Can someone explain why these echo commands doesn't output [#10] and so on?
# echo [#10]
1
# echo [#11]
1
# echo [#12]
1 2
# echo [#13]
1
# echo [#14]
1
You have a file named "1" and a file named "2" in your current directory.
The shell is performing pattern matching on the glob patterns before handing the results to echo. [#10] is a character class containing a #, a 1 and a 0.
See http://www.gnu.org/software/bash/manual/bashref.html#Pattern-Matching
If you want the literal [#10], etc, you have to enclose it in quotes, single or double doesn't matter.
(to answer the question in your last comment)
You could use the printf(1) command:
printf "Error: %s went wrong. Error code [#%d]\n" "something" $[10+2]
The $[10+2] is here to show how to do arithmetic in shell. You could replace "something" with e.g. $somevariable ...
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'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 [[ $count -gt 0 ]]; then
echo "Test"
fi
This question already has answers here:
Variables in bash seq replacement ({1..10}) [duplicate]
(7 answers)
Brace expansion with a Bash variable - {0..$foo}
(5 answers)
Closed 8 years ago.
I'm making a program in bash that creates a histoplot, using numbers I have created. The numbers are stored as such (where the 1st number is how many words are on a line of a file, and the 2nd number is how many times this amount of words on a line comes up, in a given file.)
1 1
2 4
3 1
4 2
this should produce:
1 #
2 ####
3 #
4 ##
BUT the output I'm getting is:
1 #
2 #
3 #
4 #
however the for loop is not recognising that my variable "hashNo" is a number.
#!/bin/bash
if [ -e $f ] ; then
while read line
do
lineAmnt=${line% *}
hashNo=${line##* }
#VVVV Problem is this line here VVVV
for i in {1..$hashNo}
#This line ^^^^^^^ the {1..$hashNo}
do
hashes+="#"
done
printf "%4s" $lineAmnt
printf " $hashes\n"
hashes=""
done < $1
fi
the code works if I replace hashNo with a number (eg 4 makes 4 hashes in my output) but it needs to be able to change with each line (no all lines on a file will have the same amount of chars in them.
thanks for any help :D
A sequence expression in bash must be formed from either integers or characters, no parameter substitutions take place before hand. That's because, as per the bash doco:
The order of expansions is: brace expansion, tilde expansion, parameter, variable and arithmetic expansion and command substitution (done in a left-to-right fashion), word splitting, and pathname expansion.
In other words, brace expansion (which includes the sequence expression form) happens first.
In any case, this cries out to be done as a function so that it can be done easily from anywhere, and also made more efficient:
#!/bin/bash
hashes() {
sz=$1
while [[ $sz -ge 10 ]]; do
printf "##########"
((sz -= 10))
done
while [[ $sz -gt 0 ]]; do
printf "#"
((sz--))
done
}
echo 1 "$(hashes 1)"
echo 2 "$(hashes 4)"
echo 3 "$(hashes 1)"
echo 4 "$(hashes 2)"
which outputs, as desired:
1 #
2 ####
3 #
4 ##
The use of the first loop (doing ten hashes at a time) will almost certainly be more efficient than adding one character at a time and you can, if you wish, do a size-50 loop before that for even more efficiencies if your values can be larger.
I tried this for (( i=1; i<=$hashNo; i++ )) for the for loop, it seems to be working
Your loop should be
for ((i=0; i<hashNo; i++))
do
hashes+="#"
done
Also you can stick with your loop by the use of eval and command substitution $()
for i in $(eval echo {1..$hashNo})
do
hashes+="#"
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