Hi.
If your id is not a numeric field you should change your query this way:
$result = mysql_query("SELECT * FROM employees WHERE id='$id'",$db);
<zxx10@cwru.edu> wrote in message 84db9b84fea8.84fea884db9b@cwru.edu">news:84db9b84fea8.84fea884db9b@cwru.edu...
> Hi, All:
>
> I'm a beginner of PHP. While trying the code from
> a tutorial, I encountered the following problem.
>
> The variable $id can not be transfered to my server.
> You can find the code at the end of this email. When
> I visit http://mydomain.com/test.php?id=1
> it always shows the list of the database instead
> of a perticular record.
>
> I appreciate your kind helps!
>
> Zhan Xu
> EECS Department
> Case Western Reserve University
>
>
> <html>
> <body>
> <?php
> $db = mysql_connect("localhost", "root");
> mysql_select_db("mydb",$db);
> // display individual record
> if ($id) {
>
> $result = mysql_query("SELECT * FROM employees WHERE id=$id",$db);
>
> $myrow = mysql_fetch_array($result);
>
> printf("First name: %s\n<br>", $myrow["first"]);
>
> printf("Last name: %s\n<br>", $myrow["last"]);
>
> printf("Address: %s\n<br>", $myrow["address"]);
>
> printf("Position: %s\n<br>", $myrow["position"]);
>
> } else {
>
> // show employee list
> $result = mysql_query("SELECT * FROM employees",$db);
>
> if ($myrow = mysql_fetch_array($result)) {
>
> // display list if there are records to display
>
> do {
>
> printf("<a href=\"%s?id=%s\">%s %s</a><br>\n", $PHP_SELF,
$myrow["id"], $myrow["first"], $myrow["last"]);
>
> } while ($myrow = mysql_fetch_array($result));
>
> } else {
>
> // no records to display
>
> echo "Sorry, no records were found!";
>
> }
>
> }
> ?>
>
> </body>
> </html>
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