Hi,
yes in fact it worked...I also had the $numrows with wrong mysql command, it
should be mysql_num_rows() instead of the mysql_fetch_rows().
Thx Ernest
But now that this is working, i have another big problem...my
$drop_down_list isn't assuming the value of the shown option...right here on
the below part
********
for ($i = 0; $i < $num; $i++) {
$aviao = mysql_result($resultaviao, $i, "aviao");
$sk = mysql_result($resultaviao, $i, "sk");
$drop_down_list .= "<optionvalue=$aviao>$aviao</option>";
}
**********
So he isn't getting the data from DB. He displays all records on DB
available at aviao field, but he is not making the correct mysql query...
Anyone knows why?
----- Original Message -----
From: "Ernest E Vogelsinger" <[EMAIL PROTECTED]>
To: "Miguel Brás" <[EMAIL PROTECTED]>
Cc: <[EMAIL PROTECTED]>
Sent: Monday, February 24, 2003 9:45 AM
Subject: Re: [PHP] What is wrong with this?
At 10:37 24.02.2003, Miguel Brás said:
--------------------[snip]--------------------
>//logical request
>if ($nome) {
>$sql="aviao = '$nome' ";
>}
>if ($code) {
>$sql="code = '$code' ";
>}
> $query = "SELECT * FROM ssr WHERE";
> $query .= $sql;
--------------------[snip]--------------------
Hi,
I've copied only the relevant lines of your program.
The issue here is that your string concatenation results in
SELECT * FROM ssr WHEREaviao = 'Airbus'
Note the missing space after "WHERE". Your code should better read
$query = "SELECT * FROM ssr WHERE ";
$query .= $sql;
Have an eye on the added blank after WHERE.
HTH,
--
>O Ernest E. Vogelsinger
(\) ICQ #13394035
^ http://www.vogelsinger.at/
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