I'm currently learning Linux and as an homework, we have to create a few basic shell scripts. Nothing especially complicated but this one is giving me headaches. Here's my code :
until [ "$toPrint" == 'fin' ]
do
echo "Enter file name to print out :" ; read toPrint
sh ./afficher.sh "$toPrint"
done
Basically, I have another script called afficher.sh (I'm french so don't mind the french language used) and it reads whatever file name it gets as a parameter. However, the moment I type "fin", everything is supposed to stop except it still tries to print the file called "fin". I read a bit about the until loop on Internet and once it becomes True, it should stop, which is not my case...
Personally, I'd implement this like so -- with a while loop, not an until loop, and checking for the exit condition separately and explicitly:
while true; do
echo "Enter file name to print out :" ; read toPrint
[ "$toPrint" = fin ] && break
sh ./afficher.sh "$toPrint"
done
If you really want to use the loop's condition, you can do that:
while echo "Enter file name to print out :";
read toPrint &&
[ "$toPrint" != fin ]; do
sh ./afficher.sh "$toPrint"
done
...but personally, I'm less fond of this on aesthetic grounds.
You check the condition at the top of the loop, but you enter the value in the middle of the loop. The next thing you do after reading the value is always pass it to afficher.sh and then once that is done you check its value to see if you should stop. If you don't want to run afficher.sh on the fin value, you'll need to make sure your control flow allows you to do the comparison before you invoke afficher.sh.
Related
I'm trying to combine two lists containing names (if available) and emails with a standard email text in bash (shell)
(I had to delete the irrelevant code as it contains some private info, so some of the code might look unusal.)
The first half of the code checks if there is a name list along with the email list.
The second half combines only the email address and text if no name is available, if the name list is available it also 'tries' to combine the name, email and text.
f1 = email list and f2 = name list.
As you can see in the first half of the code below, $f2 should show the names if the list is available but it does not show anything in the log file.
I been trying to sort this problem out for two days but nothing has worked. When names are available it always outputs as "Hello ..." when it should be "Hello John D..."
#FIRST HALF
if [ "$names" = "no" ]
then
text="Hello..."
elif [ "$names" = "yes" ]
then
text="Hello $f2..."
fi
#SECOND HALF
if [ "$names" = "no" ]
then
for i in $(cat $emaillist); do
echo "$text" >> /root/log
echo "$i" >> /root/log
done
elif [ "$names" = "yes" ]
then
paste $emaillist $namelist | while IFS="$(printf '\t')" read -r f1 f2
do
echo "$text" >> /root/log
echo "$f1" >> /root/log
done
fi
When you run text="Hello $f2", $f2 is looked up at the time of the assignment; an exact string is assigned to text, and only that exact string is used later, on echo "$text".
This is very desirable behavior: If shell variables' values could run arbitrary code, it would be impossible to write shell scripts that handled untrusted data safely... but it does mean that implementing your program requires some changes.
If you want to defer evaluation (looking up the value of $f2 at expansion time rather than assignment), don't use a shell variable at all: Use a function instead.
case $names in
yes) write_greeting() { echo "Hello $name..."; };;
*) write_greeting() { echo "Hello..."; };;
esac
while read -r name <&3 && read -r email <&4; do
write_greeting
echo "$email"
done 3<"$namelist" 4<"$emaillist" >>/root/log
Some enhancements in the code above:
You don't need paste to read from two streams in lockstep; you can simply open them on different file descriptors (above, FDs 3 and 4 are chosen; only 0, 1 and 2 are reserved, so larger numbers could have been selected as well) with a separate read command for each.
Opening your output sink only once for the entire loop (by putting the redirection after the done) is far more efficient than re-opening it every time you want to write a single line.
Expansions, such as "$namelist" and "$emaillist", are always quoted; this makes code more reliable if dealing with filenames with unusual characters (including spaces and glob expressions), or if IFS is at a non-default value.
I am trying to assign a variable from a prompt input choice with no luck. If the user inputs 1, I want target_db_name = "database2".
My code:
while true; do
read -p "What is the table name?" table_name
table_name=${table_name,,}
if hdfs dfs -test -e /foo/$table_name ;
then read -p "What is the target database you want to copy the
“foo.${table_name}” table to?
Your three options are:
1) database1
2) database2
3) database3
Type 1, 2, or 3: " target_db;
(((Here is where I want to state if $target_db = "1" then target_db_name
= "database1", if $target_db = "2" then target_db_name = "database2" etc...)))
read -p "Would you like to begin the HDFS copy with the following configuration:
Target Database: ${target_db_name}
Table Name: ${table_name}
Continue (Y/N):"
else echo "Please provide a valid table name.
Exiting this script" ; exit ; fi
done
I have tried creating another if statement with no luck.
"....Type 1, 2, or 3: " target_db;
else if $target_db = "1" then target_db_name = "edw_qa_history"; fi
if $target_db = "1" then won't work, because what follows if must be a command, not a test expression. Now, the most common command used in if statements is [ (yes, that's actually a command name; it's synonymous with the test command), which takes a test expression (and a close bracket) as its arguments and succeeds or fails depending on whether the expression is true or not. So the correct syntax would be something like:
if [ "$target_db" = "1" ]; then
Note that there are two other differences from what you had: I put double-quotes around the variable reference (almost always a good idea, to avoid may parsing oddities), and added a semicolon before then (needed to indicate where the arguments to [ end and shell syntax resumes). I also notice you have semicolons at the end of many lines of your script; this isn't necessary, the end-of-line is enough to indicate the end of a command. It's only if you have another command (or something like then) on the same line that you need a semicolon as a delimiter.
HOWEVER, as #Barmar pointed out in a comment, case would probably be better than a list of if and elif statements here. case is intended specifically for comparing a string against a list of other strings (or patterns), and executing different things depending on which one it matches. It looks something like this:
case "$target_db" in
1) target_db_name="database1" ;;
2) target_db_name="database2" ;;
3) target_db_name="database3" ;;
*) "Please provide a valid table name. Exiting this script" ; exit ;;
esac
Here, the double-semicolon is needed, even at the end of a line, to indicate the end of each case. Also, note that the * pattern (the last case) matches anything, so it functions like an else would in an if ... elif ... sequence.
Final note: use shellcheck.net to sanity-check your code.
You don't need an if statement to map the number to an array; you just need an array.
db_names=(
"datebase 1"
"database 2"
"database 3"
)
# ...
target_db_name=${db_names[$target_db - 1]}
if [[ -z $target_db_name ]]; then
exit
fi
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 "* $item"
fi
done
And a pointer to the great Advanced Bash-Scripting Guide here: http://tldp.org/LDP/abs/html/
I have a csv file which im trying to loop through with the purpose to find out if an User Input is found inside the csv data. I wrote the following code which sometimes works and others doesn't. It always stops working when I try to compare to a 2+ digit number. It works OK for numbers 1 through 9, but once u enter lets say 56 , or 99 or 100, it stops working.
the csv data is comma delimited, i have about 300 lines they are just like this.
1,John Doe,Calculus I,5.0
1,John Doe,Calculus II,4.3
1,John Doe,Physics II,3.5
2,Mary Poppins,Calculus I,3.7
2,Mary Poppins,Calculus II,4.7
2,Mary Poppins,Physics I,3.7
Data is just like that, all the way down until ID #100 for a total of 300 lines. Both the sh file and csv file are in the same folder, I'm using a fresh installation of Ubuntu 12.04.3, using gedit as the text editor.
I tried Echoing the variables ID and inside the IF conditionals but it doesn't behave the way it should when testing for the same value. Could someone point me out in the right direction. Thanks
Here's the code:
#s!/bin/bash
echo "enter your user ID";
read user;
INPUT_FILE=notas.csv
while IFS="," read r- ID name asignature final;
do
if [$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 "$user" ]]
I'm trying to do something like these:
while[$read!="0"];
In this program
#!/bin/sh
i=0
cont=0
while[$read!="0"]; do
read number
cont=`expr $cont + $number`
i++
done
cont=`expr $cont / $i -1`
echo
I want to stop suming the entries when I give it a 0
tnx
The variable you're reading into is $number, so reference that rather than $read in your loop.
Whitespace is significant, so make sure to include spaces before, after, and between all of the items in your loop. (Confusingly, you must not include spaces in an assignment statement like i=0. i = 0 is wrong.)
For good measure, use double quotes around the variable. That's a good practice so that if the user hits enter without typing a number your script doesn't barf on the empty string.
while [ "$number" != "0" ]; do
Also, your i++ isn't right. There are various ways to write that, the simplest being:
let i++
In this, an infinite loop would be appropriate, since you know the condition on which you want to (hint) break out of the loop. The way to get an infinite loop in sh is: while true; do ...; done
Also, read has a -p option that lets you have a prompt (so you know what you're being asked to enter): read -p "Enter a number: " number