
1
$ s2=5 s1=4
$ echo $s2 $s1
5 4
$ res= expr $s2 - $s1
1
$ echo $res
What's actually happening on the fourth line is that res is being set to nothing and exported for the expr command. Thus, when you run [ "$res" -lt 0 ] res is expanding to nothing and you see the error.
You could just use an arithmetic expression:
$ (( res=s2-s1 ))
$ echo $res
1
Arithmetic context guarantees the result will be an integer, so even if all your terms are undefined to begin with, you will get an integer result (namely zero).
$ (( res = whoknows - whocares )); echo $res
0
Alternatively, you can tell the shell that res is an integer by declaring it as such:
$ declare -i res
$ res=s2-s1
The interesting thing here is that the right hand side of an assignment is treated in arithmetic context, so you don't need the $ for the expansions.
I know this thread is WAY old at this point, but I wanted to share a function I wrote that could help with this:
abs() {
[[ $[ $# ] -lt 0 ]] && echo "$[ ($#) * -1 ]" || echo "$[ $# ]"
}
This will take any mathematical/numeric expression as an argument and return the absolute value. For instance: abs -4 => 4 or abs 5-8 => 3
A workaround: try to eliminate the minus sign.
with sed
x=-12
x=$( sed "s/-//" <<< $x )
echo $x
12
Checking the first character with parameter expansion
x=-12
[[ ${x:0:1} = '-' ]] && x=${x:1} || :
echo $x
12
This syntax is a ternary opeartor. The colon ':' is the do-nothing instruction.
or substitute the '-' sign with nothing (again parameter expansion)
x=-12
echo ${x/-/}
12
Personally, scripting bash appears easier to me when I think string-first.
I translated this solution to bash. I like it more than the accepted string manipulation method or other conditionals because it keeps the abs() process inside the mathematical section
abs_x=$(( x * ((x>0) - (x<0)) ))
x=-3
abs_x= -3 * (0-1) = 3
x=4
abs_x= 4 * (1-0) = 4
For the purist, assuming bash and a relatively recent one (I tested on 4.2 and 5.1):
abs() {
declare -i _value
_value=$1
(( _value < 0 )) && _value=$(( _value * -1 ))
printf "%d\n" $_value
}
If you don't care about the math and only the result matters, you may use
echo $res | awk -F- '{print $NF}'
The simplest solution:
res="${res/#-}"
Deletes only one / occurrence if - is at the first # character.

printing line numbers that are multiple of 5

Hi I am trying to print/echo line numbers that are multiple of 5. I am doing this in shell script. I am getting errors and unable to proceed. below is the script
#!/bin/bash
x=0
y=$wc -l $1
while [ $x -le $y ]
do
sed -n `$x`p $1
x=$(( $x + 5 ))
done
When executing above script i get below errors
#./echo5.sh sample.h
./echo5.sh: line 3: -l: command not found
./echo5.sh: line 4: [: 0: unary operator expected
Please help me with this issue.
For efficiency, you don't want to be invoking sed multiple times on your file just to select a particular line. You want to read through the file once, filtering out the lines you don't want.
#!/bin/bash
i=0
while IFS= read -r line; do
(( ++i % 5 == 0 )) && echo "$line"
done < "$1"
Demo:
$ i=0; while read line; do (( ++i % 5 == 0 )) && echo "$line"; done < <(seq 42)
5
10
15
20
25
30
35
40
A funny pure Bash possibility:
#!/bin/bash
mapfile ary < "$1"
printf "%.0s%.0s%.0s%.0s%s" "${ary[#]}"
This slurps the file into an array ary, which each line of the file in a field of the array. Then printf takes care of printing one every 5 lines: %.0s takes a field, but does nothing, and %s prints the field. Since mapfile is used without the -t option, the newlines are included in the array. Of course this really slurps the file into memory, so it might not be good for huge files. For large files you can use a callback with mapfile:
#!/bin/bash
callback() {
printf '%s' "$2"
ary=()
}
mapfile -c 5 -C callback ary < "$1"
We're removing all the elements of the array during the callback, so that the array doesn't grow too large, and the printing is done on the fly, as the file is read.
Another funny possibility, in the spirit of glenn jackmann's solution, yet without a counter (and still pure Bash):
#!/bin/bash
while read && read && read && read && IFS= read -r line; do
printf '%s\n' "$line"
done < "$1"
Use sed.
sed -n '0~5p' $1
This prints every fifth line in the file starting from 0
Also
y=$wc -l $1
wont work
y=$(wc -l < $1)
You need to create a subshell as bash will see the spaces as the end of the assignment, also if you just want the number its best to redirect the file into wc.
Dont know what you were trying to do with this ?
x=$(( $x + 5 ))
Guessing you were trying to use let, so id suggest looking up the syntax for that command. It would look more like
(( x = x + 5 ))
Hope this helps
There are cleaner ways to do it, but what you're looking for is this.
#!/bin/bash
x=5
y=`wc -l $1`
y=`echo $y | cut -f1 -d\ `
while [ "$y" -gt "$x" ]
do
sed -n "${x}p" "$1"
x=$(( $x + 5 ))
done
Initialize x to 5, since there is no "line zero" in your file $1.
Also, wc -l $1 will display the number of line counts, followed by the name of the file. Use cut to strip the file name out and keep just the first word.
In conditionals, a value of zero can be interpreted as "true" in Bash.
You should not have space between your $x and your p in your sed command. You can put them right next to each other using curly braces.
You can do this quite succinctly using awk:
awk 'NR % 5 == 0' "$1"
NR is the record number (line number in this case). Whenever it is a multiple of 5, the expression is true, so the line is printed.
You might also like the even shorter but slightly less readable:
awk '!(NR%5)' "$1"
which does the same thing.

Using bash, how to assign integer to variable using echo

I'd like to understand bash a bit better as I'm apparently horrible at it...
I'm trying to generate a sequence of constant width integers, but then test them to do something exceptional for particular values. Like so:
for n in $(seq -w 1 150)
do
# The next line does not work: doit.sh: line 9: XX: command not found
#decval= $( echo ${n} | sed 's/^0//g' | sed 's/^0//g' )
#if [[ ${decal} -eq 98 ]] ; then
if [[ $( echo ${n} | sed 's/^0//g' | sed 's/^0//g' ) -eq 98 ]] ; then
echo "Do something different for 98"
elif [[ $( echo ${n} | sed 's/^0//g' | sed 's/^0//g' ) -eq 105 ]] ; then
echo "Do something different for 98"
fi
done
This script works for my purposes, but if I try and make the assignment 'decval= $(…' I get an error 'command not found'. I don't understand this, can someone explain?
Also, is there an improvement I can make to this script if I have a large number of exceptions to prevent a long list of if ; then elif … ?
The problem is in the space between = and $:
decval= $(…
You should write without spaces:
decval=$(...
Because, if you write the space, your shell reads decval= as declval="" and treats the result of $(echo...) as the name of a command to execute, and obviously it doesn't find the command.
Also (just a small optimization), you can write:
sed 's/^0\+//'
instead of
sed 's/^0//g' | sed 's/^0//g'
Here:
0\+ means 0 one or more times;
g is removed, because g means replace all occurences in the string, and you have only one occurence (^ can be only one time in a string).
Also, you can check your variable even with leading zeros, without sed:
[[ "$n" =~ "0*98" ]]

Unexpected behaviour of for

Script:
#!/bin/bash
IFS=','
i=0
for j in `cat database | head -n 1`; do
variables[$i]=$j
i=`expr $i + 1`
done
k=0
for l in `cat database | tail -n $(expr $(cat database | wc -l) - 1)`; do
echo -n $k
k=`expr $k + 1`
if [ $k -eq 3 ]; then
k=0
fi
done
Input file
a,b,c
d,e,f
g,e,f
Output
01201
Expected output
012012
The question is why the for skips last echo? It is weird, because if I change $k to $l echo will run 6 times.
Update:
#thom's analysis is correct. You can fix the problem by changing IFS=',' to IFS=$',\n'.
My original statements below may be of general interest, but do not address the specific problem.
If accidental shell expansions were a concern, here's how the loop could be rewritten (assuming it's practical to read everything into an array variable first):
IFS=$',\n' read -d '' -r -a fields < <(echo $'*,b,c\nd,e,f\ng,h,i')
for field in "${fields[#]}"; do
  # $field is '*' in 1st iteration, then 'b', 'c', 'd',...
done
Original statements:
Just a few general pointers:
You should use a while loop rather than for to read command output - see http://mywiki.wooledge.org/BashFAQ/001; the short of it: with for, the input lines are subject to various shell expansions.
A missing iteration typically stems from the last input line missing a terminating \n (or a separator as defined in $IFS). With a while loop, you can use the following approach to address this: while read -r line || [[ -n $line ]]; do …
For instance, your 2nd for loop could be rewritten as (using process substitution as input to avoid creating a subshell with a separate variable scope):
while read -r l || [[ -n $l ]]; do …; done < <(cat database | tail -n $(expr $(cat database | wc -l) - 1))
Finally, you could benefit from using modern bashisms: for instance,
k=`expr $k + 1`
could be rewritten much more succinctly as (( ++k )) (which will run faster, too).
Your code expects after EVERY read variable a comma but you only give this:
a,b,c
d,e,f
g,e,f
instead of this:
a,b,c,
d,e,f,
g,e,f,
so it reads:
d,e,f'\n'g,e,f
and that is equal to 5 values, not 6

bashscript for changing wput uploaddirectory on every 5000 passage

how i can change the uploaddirectory of wput on every 5000 passage?
whats wrong in my code?
only for the record; i need infnite loop i want run this script 24h 365 days!
every 5000 passage one number higher in directory like incoming1-infinite
best regards
haug
#!/bin/bash
for (( ; ; ))
do
no=1
while [ $no -le 5000 ]
do
perl job.pl false false 1
cd comp
for fname in *.jpg;
do
mv -i "$fname" ${RANDOM}${RANDOM}.jpg
done
for fname in *.jpg;
do
mv "$fname" $(echo "$fname" | sha1sum | cut -f1 -d' ').jpg
done
if [ $no = '5000' ]
then
echo $no
echo $n
n=$(( n+1 )) # <-- this one not do what i need :-(
else
wput *.jpg ftp://user:pass#ip:port/incoming$n/ # <-- here is the important part 2, i need it to change one directory higher every 5000 passage for example; incoming1 until infinite
rm -rf *.jpg
cd ..
fi
no=`expr $no + 1`
done
done
cd comp
for ((no=1; ; no++))
do
n=`expr $no / 5000`
perl job.pl false false 1
for fname in *.jpg;
# ... rest of the script
done
Just calculate the remainder of division by 5000 and if it is equal 0 (or 1 or anything < 5000) do what you need.
if [ $((no%5000)) = 0 ]
then
...
fi
Just put a for-loop that executes 5,000 times inside your infinite loop (which can keep track of the directory sequence number for you):
#!/bin/bash
for (( n=0; ; n++ )); do
for (( no=1; no < 5000; no++ )); do
perl job.pl false false 1
( # Subshell so we don't have to worry about returning
# to the previous directory.
cd comp
# No need to rename each file twice.
for fname in *.jpg; do
mv "$fname" $(echo "${RANDOM}${RANDOM}.jpg" | sha1sum | cut -f1 -d' ').jpg
done
wput *.jpg ftp://user:pass#ip:port/incoming$n
)
done
done

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