declare -a result=`$ORACLE_HOME/bin/sqlplus -silent $DBUSER/$DBPASSWORD#$DB << EOF $SQLPLUSOPTIONS $roam_query exit; EOF`
I am trying to pull data from an oracle database and populate a bash variable. The select query works however it returns multiple rows and those rows are returned as a long continuous string. I want to capture each row from the database in an array index for example:
index[0] = row 1 information
index[1] = row 2 information
Please help. All suggestions are appreciated. I checked all documentation without no luck. Thank you. I am using solaris unix
If you have bash version 4, you can use the readarray -t command to do this. Any vaguely recent linux should have bash v4, but I don't know about Solaris.
BTW, I'd also recommend putting double-quotes around variable references (e.g. "$DBUSER/$DBPASSWORD#$DB" instead of just $DBUSER/$DBPASSWORD#$DB) (except in here-documents), using $( ) instead of backticks, and using lower- or mixed-case variable names (there are a bunch of all-caps names with special meanings, and if you use one of those by accident, weird things can happen).
I'm not sure I have the here-document (the SQL commands) right, but here's roughly how I'd do it:
readarray -t result < <("$oracle_home/bin/sqlplus" -silent "$dbuser/$dbpassword#$db" << EOF
$sqlplusoptions $roam_query
exit;
EOF
)
Related
count1=`cat $affected_ip|wc -l`;
echo $count1;
Will these lines of code fetch the count from the file named afffected_ip and print the output?
Need help to perform above task.
I suppose you are trying to count the number of ips affected etc, etc, whereby you have 1 ip per line.
You are not too far off actually.
First of all ( as mentioned by user simon3270 ) , you dont need thesemicolons in this snippet. Variables for assignment in bash dont need the $ too in this scenario.
In your case, running at bash shell
wc - l < affected_ips.txt
And you will get the number of lines / IPs. Grepping etc could help you if you have particular patters to discern.
Share more info if you need more help.
Is it possible to store mongo output into a variable in shell script, below is the example query which prints decimal date
ex: 1489442900000
maxdate =$( echo mongo getmaxdate.js --quiet)
Any help is highly appreciated.
Thanks,
Ashwin.
There should be no space between the variable name and the command during the assignment.
variable_name=$(command1 |command2 .....)
In your case:
maxdate=$( mongo getmaxdate.js --quiet)
Make sure to preserve the formatting of the output. if echo'ed Without double quote the whole content will show up in one line.
echo "$maxdate"
I am writing a curl bash script to test webservices. I will have file_1 which would contain the URL paths
/path/to/url/1/{dynamic_path}.xml
/path/to/url/2/list.xml?{query_param}
Since the values in between {} is dynamic, I am creating a separate file, which will have values for these params. the input would be in key-value pair i.e.,
dynamic_path=123
query_param=shipment
By combining two files, the input should become
/path/to/url/1/123.xml
/path/to/url/2/list.xml?shipment
This is the background of my problem. Now my questions
I am doing it in bash script, and the approach I am using is first reading the file with parameters and parse it based on '=' and store it in key/value pair. so it will be easy to replace i.e., for each url I will find the substring between {} and whatever the text it comes with, I will use it as the key to fetch the value from the array
My approach sounds okay (at least to me) BUT, I just realized that
declare -A input_map is only supported in bashscript higher than 4.0. Now, I am not 100% sure what would be the target environment for my script, since it could run in multiple department.
Is there anything better you could suggest ? Any other approach ? Any other design ?
P.S:
This is the first time i am working on bash script.
Here's a risky way to do it: Assuming the values are in a file named "values"
. values
eval "$( sed 's/^/echo "/; s/{/${/; s/$/"/' file_1 )"
Basically, stick a dollar sign in front of the braces and transform each line into an echo statement.
More effort, with awk:
awk '
NR==FNR {split($0, a, /=/); v[a[1]]=a[2]; next}
(i=index($0, "{")) && (j=index($0,"}")) {
key=substr($0,i+1, j-i-1)
print substr($0, 1, i-1) v[key] substr($0, j+1)
}
' values file_1
There are many ways to do this. You seem to think of putting all inputs in a hashmap, and then iterate over that hashmap. In shell scripting it's more common and practical to process things as a stream using pipelines.
For example, your inputs could be in a csv file:
123,shipment
345,order
Then you could process this file like this:
while IFS=, read path param; do
sed -e "s/{dynamic_path}/$path/" -e "s/{query_param}/$param/" file_1
done < input.csv
The output will be:
/path/to/url/1/123.xml
/path/to/url/2/list.xml?shipment
/path/to/url/1/345.xml
/path/to/url/2/list.xml?order
But this is just an example, there can be so many other ways.
You should definitely start by writing a proof of concept and test it on your deployment server. This example should work in old versions of bash too.
I'm pretty new to shell scripts and am only doing them because its about time I learnt and I need to for work.
I have been looking around and have tried multiple methods to get this working but can't seem to figure it out.
I have a script in which I want to access an SQLite database and store the result of a select statement in a variable.
What I've Tried So Far
This one just echoes whats inside the apostrophe. If I remove the dollar sign before the apostrophe I get the same outcome.
track_name=$'sqlite3 "$database_name" << EOF
select name from track where id = "$required_track";
exit;
EOF'
Here I get a syntax error near "track_name"
sqlite3 "$database_name" << EOF
track_name='select name from track where id = "$required_track";'
exit;
EOF
I have successfully executed the select statement without trying to store it in a variable but its not much use to me without being able to store it...
Any help would be much appreciated
To store the output of a command into a BASH variable you should use:
VAR_NAME=$(command);
For example, if you want to store your system current time into a variable or
the results of a list directory command ejecution:
DATE_EXAMPLE_VAR=$(date); #Stores 'date' command output into DATE_EXAMPLE_VAR
echo $DATE_EXAMPLE_VAR; #Shows DATE_EXAMPLE_VAR contents
DIRCONTENTS=$(ls); #Stores a list of your current directory contents.
Similarly, this should work for sqlite3:
track_name=$(sqlite3 "$database_name" "select name from track where id = $required_track")
I have a bash script I am having some issues with concatenating 2 variables to call a 3rd.
Here is a simplification of the script, but the syntax is eluding me after reading the docs.
server_list_all="server1 server2 server3";
var1 = "server";
var2 = "all";
echo $(($var1_list_$var2));
This is about as close as I get to the right answer, it acknowledges the string and tosses an error on tokenization.
syntax error in expression (error token is "server1 server2 server3....
Not really seeing anything in the docs for this, but it should be doable.
EDIT: Cleaned up a bit
The Bash Reference Manual explains how you can use a neat feature of parameter expansion to do some indirection. In your case, you're interested in finding the contents of a variable whose name is defined by two other variables:
server_list_all="server1 server2 server3"
var1=server
var2=all
combined=${var1}_list_${var2}
echo ${!combined}
The exclamation mark when referring to combined means "use the variable whose name is defined by the contents of combined"
The Advanced Bash Scripting Guide has the answer for you (http://tldp.org/LDP/abs/html/ivr.html). You have two options, the first is classic shell:
#!/bin/bash
server_list_all="server1 server2 server3";
var1="server";
var2="all";
server_var="${var1}_list_${var2}"
eval servers=\$$server_var;
echo $servers
Alternatively you can use the bash shortcut ${!var}
#!/bin/bash
server_list_all="server1 server2 server3";
var1="server";
var2="all";
server_var="${var1}_list_${var2}"
echo ${!server_var}
Either approach works.