bash variable loses contents in mysql statement

thedude2010 · 06-01-2006, 09:11 AM

I have a variable $id that contains an id

I can echo it without any problem but...

as soon as I do this
`$mysql "insert into webusers_db(webuserid,db) values('"${id}"','"${db}"')"`

the db gets entered but the id is the default value in my table of zero
I have tried '$id' '"$id"' '"${id}"' '"[${id}]"' '"["${id}"]"' nothing seems to work

bigearsbilly · 06-01-2006, 10:11 AM

it's the single quotes, try this:

Code:

`$mysql "insert into webusers_db(webuserid,db) values(\'${id}\',\'${db}\')"`

thedude2010 · 06-01-2006, 10:18 AM

alright then...here what I got when I escaped the single 's
ERROR at line 1: Unknown command '\''.

/me scratches his head

Hko · 06-02-2006, 04:15 AM

This should work:

Code:

$mysql "insert into webusers_db(webuserid,db) values('${id}','${db}')"

Quoting and escaping is not allways useful or needed