Jerry Schwartz wrote:
>
>> -----Original Message-----
>> From: PJ [mailto:[email protected]]
>> Sent: Thursday, February 26, 2009 12:28 PM
>> To: [email protected]; MySql
>> Subject: catch the error
>>
>> What is wrond with this file? same identical insert works from console
>> but not from this file :-(
>>
>> <html>
>> <head>
>> <title>Untitled</title>
>> </head>
>>
>> <body>
>> <?
>> //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();
>> }
>> ?>
>>
>> </body>
>> </html>
>>
>> Seems to be good to print out the error message, but that's all. db not
>> written.
>>
>>
> [JS] You need
>
> $db = mysql_connect('biggie', 'user', 'password', 'test');
>
>
>> --
>>
>> Phil Jourdan --- [email protected]
>> http://www.ptahhotep.com
>> http://www.chiccantine.com
>>
>>
>> --
>> MySQL General Mailing List
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>> To unsubscribe: http://lists.mysql.com/mysql?unsub=jschwa...@the-
>> infoshop.com
>>
>
>
>
>
>
>
I think the problem here has been that this is such a basic operation
and most of us just are too busy with more complicated stuff...that we
didn't catch it...
<?
//include ("../lib/db1.php"); // Connect to database
$db_host = 'xxxx';
$db_user = 'xxxx';
$db_pass = 'xxxx';
$db_name = 'xxxx';
$db = mysql_connect($db_host, $db_user, $db_pass);
mysql_select_db($db_name,$db);
$sql1 = "INSERT INTO test (name, age) VALUES ('Arnie Shwartz', '75')";
$result1 = mysql_query($sql1,$db);
if (!$result1) {
echo("<P>Error performing 3st query: " .
mysql_error() . "</P>");
}
echo $sql1;
echo "<br />";
echo $db_select;
exit();
?>
anyway, I am learning a lot...
thanks, guys... you're all great...
I have lots more coming... :-D
--
Phil Jourdan --- [email protected]
http://www.ptahhotep.com
http://www.chiccantine.com
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