[RESOLVED] Quick query issue

TheCase

Hi,

I cant seem to get this query to work. There is no output although when I replace the $line[\'5\'] with a static number I get the correct output so its something to do with this.


 $result = mysql_query('SELECT username FROM users, showbeta WHERE users.id = "$line[\'5\']" AND users.id = showbeta.presenter1id ') or die('Query failed: ' . mysql_error());
          while ($row = mysql_fetch_array($result, MYSQL_ASSOC))
              {
              $name1 = $row['username'];
              }

If you have a better way or writing this please show me I need to improve on my queries.

Thanks

bradgrafelman

There's a couple of problems with that string. For one, variable interpolation doesn't occur at all within single-quote delimited strings. Even if you had used double quotes as your string delimiter, however, you can't have a quoted string array index within a double-quote delimited string without using complex syntax or omitting the quotes around the index. What's more is that it's entirely possible that the index you're after isn't even a string at all but a number, in which case you shouldn't even be quoting it at all.

More information/examples on variable interpolation (including arrays) can be found on this manual page: [man]string[/man].

Also, depending upon what $line contains, don't forget about data sanitization (e.g. via [man]mysql_real_escape_string/man).

TheCase

Thanks for the reply. The string is a number and will always be a number. When I dont quote it it still doesnt work.

bradgrafelman

Two things:

Show us a fresh copy of the code you're using now that you've made modifications to it.
Store the MySQL query string in a variable (and use that variable in mysql_query() rather than the string itself) so that you can output the query string and visually examine it. If you can't spot the error, post the query string for us to see as well.

TheCase

I now have this


$line5 = $line['5']; 

	  $query = "SELECT username FROM users, showbeta WHERE users.id = $line5 AND users.id = showbeta.presenter1id ";
	  $result = mysql_query ($query) or die('Query failed: ' . mysql_error());
	 echo "$result";

      while ($row = mysql_fetch_array($result, MYSQL_ASSOC))

          {
          $name1 = $row['username'];
          }

Which just comes back as Resource id #8. Not too sure what this means

bradgrafelman

TheCase wrote:
Not too sure what this means

It means that [man]mysql_query/man returned a MySQL resource identifier, which is what it's supposed to do.

What I meant for you to do was echo the SQL query string itself, not the result of mysql_query().

TheCase

Thanks for the explination.

bradgrafelman;10959802 wrote:
What I meant for you to do was echo the SQL query string itself, not the result of mysql_query().

Oops, I echoed $query and saw the $line5 was producing number 8 which I didnt expect that. It was because I forgot that the number starts at 0.

Now I know how to debug queries better

Many Thanks