I want to update database from bash command:
#!/bin/bash
list= cat /home/wwwroot/list.txt;
mysql -u root -D dbname -e "UPDATE wp_postmeta SET meta_value = '$list' WHERE meta_key = 1";
cat /home/wwwroot/list.txt
text1
text2
text3
...
How to $list variable can be read in sql query?
Thank you.
Cat is an overkill here.
I suggest using an array if you need space delimited values.
list=($(</home/wwwroot/list.txt))
mysql -u root -D dbname -e "UPDATE wp_postmeta \
SET meta_value = '${list[#]}' WHERE meta_key = 1";
The advantage here is if you need comma separated values then you could employ bash [ parameter substitution ] to achieve your feet. Just change the query to
mysql -u root -D dbname -e "UPDATE wp_postmeta \
SET meta_value = '${list[#]/%/,}' WHERE meta_key = 1";
#Note ${list[#]/%/,} appends comma after each value in the text file
All good :-)
Related
I want to make specific .sql file for each of my routines.
#!/bin/bash
routine_names=$(mysql mydb --execute="SELECT
*
FROM
information_schema.routines
WHERE
routine_type = 'PROCEDURE' OR routine_type ='FUNCTION'
AND routine_schema = 'mydb';"|cut -f1)
for routine in "$routine_names"
do
if [ -e "${routine}.sql" ]
then
echo "ok"
else
content_procedure=$(mysql mydb--execute="SHOW CREATE PROCEDURE $routine;")
echo "$content_procedure" >> masoud.txt
fi
done
my routine_names variable is a list of my procedures. like this:
SP_ONE
SP_TWO
SP_THREE
I want to loop of these result. but I think the result is not an array. because routine variable has all content.
wrap your mysql mydb ... command with ()
routine_names=($(mysql mydb --execute="SELECT
*
FROM
information_schema.routines
WHERE
routine_type = 'PROCEDURE' OR routine_type ='FUNCTION'
AND routine_schema = 'mydb';"|cut -f1))
I was curious and got the idea from here.
Just another case how I fix my problem
#!/bin/bash
mysql -u USER -p -h localhost -D database1 -e "SELECT ID FROM prodTable WHERE display=1 AND new=1 AND exDate<DATE_SUB(CURDATE(),INTERVAL 2 YEAR)" | while read ID;
do
echo "The following product has been moved: $ID"
done
I am trying to check if a MYSQL user exists. This is as far as I have got. I fall down on capturing the answer from the output.
#!/bin/bash
echo -e "What is the MYSQL username called"
read DBUSER
if [ -z "$DBUSER" ]
then
exit
mysql -uUSER -pPASS -e "SELECT EXISTS(SELECT 1 FROM mysql.user WHERE user = '$DBUSER')";
if
yes
do this
else
do this
this is the output I am getting
+-----------------------------------------------------+
| EXISTS(SELECT 1 FROM mysql.user WHERE user = 'bob') |
+-----------------------------------------------------+
| 1 |
+-----------------------------------------------------+
Can any one help please
Thanks very much for your help. Here is the final result working.
It needs the -sse
RESULT_VARIABLE="$(mysql -uUSER -pPASS -sse "SELECT EXISTS(SELECT 1 FROM mysql.user WHERE user = '$DBUSER')")"
if [ "$RESULT_VARIABLE" = 1 ]; then
echo "TRUE"
else
echo "FALSE"
fi
Assigning the result to a variable can be done like this:
RESULT_VARIABLE="$(mysql -uUSER -pPASS -se "SELECT EXISTS(SELECT 1 FROM mysql.user WHERE user = '$DBUSER')")"
And you can also alias a column in MySQL, btw.
SELECT EXISTS(SELECT 1 FROM mysql.user WHERE user = '$DBUSER') AS does_it_exist
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
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.
My Bash script makes an array from MySQL:
info_tmp=$(mysql --batch --silent -u root -ppassword database -e "SELECT id,info1,info2 FROM table WHERE id=1")
info=($(for i in $info_tmp;do echo $i;done))
info1=${info[1]}
My problem is, that, if info1 is an empty string in the database, then $info1 became info2.
How can I put an empty string into $info array?
Mysql Database:
Id | info1 | info2
1 | | data2
2 | data3 | data4
$info_tmp
1 data2
2 data3 data4
Thank you for your answer
This is the final code that worked (#Barmar):
IFS="|"
info_tmp=$(mysql --batch --silent -u root -ppassword database -e "SELECT CONCAT_WS('|', id,info1,info2) FROM table WHERE id=1")
info=(${info_tmp// / })
info1=${info[1]}
If there's a character that shouldn't appear in any of the columns, use that as a delimiter.
IFS="|"
info_tmp=$(mysql --batch --silent -u root -ppassword database -e "SELECT CONCAT_WS('|', id,info1,info2) FROM table WHERE id=1")
This works because bash doesn't merge sequences of non-whitespace delimiters in IFS, only whitespace characters.
I'm not sure what the point of the for loop that copies $info_tmp to $info is, but you need to do the same thing there. If you use whitespace as your word delimiter, you'll never be able to get empty array values from command substitution.
What about temporarily adding a single character in your for-loop:
info_tmp=$(mysql --batch --silent -u root -ppassword database -e "SELECT id,info1,info2 FROM table WHERE id=1")
info=($(for i in $info_tmp;do echo " "$i;done))
info1=$(${info[1]} | cut -c 2-)