Does exist in linux bash something similar to the following code in PHP:
list($var1, $var2, $var3) = function_that_returns_a_three_element_array() ;
i.e. you assign in one sentence a corresponding value to 3 different variables.
Let's say I have the bash function myBashFuntion that writes to stdout the string "qwert asdfg zxcvb".
Is it possible to do something like:
(var1 var2 var3) = ( `myBashFuntion param1 param2` )
The part at the left of the equal sign is not valid syntax of course. I'm just trying to explain what I'm asking for.
What does work, though, is the following:
array = ( `myBashFuntion param1 param2` )
echo ${array[0]} ${array[1]} ${array[2]}
But an indexed array is not as descriptive as plain variable names.
However, I could just do:
var1 = ${array[0]} ; var2 = ${array[1]} ; var3 = ${array[2]}
But those are 3 more statements that I'd prefer to avoid.
I'm just looking for a shortcut syntax. Is it possible?
First thing that comes into my mind:
read -r a b c <<<$(echo 1 2 3) ; echo "$a|$b|$c"
output is, unsurprisingly
1|2|3
I wanted to assign the values to an array. So, extending Michael Krelin's approach, I did:
read a[{1..3}] <<< $(echo 2 4 6); echo "${a[1]}|${a[2]}|${a[3]}"
which yields:
2|4|6
as expected.
I think this might help...
In order to break down user inputted dates (mm/dd/yyyy) in my scripts, I store the day, month, and year into an array, and then put the values into separate variables as follows:
DATE_ARRAY=(`echo $2 | sed -e 's/\// /g'`)
MONTH=(`echo ${DATE_ARRAY[0]}`)
DAY=(`echo ${DATE_ARRAY[1]}`)
YEAR=(`echo ${DATE_ARRAY[2]}`)
Sometimes you have to do something funky. Let's say you want to read from a command (the date example by SDGuero for example) but you want to avoid multiple forks.
read month day year << DATE_COMMAND
$(date "+%m %d %Y")
DATE_COMMAND
echo $month $day $year
You could also pipe into the read command, but then you'd have to use the variables within a subshell:
day=n/a; month=n/a; year=n/a
date "+%d %m %Y" | { read day month year ; echo $day $month $year; }
echo $day $month $year
results in...
13 08 2013
n/a n/a n/a
Chapter 5 of the Bash Cookbook by O'Reilly, discusses (at some length) the reasons for the requirement in a variable assignment that there be no spaces around the '=' sign
MYVAR="something"
The explanation has something to do with distinguishing between the name of a command and a variable (where '=' may be a valid argument).
This all seems a little like justifying after the event, but in any case there is no mention of a method of assigning to a list of variables.
let var1=var2=var3=0
or
var1=var2=var3="Default value"
Related
What I have is an array with some variables. I can iterate to get the values of those vars but what I need is actually their names (values will be used elsewhere).
Going with var[i] won't work cause I will have different names. I guess I could workaround this by creating another array with the names - something similar to this:
Getting variable values from variable names listed in array in Bash
But I'm wondering if there is a better way to do this.
var1=$'1'
var2=$'2'
var3=$'3'
Array=( $var1 $var2 $var3)
for ((i=0; i<${#Array[#]}; i++))
do
echo ${Array[i]}
done
Is:
>1
>2
>3
Should be:
>var1
>var2
>var3
It sounds like you want an associative array.
# to set values over time
declare -A Array=( ) || { echo "ERROR: Need bash 4.0 or newer" >&2; exit 1; }
Array[var1]=1
Array[var2]=2
Array[var3]=3
This can also be assigned at once:
# or as just one assignment
declare -A Array=( [var1]=1 [var2]=2 [var3]=3 )
Either way, one can iterate over the keys with "${!Array[#]}", and retrieve the value for a key with ${Array[key]}:
for var in "${!Array[#]}"; do
val="${Array[$var]}"
echo "$var -> $val"
done
...will, after either of the assignments up top, properly emit:
var1 -> 1
var2 -> 2
var3 -> 3
What about this solution?
#!/bin/bash
var1=$'1'
var2=$'2'
var3=$'3'
Array=( var1 var2 var3 )
for var in "${Array[#]}"; do
echo "$var = ${!var}"
done
The idea just consists in putting your variable names in the array, then relying on the indirection feature of Bash.
But as pointed out by #CharlesDuffy, the use of associative arrays sounds better adapted to the OP's use case.
Also, this related article may be worth reading: How can I use variable variables… or associative arrays?
I'm practicing bash and honestly, it is pretty fun. However, I'm trying to write a program that compares an array's value to a variable and if they are the same then it should print the array's value with an asterisk to the left of it.
#!/bin/bash
color[0]=red
color[1]=blue
color[2]=black
color[3]=brown
color[4]=yellow
favorite="black"
for i in {0..4};do echo ${color[$i]};
if {"$favorite"=$color[i]}; then
echo"* $color[i]"
done
output should be *black
There's few incorrect statements in your code that prevent it from doing what you ask it to. The comparison in bash is done withing square brackets, leaving space around them. You correctly use the = for string comparison, but should enclose in " the string variable. Also, while you correctly address the element array in the echo statement, you don't do so inside the comparison, where it should read ${color[$i]} as well. Same error in the asterisk print. So, here a reworked code with the fixes, but read more below.
#!/bin/bash
color[0]=red
color[1]=blue
color[2]=black
color[3]=brown
color[4]=yellow
favorite=black
for i in {0..4};do
echo ${color[$i]};
if [ "$favorite" = "${color[$i]}" ]; then
echo "* ${color[$i]}"
fi
done
While that code works now, few things that probably I like and would suggest (open to more expert input of course by the SO community): always enclose strings in ", as it makes evident it is a string variable; when looping an array, no need to use index variables; enclose variables always within ${}.
So my version of the same code would be:
#!/bin/bash
color=("red" "blue" "black" "brown" "yellow")
favorite="black"
for item in ${color[#]}; do
echo ${item}
if [ "${item}" = "${favorite}" ]; then
echo "* $item"
fi
done
And a pointer to the great Advanced Bash-Scripting Guide here: http://tldp.org/LDP/abs/html/
Hopefully I can make it clear. I would like to create a filename out of different strings in bash. For example hmd.sh so h, m, d are different values (number 0..9 or letter aA..zZ). So for example I want to convert
h=1 m=11 and d=12 to 1aA.sh. h=> 1, m=>a and d=>A
To declare variables like
a01=1; a02=2 .. a09=9, a10=0; a11=a; a12=b and so on. h(1)=a01=1 m(11)=a11=a
and
d(12)=a12=A.
To test it I wrote this:
#!/bin/bash
dd01="1"
aa="01"
bb="dd$aa"
echo $bb
But of course $bb is dd01 and not its value. How can I make $bb its value of 1?
Associative arrays make this kind of thing much more readable.
However your answer is "variable indirection"
$ echo $bb
dd01
$ echo ${!bb}
1
Do not listen to any advice suggesting eval -- you open yourself up to all kinds of code injection.
The only way to expand a variable inside another is in an array when the variable is enclosed in the key's brackets like [$var].
You could store your values in an associative array, and reference them like so:
declare -A arr
arr[dd01]="1"
arr[aa]="01"
arr[bb]="dd${arr[aa]}"
echo ${arr[${arr[bb]}]}
Using arrays like this may be more convoluted for this example than referencing the variable name using ${!bb} syntax, but if you need to do this while keeping different sets of variables that may need to reference each other, creating an associative array may make more organizational sense.
I rewrote your code example as follows which gives you the value of 1 which is what you are trying to achieve.
#!/bin/bash
dd01="1"
aa="01"
bb="dd$aa"
echo $[$bb]
This did the trick:
#!/bin/bash
dd=1234567890aAbBcC
aa="11"
echo ${dd:(aa-1):1}
Appearantly 1 is the 0 position and aa can also be like 01 and still work!
Thank for all the advices.
I found my answer here:
http://tldp.org/LDP/abs/html/string-manipulation.html
This should do the job:
#!/bin/bash
dd01="1"
aa="01"
bb="dd$aa"
eval echo \$$bb
You can use the '$' operator to access the value of a variable. For example...
d = 'Hi'
e = ' there '
f = 'friend.'
foo = '$d$e$f'
This would cause the value of foo to be 'Hi there friend.'
Hope this helps.
Use eval:
#!/bin/bash
dd01="1"
aa="01"
eval bb=\$dd$aa
echo $bb
This script outputs the expected 1.
I have a csv file which im trying to loop through with the purpose to find out if an User Input is found inside the csv data. I wrote the following code which sometimes works and others doesn't. It always stops working when I try to compare to a 2+ digit number. It works OK for numbers 1 through 9, but once u enter lets say 56 , or 99 or 100, it stops working.
the csv data is comma delimited, i have about 300 lines they are just like this.
1,John Doe,Calculus I,5.0
1,John Doe,Calculus II,4.3
1,John Doe,Physics II,3.5
2,Mary Poppins,Calculus I,3.7
2,Mary Poppins,Calculus II,4.7
2,Mary Poppins,Physics I,3.7
Data is just like that, all the way down until ID #100 for a total of 300 lines. Both the sh file and csv file are in the same folder, I'm using a fresh installation of Ubuntu 12.04.3, using gedit as the text editor.
I tried Echoing the variables ID and inside the IF conditionals but it doesn't behave the way it should when testing for the same value. Could someone point me out in the right direction. Thanks
Here's the code:
#s!/bin/bash
echo "enter your user ID";
read user;
INPUT_FILE=notas.csv
while IFS="," read r- ID name asignature final;
do
if [$ID = $user]; then
userType=1;
else
userType=2;
fi
done < notas.csv
Well, your code as written has a few issues.
You have r- instead of -r on the read line - I assume that's a typo not present in your actual code or you wouldn't get very far.
Similarly, you need space around the [...] brackets: [$ID is a syntax error.
You need to quote the parameter expansions in your if clause, and/or switch bracket types. You probably make it a numeric comparison as #imp25 suggested, which I would do by using ((...)).
You probably don't want to set userType to 2 in an else clause, because that will set it to 2 for everyone except whoever is listed last in the file (ID 100, presumably). You want to set it to 2 first, outside the loop. Then, inside the loop when you find a match, set it to 1 and break out of the loop:
userType=2
while IFS=, read -r ID name asignature final; do
if (( $ID == $user )); then
userType=1;
break
fi
done < notas.csv
You could also just use shell tools like awk:
userType=$(awk -F, -vtype=2 '($1=="'"$user"'") {type=1}; END {print type}' notas.csv)
or grep:
grep -q "^$user," notas.csv
userType=$(( $? + 1 ))
etc.
You should quote your variables in the if test statement. You should also perform a numeric test -eq rather than a string comparison =. So your if statement should look like:
if [[ "$ID" -eq "$user" ]]
It is my understanding that when writing a Unix shell program you can iterate through a string like a list with a for loop. Does this mean you can access elements of the string by their index as well?
For example:
foo="fruit vegetable bread"
How could I access the first word of this sentence? I've tried using brackets like the C-based languages to no avail, and solutions I've read online require regular expressions, which I would like to avoid for now.
Pass $foo as argument to a function. Than you can use $1, $2 and so on to access the corresponding word in the function.
function try {
echo $1
}
a="one two three"
try $a
EDIT: another better version is:
a="one two three"
b=( $a )
echo ${b[0]}
EDIT(2): have a look at this thread.
Using arrays is the best solution.
Here's a tricky way using indirect variables
get() { local idx=${!#}; echo "${!idx}"; }
foo="one two three"
get $foo 1 # one
get $foo 2 # two
get $foo 3 # three
Notes:
$# is the number of parameters given to the function (4 in all these cases)
${!#} is the value of the last parameter
${!idx} is the value of the idx'th parameter
You must not quote $foo so the shell can split the string into words.
With a bit of error checking:
get() {
local idx=${!#}
if (( $idx < 1 || $idx >= $# )); then
echo "index out of bounds" >&2
return 1
fi
echo "${!idx}"
}
Please don't actually use this function. Use an array.