Reply to Re: php variable in sql string

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Posted by Norman Peelman on 01/19/07 02:27

user wrote:
> Have require file with several query stings in it.
>
> Depending on user input one of strings is selected. Everything going
> along smoothly until I wanted to also input a variable in string. If I
> put string in program works ok, but, if I use string from require file I
> can not seem to insert string.
>
> $cccb_id is sting..... to be inserted into $query4 and changes depending
> on user input.
>
> $query4 = "select cccb.cccb_name as 'cccb', CONCAT(member_fname,'
> ',member_lname) as 'member' from member_cccb_lnk join member on
> (member.member_no = member_cccb_lnk.member_no) join cccb on
> member_cccb_lnk.cccb_id = cccb.cccb_id and cccb.cccb_id = "$cccb_id"
> order by member";
>
> output is: select cccb.cccb_name as 'cccb', CONCAT(member_fname,'
> ',member_lname) as 'member' from member_cccb_lnk join member on
> (member.member_no = member_cccb_lnk.member_no) join cccb on
> member_cccb_lnk.cccb_id = cccb.cccb_id and cccb.cccb_id = order by
> memberError 1064
>
> as you can see, "$cccb_id" is not in query string.
>
> any help will be appreciated.

Try:

$query4 = "select cccb.cccb_name as cccb, CONCAT(member_fname,'
',member_lname) as member from member_cccb_lnk join member on
(member.member_no = member_cccb_lnk.member_no) join cccb on
member_cccb_lnk.cccb_id = cccb.cccb_id and cccb.cccb_id = '$cccb_id'
order by member";

If $cccb_id has a value in it it will show up in the output. Notice I
removed the single quotes from around cccb and member as they are not
needed. Use backticks (`) if you must. Always remember, to include
variables whithin strings the entire string must be enclosed in double
quotes or the variables won't get parsed.

Norm

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