Whenever I am having problems with SQL queries I always display the query on
the page with all the replaced variables so I can make sure that the query
makes sense. If it passes the MySQL parser in my brain and still doesn't
work, I will copy the displayed query into the command line mysql monitor (or
phpmyadmin) and try the command manually to see what happens.
I also found it easier to solve my SQL problems by not using "or die " and
just displaying some information about the problem query. This way I know
something is wrong if the third line of the echo is not at least 1.
$query="INSERT INTO test (test1, test2) VALUES
('$_POST['test1']','$_POST['test2']')";
$result = mysql_query($query);
echo(mysql_error()."<BR>$query<BR>".mysql_affected_rows($result));
James
On Sunday 15 June 2003 09:39 am, Frank Keessen wrote:
> Dear all,
>
> Refering to my first e-mail with the same subject; I've reworked the code
> to this:
>
> But the only problem is; Nothing is written into the mysql database! And
> there are no error displayed..
>
> Can someone have a look at my code?
>
> <?
> $_REQUEST["submit"]=isset($_REQUEST["submit"])?$_REQUEST["submit"]:"";
> if($_REQUEST['submit']!="")
> {
> for($i=0;$i<count($_POST['test1[]']);$i++) {
> $query="INSERT INTO test (test1, test2) VALUES
> ('$_POST['test1']','$_POST['test2']')"; $result = mysql_query($query) or
> die ("Error in query: $query. " . mysql_error()); }
> }
> ?>
> <form name="form1" method=post id=form1 enctype=multipart/form-data
> action="<? $_SERVER['PHP_SELF']?>"> <?
> for ($i=1; $i<=2; $i++)
> {
> echo "# $i<input type=\"text\" name=\"test1[]\"><br>";
> echo "# $i<input type=\"text\" name=\"test2[]\"><br>";
>
> }
>
> ?>
> <input type=submit name=submit value=submit>
> </form>
>
> Thanks very much,
>
> Frank
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