what's wrong with my simple code?

[johnnytag]

Hi newbie here, I can't get this simple code to work. I would really appreciate any help you can give me. Thanks

$result = mysql_query("select max(num) from results");
$row = mysql_fetch_array($result);
echo $row['num'];

The num field is an auto incremented unsigned int in table results;
The connection to the database is good. I'm running php 4.3.1, mysql 3.23.55 on windows xp.

[WzL]

Originally posted by johnnytag
$result = mysql_query("select max(num) from results");
$row = mysql_fetch_array($result);
echo $row['num'];

I think your problem is:
echo $row['num'];

i would do it like this: echo $row[1];
or if you want to use var: echo $row[$num];

[johnnytag]

I don't think the query is even successful. When I test $result with

if(!$result)
echo'failed';

failed is printed.

[WzL]

Heh, actually i didnt try your query =) I just noticed that little thing with that echo. 🙂

I don't know if this helps you out, but here is one example:

For example take all from sometable

$SQL="SELECT * FROM sometable";
$rs=mysql_query($SQL);
$output=mysql_fetch_row($rs);

Now we have first line from query in var $output

Then we can print it with echo $output[0] (where 0 is first column and so on...)

But before you do anything like this, have you made connection to your database?

$db_bost="yourhost";
$db_user="username";
$db_pw="password";

mysql_connect( $db_host, $db_user, $db_pw );
mysql_select_db( $db_base );

This has to be done before any queries.

[johnnytag]

Thank you for you help. I forgot to include

mysql_select_db('survey');

then I used your first suggestion

$result = mysql_query("select max(num) from results");
$row = mysql_fetch_row($result);
echo $row[0];

and it works. Thanks again for your help. 🙂