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up vote 0 down vote favorite

I am new to shell scripting world.

I wrote a bash shell script.

Name of the script is new.sh 

Select T1.date as dateFact,
T2.idmfg as idMfg,
T1.id as userId,
FROM Table1 T1
JOIN 
Table2 T2
ON 
T1.id=T2.id 
WHERE 
T1.date ='${Date}'
T2.idmfg='${idMfg}';

While running new.sh, I want to enter Date and idMfg manually. E.g. 

sh new.sh -d 2013-03-20 -i 201

Where, Date is 2013-03-20 and idMfg is 201.

How can I do it?

Thank you in advance.

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2 Answers 2

up vote 3 down vote accepted 
#!/bin/bash

Date="$1"
idMfg="$2"

mysql<<EOF
SELECT T1.date AS dateFact, T2.idmfg AS idMfg, T1.id AS userId
FROM Table1 T1
JOIN Table2 T2
ON T1.id=T2.id 
WHERE T1.date = "${Date}", T2.idmfg = "${idMfg}";
EOF

Usage : 

./new.sh 2013-03-20 201

To go further if you want named switches, see the tutorial: http://wiki.bash-hackers.org/howto/getopts_tutorial Examples: http://mywiki.wooledge.org/BashFAQ/035

share|improve this answer
    
+1 for the script, but he never mentioned mysql :) –  tink Mar 21 '13 at 22:44
    
How can I put it inside case e.g. case $Option in d) date="${date}" i) idmfg="${idMfg}" esac; so I can run as: new.sh -d 2013-03-20 -i 201 –  Mario Mar 21 '13 at 23:49
    
Use the getopts tutorial mentioned in the answer –  dtmilano Mar 22 '13 at 5:44
    
@tink : MySQL is just an example ;) –  sputnick Mar 23 '13 at 3:49
up vote 0 down vote 
 new.sh 2013-03-20 201

Inside the script the first (date) param is $1, the second $2.

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