Get return code of the mysql command

Question

I'm writing a shell script to import a csv file in mysql, this is my code:

DATABASE=*mydatabase*
USER=*myuser*
PASS=*mypassword*

file0="myfile.csv"
if [ -e "$file0" ]
then
  mysql --user=${USER} --password=${PASS} --database= ${DATABASE} --local-infile=1 < myquery.sql
else
  echo "$file0 does not exist"
fi

The query to execute is in myquery.sql and this is its content:

LOAD DATA LOCAL INFILE 'myfile.csv' replace INTO TABLE imported_table
FIELDS TERMINATED BY ';' 
ENCLOSED BY '' 
LINES TERMINATED BY '\n' 
IGNORE 1 LINES 
(Field1, Field2, Field3, Field4, Field5);

This code works just fine, but how do I check if the query is executed? My goal is to delete the file after importation if it is successful or write to a custom log file in case of error. Any suggestion?

Answer 1

Whenever you run a command with mysql (or any other in Unix), you get a return code that you can check with $?. In case it was successful, it will be 0. Otherwise, it will be a different number.

So you can do:

mysql --user=${USER} --password=${PASS} --database= ${DATABASE} --local-infile=1 < myquery.sql
if [ $? -eq 0 ]; then
   # remove file
else
   # write to log
fi

Test

$ mysql -u me -pXXXX -e "SELECT COUNT(*) FROM YYY.ZZZ;"


+----------+
| COUNT(*) |
+----------+
|   123 |
+----------+
$ echo $?
0
$ mysql -u meASDFASFASD -pXXXX -e "SELECT COUNT(*) FROM YYY.ZZZ;"
ERROR 1045 (28000): Access denied for user 'meASDFASFASD'@'local' (using password: YES)
$ echo $?
1