In ksh , pass parameters to mysql select query - mysql

I have a linux script that gets a variable and I store it to var JOB_EXEC_ID
I am trying to pass the value of this to a MySQL query
Here is MySQL query set-up
print "JOB EXEC ID value for DataMeer Job ${LOADJOB} is : ${JobExecId} " |
tee -a ${LOGDIR}/${LOGFILE}
#Log on to MySQL to get the DataId
#Remove first the output file that would house the dataid
rm -f ${SCRDIR}/list_dataid.csv
mysql -u root -pmonday1 ${DAPDBNAME} < ${SCRDIR}/dataid_query.nosql
SQLRTN=$?
if [[ ${SQLRTN} != 0 ]]
then
print "Return code from sqlcall - DAP : ${SQLRTN}" |
tee -a ${LOGDIR}/${LOGFILE}
print "Exiting $Script - 55 " |
tee -a ${LOGDIR}/${LOGFILE}
exit 55
fi
The file dataid_query.nosql looks like this:
set #job_exec_id=10151
select d.id DataId
from data d inner join dap_job_configuration djc on d.dap_job_configuration__id = djc.id
left outer join dap_job_execution dje on djc.id = dje.dap_job_configuration__id and dje.created_data__id = d.id
where dje.id=#job_exec_id
into OUTFILE "/home/app1ebb/cs/list_dataid.csv"
I want to pass the value of JOB_EXEC_ID to the set command that is currently hardcoded right now with a value of 10151

in place of
mysql -u root -pmonday1 ${DAPDBNAME} < ${SCRDIR}/dataid_query.nosql
SQLRTN=$?
this lines
sed "1 s/[0-9]*$/${JOB_EXEC_ID}/" > /tmp/dataid_query.nosql
mysql -u root -pmonday1 ${DAPDBNAME} < /tmp/dataid_query.nosql
SQLRTN=$?
rm /tmp/dataid_query.nosql

Related

Store multiple values from sql select in bash

I am trying to store each value from the following sql select statement and store them in separate variables using bash.
#!/bin/bash
mysqlhost="thehost"
mysqldb="thedb"
mysqlun="theusername"
mysqlpw="thepassword"
mysqlconnection="--disable-column-names --host=$mysqlhost --user $mysqlun --password=$mysqlpw --database=$mysqldb"
declare -a pinIDs=$(mysql $mysqlconnection -e "SELECT pinID FROM somewhere WHERE something = something";)
I get the following result when I use code
echo $pinIDs
8 11 23 26
I need to store each of those values into their own variable.
Add brackets to put output in array pinIDs. Replace
declare -a pinIDs=$(mysql $mysqlconnection -e "SELECT pinID FROM somewhere WHERE something = something";)
by
declare -a pinIDs=( $(mysql $mysqlconnection -e "SELECT pinID FROM somewhere WHERE something = something";) )
Then see output of: declare -p pinIDs
For others that are using this to learn. What I did after fixing the brackets in order to put the result into an array such as (8 11 23 26) instead of 8 11 23 26, was this:
cnt=${#pinIDs[#]}
for (( i=0 ; i<cnt ; i++ ))
do
echo "pinId: ""${pinIDs[$i]}"
done

How to get a value without characters like "- , + "?

This is the normal output:
mysql> select module_id from Modules where Module_name = 'STP_XENA';
+-----------+
| module_id |
+-----------+
| 3 |
+-----------+
1 row in set (0.00 sec)
Can I get answer for the query as only "3"
I need something like,
mysql> select module_id from Modules where Module_name = 'STP_XENA';
3
mysql>
But not from bash or console. Is there any option to do this ?
You cannot do it inside MySQL editor, AFAIK. If you execute the script from console, then adding -B switch can get you desired result.
> mysql -B -u username -p password -e "select module_id from Modules where Module_name = 'STP_XENA';" DBNAME
will yield value with column name:
module_id
3
Also, if you add --skip-column-names
> mysql -B --skip-column-names -u username -p password -e "select module_id from Modules where Module_name = 'STP_XENA';" DBNAME
will yield only value (minus column name):
3
HTH
EDIT: You may start mysql with --skip-column-names switch. I am not sure about -B though. If you are able to start with -B, then great.
You want to use the query result in something like a bash script?
If so, you could do with:
mysql -u{user} -p{password} -s -N -e "select module_id from Modules where Module_name = 'STP_XENA'" database_name
Example:
module_id = `mysql -u{user} -p{password} -s -N -e "select module_id from Modules where Module_name = 'STP_XENA'" database_name`
echo $module_id

Pass MySQL variables to script from command line

I have a MySQL update script I'd like to run from the command line, but I want to be able to pass a stage domain variable to the script.
I know this won't work, but it's the best way I can describe what I'm trying to do:
$ -uroot -hlocalhost mydatabase --execute "SET #domain = 'mydomain.dev' " < ./sql/update_domain.sql
Inside the script, I'm using the #domain variable, to update some configuration variables in a config table, using commands like this:
UPDATE my_cfg SET value = #domain WHERE name = 'DOMAIN';
Basically I want to prefix the SET #domain on the update_domain.sql file.
Any ideas how I can rectify my approach?
In your BATCH File :
mysql -e "set #domain=PARAMVALUE;source ./sql/update_domain.sql"
And in you SQL file :
UPDATE my_cfg SET value = #domain WHERE name = 'DOMAIN';
you can do that with sed like this:
echo "UPDATE my_cfg SET value = '#domain#' WHERE name = 'DOMAIN'" | sed 's/#domain#/mydomain.dev/' | mysql -uusername -ppassword dbname
or update.sql has UPDATE:
cat update.sql | sed 's/#domain#/mydomain.dev/' | mysql -uusername -ppassword dbname
This works for me:
system("(echo \"SET #domain = 'newstore.personera.abc';\"; cat sql/set_domain.sql) > /tmp/_tmp.sql")
system("mysql -uroot -hlocalhost newstore.personera.dev < /tmp/_tmp.sql")
system("rm /tmp/_tmp.sql")
...calling with system() from Capistrano.
I've found a better solution.
--init-command=name SQL Command to execute when connecting to MariaDB server.
mysql --init-command="SET #foo = 1; SET #bar = 2" -e "SELECT #foo, #bar, VERSION()"
Output:
+------+------+-------------------------------------+
| #foo | #bar | VERSION() |
+------+------+-------------------------------------+
| 1 | 2 | 10.6.3-MariaDB-1:10.6.3+maria~focal |
+------+------+-------------------------------------+
It also works with file redirection.

bash - SQL Query Outputs to variable

Im new in bash scripting.
I want to save sql-query outputs in variable, but
actually I must connect for every query to mysql with:
mysql -u $MYUSER -p$MYPASS -D database
and want to save every output in seperatly variable
sample query is: SELECT domain FROM domains WHERE user='$USER'
to
$variable1 = FIRST_OUTPUT
$variable2 = 2ND_OUTPUT
thank you
Taken from bash script - select from database into variable, you can read the query result into a variable.
Example
mysql> SELECT * FROM domains;
+-------+---------+
| user | domain |
+-------+---------+
| user1 | domain1 |
| user2 | domain2 |
| user3 | domain3 |
+-------+---------+
Usage
$ myvar=$(mysql -D$MYDB -u$MYUSER -p$MYPASS -se "SELECT domain FROM domains")
$ echo $myvar
domain1 domain2 domain3
echo is the bash command for output. You can then split $myvar into separate variables:
$ read var1 var2 var3 <<< $myvar
$ echo $var1
domain1
$ echo $var2
domain2
You can combine these two commands into a single one:
read var1 var2 var3 <<< $(mysql -D$MYDB -u$MYUSER -p$MYPASS -se "SELECT domain FROM domains")
It is possible to store the results into arrays (useful if you don't know how many records there):
$ read -ra vars <<< $(mysql -D$MYDB -u$MYUSER -p$MYPASS -se "SELECT domain FROM domains")
$ for i in "${vars[#]}"; do
$ echo $i
$ done
domain1
domain2
domain3
Another way of doing is:
dbquery=`mysql -D$MYDB -u$MYUSER -p$MYPASS -se "SELECT domain FROM domains"`
dbquery_array=( $( for i in $dbquery ; do echo $i ; done ) )
The first line stores all the output from the query in a varriable dbquery in a array-like-way. The second line converts the dbquery into an array dbquery_array with a simple for loop.
I did this
variable=mysql -u root -ppassworrd database << EOF
select MAX(variable) AS a from table where variable2 = 'SOMETEXT' AND day(datevalue) >= 22;
EOF
I hope it helps

Easiest way to get count val from mysql in bash

Maybe i should use python or perl but i dont know any.
I have 4 statements and i would like to check if there are any errors longer then an hour. My user is setup so i dont need to enter a mysql user/pass. This statement is in mysql_webapp_error_check.sh
#!/bin/bash
mysql testdb -e "select count(*) from tbl where last_error_date < DATE_SUB(NOW(), INTERVAL 1 HOUR);"
How do i make it give me the return value (count(*)) instead of printing to screen?
Then i'll write an if statement and output to stdout/err for cron to use to email me (otherwise i want the script to be silent so nothing is emailed unless theres a problem)
Searched the same, -s for silent works exactly for me.
#!/bin/bash
result=`mysql testdb -s -e "select count(*) from tbl where last_error_date < DATE_SUB(NOW(), INTERVAL 1 HOUR);"`
echo result = .$result.
PS.: There is also a --batch parameter in my mysql Ver 14.14 Distrib 5.1.49 which "Write fields without conversion. Used with --batch" so its a little off-topic here, but should be mentioned here.
in bash, you use $() syntax.
#!/bin/bash
ret=$(mysql testdb -e "select count(*) from tbl where last_error_date < DATE_SUB(NOW(), INTERVAL 1 HOUR);")
if [[ "$ret" > 0 ]];then
echo "there is count"
else
echo "no count"
fi
I usually do this:
var=`mysql -e "SELECT COUNT(*) FROM ...\G" | awk '/COUNT/{print $2}/'`
For my part I simply use grep -v to exclude the line printing count(*) from the return of MySQL.
So I get the counter like that:
db_name="NAME_DB";
db_user="USER_DB";
db_pwd="PWD_DB";
counter=`mysql -u${db_user} -p${db_pwd} ${db_name} -e "SELECT count(*) FROM my_table WHERE something = '1';" | grep -v "count"`;
echo "Count for request: $counter";
I use it for some Wordpress stuff this way, reading databases infos from the wp-config.php file:
wp_db_infos="wp-config.php";
wp_db=`cat ${wp_db_infos} | grep "DB_NAME" | awk -F ', ' '{print $2}' | awk -F "'" '{print $2}'`;
wp_user=`cat ${wp_db_infos} | grep "DB_USER" | awk -F ', ' '{print $2}' | awk -F "'" '{print $2}'`;
wp_pwd=`cat ${wp_db_infos} | grep "DB_PASSWORD" | awk -F ', ' '{print $2}' | awk -F "'" '{print $2}'`;
img_to_update=`mysql -u${wp_user} -p${wp_pwd} ${wp_db} -e "SELECT count(*) FROM wp_offres WHERE maj_img = '1';" | grep -v "count"`;
#!/bin/bash
echo show databases\; | mysql -u root | (while read x; do
echo "$x"
y="$x"
done
echo "$y"
)
local count=$(mysql -u root --disable-column-names --batch --execute "SELECT COUNT(*) FROM mysql.user WHERE user = '$DstDbName'")
if [[ "$count" > 0 ]]
then
fi
--batch - do clear output w/o borders
--disable-column-names - prints only row with value
no creasy AWK used :)