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Linux parameters in SQL command

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# 1  
Old 08-07-2015
Linux parameters in SQL command
Hello.
It's my first steps in creat bash skript.
This is my skript :
Code:
mysql -e "  
UPDATE datebase.table
SET U_O_ID=NULL 
WHERE U_O_ID LIKE '"$w4"'
AND N_U != '"$w1"'
   " -u admin -p""Password"" ;

It doesn't work.
I find a lot of topic about it, but I didn't find simple answer.
Thenks for any help.Smilie
# 2  
Old 08-07-2015
Possibly you are single quoting variables $w[14] add backslash in front of single quotes to remove its special meaning
Last edited by rbatte1; 08-07-2015 at 07:36 AM..
# 3  
Old 08-07-2015
Can you give me example ?

---------- Post updated at 06:43 AM ---------- Previous update was at 04:12 AM ----------

But w1 and w2 aren't a table items. They are like A and B.
# 4  
Old 08-07-2015
Instead of '"$w4"' use '$w4'. What is inside the shell variables w1 and w4 ? What is with the double double quotes around Password ?

--
The SQL Syntax aside, usually how this is done is, something like:
Code:
mysql -u admin ....   << EOF
UPDATE datebase.table
SET U_O_ID=NULL 
WHERE U_O_ID LIKE '$w4'
AND N_U != '$w1'
EOF

Last edited by Scrutinizer; 08-07-2015 at 09:40 AM..
# 5  
Old 08-07-2015
You put first
Code:
"

. in
Code:
mysql "

Where is the last one ?
# 6  
Old 08-07-2015
Yes that quote should not be there. I corrected in my post.
# 7  
Old 08-07-2015
Code:
mysql -u admin ....   << EOF
UPDATE datebase.table
SET U_O_ID=NULL 
WHERE U_O_ID LIKE '$w4'
AND N_U != '$w1'
EOF

It made without error, but it didn't do anythink in SQL table.
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