I found out that in fact PHP is creating a variable with the name and value I'm
passing through a URL from the querystring. But it's still not working as planned.
The url server/test.php?id=1 creates the following results in my code:
printf("Variables: %s\n<br>", $HTTP_GET_VARS["id"]);
This line works - there IS a variable named 'id' in my page and it has the correct
value, 1.
if ($id) {}
This fails. If I test 'id' instead of '$id' then it works but my page doesn't seem to
equate 'id=1' with the presence of $id.
$result = mysql_query("SELECT * FROM employees WHERE id=$id",$db);
This doesn't work - again it seems $id isn't being treated properly. I get this error:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource
If I hardwire my page with the line '$id=1;' before the if statement and the query
everything works.
So why isn't the variable from my URL being treated properly?
Jesse
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