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help on sql statement


brown2005

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$sql = "SELECT LEFT(websites_url, 1) AS 'websites_url_letter', websites_url
         FROM websites
        [b] WHERE websites_url_letter='$letter' [/b]
         ORDER BY websites_url ASC;";

hi it says

#1054 - Unknown column 'websites_url_letter' in 'where clause'

is there any way round this please?
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You have to use the original column name.

$sql = "SELECT LEFT(websites_url, 1) AS 'websites_url_letter', websites_url
          FROM websites
        WHERE websites_url='$letter'
          ORDER BY websites_url ASC;";

You might be able to use the HAVING clause. I don't know the specifics of it, but someone else might be able to let us know if it will work.

Cheers,
Dave.
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[quote author=brown2005 link=topic=123062.msg508214#msg508214 date=1169162562]
$sql = "SELECT LEFT(websites_url, 1) AS 'websites_url_letter', websites_url
          FROM websites
        [b] WHERE websites_url_letter='$letter' [/b]
          ORDER BY websites_url ASC;";

hi it says

#1054 - Unknown column 'websites_url_letter' in 'where clause'

is there any way round this please?
[/quote]
$sql = "SELECT LEFT(websites_url, 1) AS 'websites_url_letter', websites_url
          FROM websites
        [b] WHERE LEFT(websites_url, 1)='$letter' [/b]
          ORDER BY websites_url ASC;";

or something like:

$sql = "SELECT LEFT(websites_url, 1) AS 'websites_url_letter', websites_url
          FROM websites
        [b] WHERE websites_url LIKE '$letter%' [/b]
          ORDER BY websites_url ASC;";
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yeah cool that works, right but now i have a function called

StripUrl();

which removes the first part of a url say

http://www.url.com

and function leaves

url.com

now i want to do the LEFT(websites_url, 1) as 'websites_url_letter' on url.com and not http://www.url.com

any ideas please?
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No No No.
The error is here.

$sql = "SELECT LEFT(websites_url, 1) AS [b]'websites_url_letter'[/b], websites_url
          FROM websites
        WHERE websites_url_letter='$letter'
          ORDER BY websites_url ASC;";

Drop those quotes.
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