I'm using .conf which contain keys and values.
Some keys contains numbers like
deployment.conf
EAR_COUNT=2
EAR_1=xxx.ear
EAR_2=yyy.ear
When I try to retrieve that value using particular key and compare with integer value i.e. natural number.
But Whatever I retrieved values from .conf ,it is should be String datatype.
How should I compare both value in Linux Bash script.
Simply : How should I compare two values in Linux.?
Ex :
. ./deployment.conf
count=$EAR_COUNT;
echo "count : $count";
if [ $count -gt 0 ]; then
echo "Test"
fi
I'm getting following error :
count : 2
: integer expression expected30: [: 2
They're all strings in bash, notwithstanding your ability to do typeset-type things to flag them differently.
If you want to do numeric comparisons, just use -eq (or its brethren like -gt, -le) rather than ==, != and so on:
if [[ $num -eq 42 ]] ; then
echo Found the answer
fi
The full range of comparison operators can be found in the bash manpage, under CONDITIONAL EXPRESSIONS.
If you have something that you think should be a number and it's not working, I'll warrant it's not a number. Do something like:
echo "[$count]"
to make sure it doesn't have a newline at the end or, better yet, get a hex dump of it in case it holds strange characters, like Windows line endings:
echo -n $count | od -xcb
The fact that you're seeing:
: integer expression expected30: [: 2
with the : back at the start of the line, rather than the more usual:
-bash: [: XX: integer expression expected
tends to indicate the presence of a carriage return in there, which might be from deployment.conf having those Windows line endings (\r\n rather than the UNIXy \n).
The hex dump should make that obvious, at which point you need to go and clean up your configuration file.
Ref : http://linux.die.net/man/1/bash
-eq, -ne, -lt, -le, -gt, or -ge
These are arithmetic binary operators in bash scripting.
I have checked your code,
deployment.conf
# CONF FILE
EAR_COUNT=5
testArithmetic.sh
#!/bin/bash
. ./deployment.conf
count=$EAR_COUNT;
echo "count : $count";
if [ $count -gt 0 ]; then
echo "Test"
fi
running the above script evaluates to numeric comparison for fine. Share us your conf file contents, if you are facing any issues. If you are including the conf file in your script file, note the conf file must have valid BASH assignments, which means, there should be no space before and after '=' sign.
Also, you have mentioned WAR_COUNT=3 in conf part and used 'count=$EAR_COUNT;' in script part. Please check this too.
Most likely you have some non-integer character like \r in your EAR_COUNT variable. Strip all non-digits while assigning to count like this:
count=${EAR_COUNT//[^[:digit:]]/}
echo "count : $count";
if [[ $count -gt 0 ]]; then
echo "Test"
fi
Related
Below is my requirement. I have a text file that has following content
File name - abc.txt
Content -
apple=0
mango=1
strawberry=10
I need to kick off the subsequent process only if any of the above stated variable has non zero values.
In this case, As two variables have values 1 and 10 respectively, I need to update an indicator - SKIP INDICATOR=N
If all variables have 0 as value, I need to update SKIP INDICATOR=Y
How to achieve this functionality in Linux. Kindly advise.
with very simple greps :
if [ $(grep '=' your_file | grep -v '=0') ]
then
echo "non zero values detected"
SKIP_INDICATOR=N
else
echo "all are zeroes"
SKIP_INDICATOR=Y
fi
Just note that this is a quick and dirty solution and it would NOT work properly if you have for example a=01 or a= 0 (eg with space)
Try:
grep -q '=0*[1-9]' textfile && skip_indicator=N || skip_indicator=Y
=0*[1-9] matches an '=' character followed by zero or more '0' characters followed by a digit in the range 1 to 9.
See Correct Bash and shell script variable capitalization for an explanation of why I changed SKIP_INDICATOR to skip_indicator.
#!/bin/bash
flag=`awk -F'=' '$NF!="0"{print;exit}' input`
if [ ! -z $flag ] ; then
SKIP_INDICATOR=N
echo "some variable value is different from 0. do something"
else
SKIP_INDICATOR=Y
echo "all variables have 0 as value. do another thing."
fi
exit 0
I have a situation where I have large number of numbered variables. I want to evaluate each variable and set variable to a specific string if the condition is matched.
#!/bin/bash
var1=""
var2="1233123213"
var3="22332323222324242"
var4=""
var5=""
for i in 1 2 3 4 5
do
if [ -z "$var{$}i" ]
then
var{$}i="None"
fi
echo "var{$}i \r"
done
but the problem is when I run the script I get following.
{1} \r
{2} \r
{3} \r
{4} \r
{5} \r
How I can fix this.
Use indirect variable expansion in bash with syntax {!var}.
From the man bash page,
If the first character of parameter is an exclamation point (!), a level of variable indirection is introduced. Bash uses the value of the variable formed from the rest of parameter as the name of the variable; this variable is then expanded and that value is used in the rest of the substitution, rather than the value of parameter itself. This is known as indirect expansion. The exclamation point must immediately follow the left brace in order to introduce indirection.
Modify your code to something like below,
for i in 1 2 3 4 5
do
var="var$i"
[ -z "${!var}" ] && declare "var$i"="none"
done
printf "var1=%s\n" "$var1"
printf "var2=%s\n" "$var2"
printf "var3=%s\n" "$var3"
printf "var4=%s\n" "$var4"
printf "var5=%s\n" "$var5"
The syntax "${!var}" in this case evaluates the value of the variable within the string var which is var1, var2, var3... and the declare syntax sets the variable value at run-time, only for those variables that are empty.
Now on printing those variables produces,
var1=none
var2=1233123213
var3=22332323222324242
var4=none
var5=none
Indirect assignment will work here, but in this specific case arrays seem like a good fit :
#!/bin/bash
declare -a var=()
var+=("")
var+=(1233123213)
var+=(22332323222324242)
var+=("")
var+=("")
for i in "${!var[#]}"
do
[[ "${var[$i]}" ]] || var[$i]="None"
echo "Index: $i - Value: ${var[$i]}"
done
Consider using an array instead of numbered variables:
#!/bin/bash
var[1]=""
var[2]="1233123213"
var[3]="22332323222324242"
var[4]=""
var[5]=""
for i in 1 2 3 4 5
do
if [ -z "${var[i]}" ]
then
var[i]="None"
fi
echo "${var[i]} \r"
done
I wrote simple script as follow
#!/bin/bash
auth_type=""
SM_Read-only="Yes"
SM_write-only="No"
echo -e ${SM_Read-only}
echo -e ${SM_Write-only}
if [ "${SM_Read-only}" == "Yes" ] && [ "${SM_Write-only}" == "Yes" ]
then
auth_type="Read Write"
else
auth_type="Read"
fi
echo -e $auth_type
And when i execute it i got following output with errors.
./script.bash: line 5: SM_Read-only=Yes: command not found
./script.bash: line 6: SM_write-only=No: command not found
only
only
Read
Any one know correct way to declare the variable with "-" (dash)?
EDIT:
have getting response from c code and evaluate the variables for example
RESP=`getValue SM_ Read-only ,Write-only 2>${ERR_DEV}`
RC=$?
eval "$RESP"
from above scripts code my c binary getValue know that script want Read-only and Write-only and return value to script.So during eval $RESP in cause error and in my script i access variable by
echo -e ${SM_Read-only}
echo -e ${SM_Write-only}
which also cause error.
Rename the variable name as follows:
SM_Read_only="Yes"
SM_write_only="No"
Please, don't use - minus sign in variable names in bash, please refer to the answer, on how to set the proper variable name in bash.
However if you generate the code, based on others output, you can simply process their output with sed:
RESP=$(getValue SM_ Read-rule,Write-rule 2>${ERR_DEV}|sed "s/-/_/g")
RC=$?
eval "$RESP"
- is not allowed in shell variable names. Only letters, numbers, and underscore, and the first character must be a letter or underscore.
I think you cant have a dash in your variables names, only letters, digits and "_"
Try:
SM_Read_only
Or
SM_ReadOnly
I am trying to verify that the format of a variable is a number and is at least 10 digits long with leading zeros, inside of an expect script.
In a bash script it would look something like this:
[[ "$var" != +([0-9]) ]] && echo "bad input" && exit
while [[ $(echo -n ${var} | wc -c) -lt 10 ]] ; do var="0${var}" ; done
For the following input:
16
I am trying to achieve the following output:
0000000016
The simplest way to check whether a variable has just digits is to use a regular expression. Expect's regular expressions are entirely up to the task:
if {![regexp {^\d+$} $var]} {
puts "bad input"
exit
}
Padding with zeroes is best done by formatting the value; if you know C's printf(), you'll recognize the format:
set var [format "%010d" $var]
Expect is actually just an extension of TCL, so you can use any facility that TCL provides. TCL is an unusual language, but it's not hard to do what you want.
# Set a test string.
set testvar 1234567890
# Store the match (if any) in matchvar.
regexp {\d{10,}} $testvar matchvar
puts $matchvar
# Test that matchvar holds an integer.
string is integer $matchvar
The string is command is relatively new, so you might have to rely on the return value of regexp if your TCL interpreter doesn't support it.
The following code gives
[: -ge: unary operator expected
when
i=0
if [ $i -ge 2 ]
then
#some code
fi
why?
Your problem arises from the fact that $i has a blank value when your statement fails. Always quote your variables when performing comparisons if there is the slightest chance that one of them may be empty, e.g.:
if [ "$i" -ge 2 ] ; then
...
fi
This is because of how the shell treats variables. Assume the original example,
if [ $i -ge 2 ] ; then ...
The first thing that the shell does when executing that particular line of code is substitute the value of $i, just like your favorite editor's search & replace function would. So assume that $i is empty or, even more illustrative, assume that $i is a bunch of spaces! The shell will replace $i as follows:
if [ -ge 2 ] ; then ...
Now that variable substitutions are done, the shell proceeds with the comparison and.... fails because it cannot see anything intelligible to the left of -gt. However, quoting $i:
if [ "$i" -ge 2 ] ; then ...
becomes:
if [ " " -ge 2 ] ; then ...
The shell now sees the double-quotes, and knows that you are actually comparing four blanks to 2 and will skip the if.
You also have the option of specifying a default value for $i if $i is blank, as follows:
if [ "${i:-0}" -ge 2 ] ; then ...
This will substitute the value 0 instead of $i is $i is undefined. I still maintain the quotes because, again, if $i is a bunch of blanks then it does not count as undefined, it will not be replaced with 0, and you will run into the problem once again.
Please read this when you have the time. The shell is treated like a black box by many, but it operates with very few and very simple rules - once you are aware of what those rules are (one of them being how variables work in the shell, as explained above) the shell will have no more secrets for you.
Judging from the error message the value of i was the empty string when you executed it, not 0.
I need to add my 5 cents. I see everybody use [ or [[, but it worth to mention that they are not part of if syntax.
For arithmetic comparisons, use ((...)) instead.
((...)) is an arithmetic command, which returns an exit status of 0 if
the expression is nonzero, or 1 if the expression is zero. Also used
as a synonym for "let", if side effects (assignments) are needed.
See: ArithmeticExpression
Your piece of script works just great. Are you sure you are not assigning anything else before the if to "i"?
A common mistake is also not to leave a space after and before the square brackets.