Problem with logical operator in bash script - bash

I read in tldp.com that
if [ $condition1 ] && [ $condition2 ]
Same as: if [ $condition1 -a $condition2 ]
Returns true if both condition1 and condition2 hold true..."
but when I tried
if [ $a == 2 ] || [ $b == 4 ]
then
echo "a or b is correct"
else
echo "a and b are not correct"
fi
it gives error. I'm using bash.

Your logic is ok but your comparison operators are incorrect, you should use the '-eq' for comparing integers and '==' for strings. See 'man test' for quick reference, though it's also documented in 'man bash'.
When using integer comparison it is always best to initialise variables to 0 as well otherwise if they remain unset you will get errors.
As mentioned by c00k, use [[ rather than [ if using bash as it is a builtin so bash will not need to shell out to use the /usr/bin/[ command.
i.e.
a=0;b=0
# do something else with a or b
if [[ $a -eq 2 ]] || [[ $b -eq 4 ]]
then
echo "a or b is correct"
else
echo "a and b are not correct"
fi

If you're using Bash, then drop the single [ and use double ones [[.
For arithmetic operations, use ((.
So you'd want to write this:
if (( a == 2 )) || (( b == 4 )); then
echo "foo"
fi # etc

Did you assign a value to a and b? If not you have to (otherwise you definitely should) quote your variables with double quotes:
if [ "$a" == "2" ] || [ "$b" == "4" ]; then
echo "a or b is correct";
else
echo "a and b are not correct";
fi

Correct me if I'm wrong. && and || are bash comparison
#!/usr/bin/ksh
set -x ###### debug mode on
while :
do
dt=`date '+%M'`
if ([ "$dt" -ge "15"] && [ "$dt" -le "17" ]) || ([ "$dt" -ge "25" ] && [ "$dt" -le "27" ])
then
echo "Time-->minutes between 15 to 17 OR 25 to 27"
else
echo "Time--> minutes out of range"
fi
sleep 300 ##### sleep for 5 minutes
done
------------- below lines are debug output
+ + date +%M
dt=17
+ [ 17 -ge 15 ]
+ [ 17 -le 17 ]
+ echo Time--> minutes between 15 to 17 OR 25 to 27
Time--> minutes between 15 to 17 OR 25 to 27
+ date
Sat Mar 3 15:17:23 AST 2012
+ sleep 300
>+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
+ :
+ + date +%M
dt=22
+ [ 22 -ge 15 ]
+ [ 22 -le 17 ]
+ [ 22 -ge 25 ]
+ echo Time--> minutes out of range
Time--> minutes out of range
+ sleep 300
^C$ ######## breaking out of loop(terminating the execution).
$
$ env |grep -i shell
+ grep -i shell
+ env
SHELL=/bin/ksh
$ set +x ##### ending debug mode.

Related

Merge sort recursive algorithm in Bash cannot exit and return value

I'm implementing a merge sort algorithm in bash, but looks like it loops forever and gives error on m1 and m2 subarrays. It's a bit hard to stop loop in conditions since I have to use echo and not return. Anyone have any idea why this happens?
MergeSort (){
local a=("$#")
if [ ${#a[#]} -eq 1 ]
then
echo ${a[#]}
elif [ ${#a[#]} -eq 2 ]
then
if [ ${a[0]} -gt ${a[1]} ]
then
local t=(${a[0]} ${a[1]})
echo ${t[#]}
else
echo ${a[#]}
fi
else
local p=($(( ${#a[#]} / 2 )))
local m1=$(MergeSort "${a[#]::p}")
local m2=$(MergeSort "${a[#]:p}")
local ret=()
while true
do
if [ "${#m1[#]}" > 0 ] && [ "${#m2[#]}" > 0 ]
then
if [ ${m1[0]} <= ${m2[0]} ]
then
ret+=(${m1[0]})
m1=${m1[#]:1}
else
ret+=(${m2[0]})
m2=${m2[#]:1}
fi
elif [ ${#m1[#]} > 0 ]
then
ret+=(${ret[#]} ${m1[#]})
unset m1
elif [ ${#m2[#]} > 0 ]
then
ret+=(${ret[#]} ${m2[#]})
unset m2
else
break
fi
done
fi
echo ${ret[#]}
}
a=(6 5 6 4 2)
b=$(MergeSort "${a[#]}")
echo ${b[#]}
There are multiple issues in your shell script:
you should use -gt instead of > for numeric comparisons on array lengths
<= is not a supported string comparison operator. You should use < and quote it as '<', or better use '>' and transpose actions to preserve sort stability.
there is no need for local t, and your code does not swap the arguments. Just use echo ${a[1]} ${a[0]}
you must parse the result of recursive calls to MergeSort as arrays: local m1=($(MergeSort "${a[#]::p}"))
when popping initial elements from m1 and m2, you must reparse as arrays: m1=(${m1[#]:1})
instead of ret+=(${ret[#]} ${m1[#]}) you should just append the elements with ret+=(${m1[#]}) and instead of unset m1, you should break from the loop. As a matter of fact, if either array is empty you should just append the remaining elements from both arrays and break.
furthermore, the while true loop should be simplified as a while [ ${#m1[#]} -gt 0 ] && [ ${#m2[#]} -gt 0 ] loop followed by the tail handling.
the final echo ${ret[#]} should be moved inside the else branch of the last if
to handle embedded spaces, you should stringize all expansions but as the resulting array is expanded with echo embedded spaces that appear in the output are indistinguishable from word breaks. There is no easy workaround for this limitation.
Here is a modified version:
#!/bin/bash
MergeSort (){
local a=("$#")
if [ ${#a[#]} -eq 1 ]; then
echo "${a[#]}"
elif [ ${#a[#]} -eq 2 ]; then
if [ "${a[0]}" '>' "${a[1]}" ]; then
echo "${a[1]}" "${a[0]}"
else
echo "${a[#]}"
fi
else
local p=($(( ${#a[#]} / 2 )))
local m1=($(MergeSort "${a[#]::p}"))
local m2=($(MergeSort "${a[#]:p}"))
local ret=()
while [ ${#m1[#]} -gt 0 ] && [ ${#m2[#]} -gt 0 ]; do
if [ "${m1[0]}" '>' "${m2[0]}" ]; then
ret+=("${m2[0]}")
m2=("${m2[#]:1}")
else
ret+=("${m1[0]}")
m1=("${m1[#]:1}")
fi
done
echo "${ret[#]}" "${m1[#]}" "${m2[#]}"
fi
}
a=(6 5 6 4 2 a c b c aa 00 0 000)
b=($(MergeSort "${a[#]}"))
echo "${b[#]}"
Output: 0 00 000 2 4 5 6 6 a aa b c c

Using if and controlling numbers in ksh

I would like to download file with the format cars000.txt, cars003.txt,cars006.txt, till cars105.txt...interval of 3 as you can see
I use the following code in ksh, but after downloading cars012.txt, it fails, it begins to download cars13.txt,...and I don't wish it. What does it fails in the code?
FHR=000
while [ $FHR -le 105 ]
do
file=cars${FHR}.txt
wget http://${dir_target}/${file}
(( FHR = $FHR + 03 ))
echo $FHR
if [[ $FHR -le 10 ]]; then FHR="00"$FHR
else FHR="0"$FHR
fi
done
You should decide: is FHR a string, a decimal or an octal.
You are mixing them currently.
Try the next improvement:
FHR=0
while [ ${FHR} -le 105 ]; do
file=cars${FHR}.txt
(( FHR = FHR + 3 ))
echo Without leading zero: FHR=${FHR}
if [[ $FHR -le 10 ]]; then
echo "FHR=00${FHR}"
else
echo "FHR=0${FHR}"
fi
sleep 1
done
(The next improvement might be using printf or awk and no zero for 102/105)

shell program does not show the value of the character given as input?

I am inputting a single character from the user and trying to print the ascii value of the character if it's value is >=97 and <=121
This is my code and it does not work.
echo "Enter a character"
read n
if ["'${n}" -ge 97 and "'${n}" -le 121]
then
print "%d","'$n"
fi
Error:
ascii.sh: 3: ascii.sh: ['a: not found
[ is a command in shell, aka test command.
You need spaces around [ and ].
Moreover, in order to compare integers you need to use < and >.
EDIT: In order to fix, you could say:
read n
asc=$(printf "%d" "'$n")
[[ "$asc" > 97 ]] && [[ "$asc" < 122 ]] && echo $asc
If you're using sh, you could change the last line to:
[ "$asc" -gt 97 ] && [ "$asc" -le 121 ] && echo $asc
Something like this will make it:
#!/bin/sh
min=97
max=121
echo "Enter a character"
read n
value=$(printf "%d" "'$n")
printf "The ASCII character of %s is %d.\n" "$n" "$value"
if [ "${value}" -le $min ] && [ "${value}" -le $max ]
then
printf "%d not in the range.\n" "$value"
else
printf "%d in the range.\n" "$value"
fi
Things that I changed:
- printf instead of print.
- ["'${n}" -ge 97 and "'${n}" -le 121] condition needed to be splited in two blocks:
if [ "${n}" -le 97 ] && [ "${n}" -le 121 ]

Bash if statement with multiple conditions throws an error

I'm trying to write a script that will check two error flags, and in case one flag (or both) are changed it'll echo-- error happened. My script:
my_error_flag=0
my_error_flag_o=0
do something.....
if [[ "$my_error_flag"=="1" || "$my_error_flag_o"=="2" ] || [ "$my_error_flag"="1" && "$my_error_flag_o"="2" ]]; then
echo "$my_error_flag"
else
echo "no flag"
fi
Basically, it should be, something along:
if ((a=1 or b=2) or (a=1 and b=2))
then
display error
else
no error
fi
The error I get is:
line 26: conditional binary operator expected
line 26: syntax error near `]'
line 26: `if [[ "$my_error_flag"=="1" || "$my_error_flag_o"=="2" ] || [ "$my_error_flag"="1" && "$my_error_flag_o"="2" ]]; then'
Are my brackets messed up?
Use -a (for and) and -o (for or) operations.
tldp.org/LDP/Bash-Beginners-Guide/html/sect_07_01.html
Update
Actually you could still use && and || with the -eq operation. So your script would be like this:
my_error_flag=1
my_error_flag_o=1
if [ $my_error_flag -eq 1 ] || [ $my_error_flag_o -eq 2 ] || ([ $my_error_flag -eq 1 ] && [ $my_error_flag_o -eq 2 ]); then
echo "$my_error_flag"
else
echo "no flag"
fi
Although in your case you can discard the last two expressions and just stick with one or operation like this:
my_error_flag=1
my_error_flag_o=1
if [ $my_error_flag -eq 1 ] || [ $my_error_flag_o -eq 2 ]; then
echo "$my_error_flag"
else
echo "no flag"
fi
You can use either [[ or (( keyword. When you use [[ keyword, you have to use string operators such as -eq, -lt. I think, (( is most preferred for arithmetic, because you can directly use operators such as ==, < and >.
Using [[ operator
a=$1
b=$2
if [[ a -eq 1 || b -eq 2 ]] || [[ a -eq 3 && b -eq 4 ]]
then
echo "Error"
else
echo "No Error"
fi
Using (( operator
a=$1
b=$2
if (( a == 1 || b == 2 )) || (( a == 3 && b == 4 ))
then
echo "Error"
else
echo "No Error"
fi
Do not use -a or -o operators Since it is not Portable.
Please try following
if ([ $dateR -ge 234 ] && [ $dateR -lt 238 ]) || ([ $dateR -ge 834 ] && [ $dateR -lt 838 ]) || ([ $dateR -ge 1434 ] && [ $dateR -lt 1438 ]) || ([ $dateR -ge 2034 ] && [ $dateR -lt 2038 ]) ;
then
echo "WORKING"
else
echo "Out of range!"
You can get some inspiration by reading an entrypoint.sh script written by the contributors from MySQL that checks whether the specified variables were set.
As the script shows, you can pipe them with -a, e.g.:
if [ -z "$MYSQL_ROOT_PASSWORD" -a -z "$MYSQL_ALLOW_EMPTY_PASSWORD" -a -z "$MYSQL_RANDOM_ROOT_PASSWORD" ]; then
...
fi

Shell Script Too Many Arguments for if condition

My current script does the following;
It takes integer as a command line argument and starts from 1 to N , it checks whether the numbers are divisible by 3, 5 or both of them. It simply prints out Uc for 3, Bes for 5 and UcBes for 3,5. If the command line argument is empty, it does the same operation but the loop goes to 1 to 20.
I am having this error "Too many arguments at line 11,15 and 19".
Here is the code:
#!/bin/bash
if [ ! -z $1 ]; then
for i in `seq 1 $1`
do
if [ [$i % 3] -eq 0 ]; then
echo "Uc"
elif [ i % 5 -eq 0 ]; then
echo "Bes"
elif [ i % 3 -eq 0 ] && [ i % 5 -eq 0 ]
then
echo "UcBes"
else
echo "$i"
fi
done
elif [ -z $1 ]
then
for i in {1..20}
do
if [ i % 3 -eq 0 ]
then
echo "Uc"
elif [ i % 5 -eq 0 ]
then
echo "Bes"
elif [ i % 3 -eq 0 ] && [ i % 5 -eq 0 ]
then
echo "UcBes"
else
echo "$i"
fi
done
else
echo "heheheh"
fi
Note that [ is actually synonym for the test builtin in shell (try which [ in your terminal), and not a conditional syntax like other languages, so you cannot do:
if [ [$i % 3] -eq 0 ]; then
Moreover, always make sure that there is at least one space between [, ], and the variables that comprise the logical condition check in between them.
The syntax for evaluating an expression such as modulo is enclosure by $((...)), and the variable names inside need not be prefixed by $:
remainder=$((i % 3))
if [ $remainder -eq 0 ]; then
You should probably use something like :
if [ $(($i % 3)) -eq 0 ]
instead of
if [ $i % 3 -eq 0 ]
if [ [$i % 3] -eq 0 ]
Your script could be greatly simplified. For example:
#!/bin/sh
n=0
while test $(( ++n )) -le ${1:-20}; do
t=$n
expr $n % 3 > /dev/null || { printf Uc; t=; }
expr $n % 5 > /dev/null || { printf Bes; t=; }
echo $t
done
gives slightly different error messages if the argument is not an integer, but otherwise behaves the same.

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