Ricardo Dias Marques wrote:
> Hi PJ,
>
> On Thu, Feb 26, 2009 at 17:28, PJ <af.gourmet@xxxxxxxxxxxx> wrote:
>
>
>> What is wrond with this file? same identical insert works from console
>> but not from this file :-(
>>
>> [snip]
>>
>> <?
>> //include ("lib/db1.php"); // Connect to database
>> mysql_connect('biggie', 'user', 'password', 'test');
>> $sql1 = "INSERT INTO example (name, age) VALUES ('Joe Blow', '69')";
>> $result1 = mysql_query($sql1,$db);
>> if (!$result1) {
>> echo("<P>Error performing 1st query: " .
>> mysql_error() . "</P>");
>> exit();
>> }
>> ?>
>>
>
> I haven't coded in PHP for a long time, but I think that your problem
> is in this line:
>
> $result1 = mysql_query($sql1,$db);
>
> Up to that point, $db (that should point to a database link
> identifier) is not defined. You probably want to assign the
> "mysql_connect" result to that $db variable.
>
>
> So, I think that you will solve your problem by changing your
> "mysql_connect" line FROM the current form:
>
> mysql_connect('biggie', 'user', 'password', 'test');
>
> .. TO this one:
>
> $db = mysql_connect('biggie', 'user', 'password', 'test');
>
>
> Am I right?
Partly. I had an error in the location of the include. Ashley corrected
the rest but it only works with the include. Not as whown below
<?
//include ("../lib/db1.php"); // Connect to database
$db_host = 'biggie';
$db_user = 'root';
$db_pass = 'gugus@#$';
$db_name = 'biblane';
$db_connect = mysql_connect($db_host, $db_user, $db_pass);
$db_select = mysql_select_db($db_name, $db_connect);
$sql1 = "INSERT INTO test (name, age) VALUES ('Arnie Shwartz', '75')";
$result1 = mysql_query($sql1,$db);
if (!$result1) {
echo("<P>Error performing 1st query: " .
mysql_error() . "</P>");
exit();
}
?>
--
Phil Jourdan --- pj@xxxxxxxxxxxxx
http://www.ptahhotep.com
http://www.chiccantine.com
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