Executing while loop issues KSH - unix

I'm trying to execute a ksh where in the file it contains
a = 0
max = 25
while [[$a -ne $max]];
do
echo "$a"
a = $((a+5))
done
It's giving me an error
"[[0: not found [No such file or directory]"
I will check the while loop and it says 2 brackets are needs and ne is for numerical which I did.....I don't know why it is not working....

Try this code,
a=0
max=25
while [ $a -ne $max ]; #Changed here
do
echo "$a"
a=$((a+5))
done

For integers you can use
max=25
for (( a=0; a<max; a+=5 )); do
echo "a=$a"
done

Related

Break the nested while loops in unix scripting

Have two files:
file1 is having the key words - INFO ERROR
file2 is having the list of log files path - path1 path2
I need to exit out of the script if any of the condition in any of the loops failed.
Here is the Code:
#!/bin/bash
RC=0
while read line
do
echo "grepping from the file $line
if [ -f $line ]; then
while read key
do
echo "searching $key from the file $line
if [ condition ]; then
RC=0;
else
RC=1;
break;
fi
done < /apps/file1
else
RC=1;
break;
fi
done < apps/file2
exit $RC
Thank you!
The ansewer to your question is using break 2:
while true; do
sleep 1
echo "outer loop"
while true; do
echo "inner loop"
break 2
done
done
I never use this, it is terrible when you want to understand or modify the code.
Already better is using a boolean
found_master=
while [ -n "${found_master}" ]; do
sleep 1
echo "outer loop"
while true; do
echo "inner loop"
found_master=true
break
done
done
When you do not need the variable found_master it is an ugly additional variable.
You can use a function
inner_loop() {
local i=0;
while ((i++ < 5)); do
((random=$RANDOM%5))
echo "Inner $i: ${random}"
if [ ${random} -eq 0 ]; then
echo "Returning 0"
return 0
fi
done;
return 1;
}
j=0
while ((j++ < 5 )); do
echo "Out loop $j"
inner_loop
if [ $? -eq 0 ]; then
echo "inner look broken"
break
fi
done
But your original problem can be handles without two while loops.
You can use grep -E "INFO|ERROR" file2 or combining the keywords. When the keywords are on different lines in file1, you can use grep -f file1 file2.
Replace condition with $(grep -c ${key} ${line}) -gt 0 like this:
echo "searching $key from the file $line
if [ $(grep -c ${key} ${line}) -eq 0 ]; then
It will count the each key-word in your log-file. If count=0 (pattern didn't found), running then. If found at least 1 key, running else, RC=1 and exit from loop.
And be sure, that your key-words can't be substrings of the longest words, or you will get an error.
Example:
[sahaquiel#sahaquiel-PC Stackoverflow]$ cat file
correctstringERROR and more useless text
ERROR thats really error string
[sahaquiel#sahaquiel-PC Stackoverflow]$ grep -c ERROR file
2
If you wish to avoid count 2 (because counting first string, obliviously, bad way), you should also add two keys for grep:
[sahaquiel#sahaquiel-PC Stackoverflow]$ grep -cow ERROR file
1
Now you have counted only the words equal to your key, not substrings in any useful strings.

Script Unix (Not equal to a string)

I wrote the following code :
echo "Choose between the following options:"
echo "1 - Create a new file"
echo "2 - Write in an existing file"
echo "3 - Change the path of a file"
echo "4 - Display a file"
echo "5 - Exit"
read number
while [ $number -ne 1 -o $number -ne 2 -o $number -ne 3 -o $number -ne 4 -o $number -ne 5 ]
do
echo "Enter a number between 1 and 5"
read number
done
if [ $number -eq 1 ]; then
echo "Enter a folder name"
read name
while [ -e $name ]
do
echo "The file name already exists enter a new name:"
read name
done
touch $name
fi
if [ $number -eq 2 ]; then
echo "Enter the folder name you want to edit :"
read name
while [ ! -f $name ]
do
echo "The file you are looking does not exist. Enter another file name :"
read name
done
echo "Enter what you want to put in the file :"
read input
echo $input >> $name
fi
if [ $number -eq 3 ]; then
echo "Enter the folder name :"
read name
while [ ! -f $name ]
do
echo "The file you are looking does not exist. Enter another file name :"
read name
done
if [ $number -eq 4 ]; then
echo "Enter the folder name you want to see :"
read name
while [ ! -f $name ]
do
echo "The file you are looking does not exist. Enter another file name :"
read name
done
cat $name
fi
if [ $number -eq 5 ]; then
exit 0
fi
The code works just fine, but on the first condition :
while [ $number -ne 1 -o $number -ne 2 -o $number -ne 3 -o $number -ne 4 -o $number -ne 5 ]
I would like it to work to whatever number or string I put.
For example, if I put hello the program will crash.
Can someone tell me what should my first condition be?
Thank you for your help. And forgive me if my question is not in the rules of the forum (I just subscribed).
Your script says:
while [ $number -ne 1 -o $number -ne 2 -o $number -ne 3 -o $number -ne 4 -o $number -ne 5 ]
So, if $number is 1, it's not equal to 2. And if it's 2, it's not equal to 1. This will always evaluate as true, so you'll never exit the loop.
A variety of options exist that will be compatible with a numeric value and still gracefully handle non-numeric input. The following uses a basic regular expression to determine whether input is a digit from 1 to 5:
while read number && ! expr "$number" : '[1-5]$' >/dev/null; do
echo "Try again" >&2
done
You would probably be better off, though, using a case statement:
while read number; do
case "$number" in
1) function_1 ;;
2) function_2 ;;
... etc
*) echo "Invalid input, please try again." >&2; continue ;;
esac
break
done
The case statement is a more graceful way of expressing what might otherwise be achieved using if..elif..elif..fi:
while read number && ! expr "$number" : '[1-5]$' >/dev/null; do
echo "Try again" >&2
done
if [ "$number" = 1 ]; then
: do_something
elif [ "$number" = 2 ]; then
: do_something
elif [ "$number" = 3 ]; then
: do_something
elif [ "$number" = 4 ]; then
: do_something
elif [ "$number" = 5 ]; then
: do_something
else
echo "What am I doing?" >&2
fi
While this construct technically works, it's inelegant and harder to read. Much better to put your functionality into functions which get called from a case statement.
Note that it's always a good idea to quote your variables within a script like this. Do you know what happens with unquoted variables? If not, then quote your variables. :)
I would strongly suggest structuring your code with a case statement:
case $number in
1) code_for_1;;
2) code_for_2;;
...
*) echo "Invalid number" >&2;;
esac
If you wish to repeat (I would actually recommend taking the number as a command line argument instead of reading from stdin and aborting on an error), you could simply do:
while read number; do
case $number in
...
*) echo 'Invalid number' >&2; continue;;
esac
break
done
And write code_for_{1,2,3,4,5} as functions to factor out the logic. For example:
create_file() {
echo "Enter a file name"
while read name; do
if test -e "$name"; then
echo "The file $name already exists enter a new name" >&2
continue
fi
touch "$name"
break
done
}
while read number; do
case $number in
1) create_file;;
...
*) echo 'Invalid number' >&2; continue;;
esac
break
done

Version Comparison using KSH

I'm trying to write a function to compare the versions of the products.
my versions can be XX.XX.XX or xx-xx-xx either it's separated with "." or "-"
and number of fields can be different either xx.xx or xx.xx.xx or xx.xx.xx.xx
the versions which im gonna compare will identical in delimiters and with the fields
#!/bin/ksh
set -x
compareVersions ()
{
typeset IFS='.'
typeset -a v1=( $1 )
typeset -a v2=( $2 )
typeset n diff
for (( n=0; n<4; n+=1 )); do
diff=$((v1[n]-v2[n]))
if [ $diff -ne 0 ] ; then
[ $diff -le 0 ] && echo '-1' || echo '1'
return
fi
done
echo '0'
} # ---------- end of function compareVersions ----------
#compareVersions "6100-09-03" "6100-09-02"
compareVersions "6100.09.03" "6100.09.02"
Please check and give me suggestions
I have tried with the below thing which i have got a other post.. but there is no luck.. hope there should some modification should be done. I have to use across platforms ( linux, solaris, AIX ) so i have preferred KSH, i have idea only in shell scripting though.
Create arrays from version strings, then loop through them comparing elements one by one and return values accordingly. The following example will compare two version strings and returns either 0 (versions are equal), 1 (the first version string is greater) or 2 (the second version string is greater).
#!/bin/ksh
function vertest {
set -A av1 `echo $1 | sed -e 's/\'$3'/ /g'`
set -A av2 `echo $2 | sed -e 's/\'$3'/ /g'`
for (( i=0; i < ${#av1[#]}; i++ )) ; do
[[ ${av1[$i]} -eq ${av2[$i]} ]] && continue
[[ ${av1[$i]} -gt ${av2[$i]} ]] && return 1
[[ ${av1[$i]} -lt ${av2[$i]} ]] && return 2
done
return 0
}
v1="2-7-2-1"
v2="1-8-0-1"
vertest $v1 $v2 '-'
exit $?
# end of file.
This example will exit to shell with exit code 1. Should you change $v1 to 1-7-2-1, it will exit to shell with exit code 2. And so on, and so forth.
The separator escaping is not complete, but this works with most reasonable separators like a period (.) and a dash (-). This, as well as parameter checking for the vertest() is left as an exercise for the reader.
When the format of both numbers is equal (leading zero as your example), you can use
compareVersions ()
{
val1=$(echo $1| tr -d ".-")
echo ${val1}
val2=$(echo $2| tr -d ".-")
echo ${val2}
if [ ${val1} -gt ${val2} ] ; then
echo 1
return
fi
if [ ${val1} -eq ${val2} ] ; then
echo 0
return
fi
echo '-1'
} # ---------- end of function compareVersions ----------

if [ $? -ne 0 ] then syntax error then unexpected

I have been trying to execute the following UNIX shell script which is not working.
I am running it by KornShell (ksh).
echo $?;
if [ $? -ne 0 ]
then
failed $LINENO-2 $5 $6
fi
failed()
{
echo "$0 failed at line number $1";
echo "moving $2 to failed folder"
}
This is giving an error saying Syntax error:then unexpected.. Basically I have to check for the last executed ksh script's highest/last statement's return code and if it is not equal to zero I have to call function failed with the given parameters. I tried putting semicolon before then but that also did not work.
Can you please help?
Edit1: Based on the inputs I changed code. Still the same problem exists.
ksh ../prescript/Pre_process $1 $2 $3
rc=$?;
if [[ $rc -ne 0 ]];then
echo "failed";
exit 1;
Edit2:
It is working for the then part by using double squared brackets. I feel I used code of bash script for ksh. I am facing problem in function call of failed. Please let me know appropriate way of function call in ksh for this example
This looks like bash rather than ksh
failed() {
echo "$0 failed at line number $1";
echo "moving $2 to failed folder"
}
if [[ $? -ne 0 ]]
then
failed $LINENO-2 $5 $6
fi
You need to be careful. The first operation on $? will usually clear it so that your if won't work anyway.
You would be better off using:
rc=$?
echo $rc
if [ $rc -ne 0 ]
:
Other than that, it works fine for me:
$ grep 1 /dev/null
$ if [ $? -ne 0 ]
> then
> echo xx
> fi
xx
$ grep 1 /dev/null
$ echo $?;
1
$ if [ $? -ne 0 ]
> then
> echo yy
> fi
$ _
Note the lack of output in the last one. That's because the echo has sucked up the return value and overwritten it (since the echo was successful).
As an aside, you should let us know which UNIX and which ksh you're actually using. My working version is ksh93 under Ubuntu. Your mileage may vary if you're using a lesser version.
It looks like, from your update, your only problem now is the function call. That's most likely because you're defining it after using it. The script:
grep 1 /dev/null
rc=$?
if [ $rc -ne 0 ]
then
failed $rc
fi
failed()
{
echo Return code was $1
}
produces:
qq.ksh[6]: failed: not found
while:
failed()
{
echo Return code was $1
}
grep 1 /dev/null
rc=$?
if [ $rc -ne 0 ]
then
failed $rc
fi
produces
Return code was 1
you are missing semicolons at the end of the lines:
if [ $? -ne 0]; then
# …

How to use loops statements in unix shell scripting

How to use loop statements in unix shell scripting for eg while ,for do while. I'm using putty server.
for: Iterate over a list.
$for i in `cat some_file | grep pattern`;do echo $i;done
while loop looks pretty much like C's.
$ i=0;while [ $i -le 10 ];do echo $i;i=`expr $i + 1` ;done
If you are going to use command line only, you could use perl, but I guess this is cheating.
$perl -e '$i=0;while ($i < 10){print $i;$i++;}'
More data
http://www.freeos.com/guides/lsst/
#!/bin/sh
items=(item1 item2 item3)
len=${#items[*]}
i=0
while [ $i -lt $len ]; do
echo ${items[$i]}
let i++
done
exit 0
As well as the 'for' and 'while' loops mentioned by Tom, there is (in classic Bourne and Korn shells at least, but also in Bash on MacOS X and presumably elsewhere too) an 'until' loop:
until [ -f /tmp/sentry.file ]
do
sleep 3
done
This loop terminates when the tested command succeeds, in contrast to the 'while' loop which terminates when the tested command fails.
Also note that you can test a sequence of commands; the last command is the one that counts:
while x=$(ls); [ -n "$x" ]
do
echo $x
done
This continues to echo all the files in the directory until they're all deleted.
to the OP, to iterate over files
for file in *
do
echo "$file"
done
to generate counters
for c in {0..10}
do
echo $c
done
using for loop
max=10
for (( i=0; i<=$max; i++ ));
do
echo $i
done
to iterate through a file in KSH
while read line ; do
echo "line from file $line"
done < filename.txt
echo "sample while loop"
i=0;
while [ $i -le 10 ]
do
echo $i
(( i++ ))
done

Resources