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From:Darryle Steplight Date:February 26 2009 6:45pm
Subject:Re: [PHP] Re: catch the error
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Additionally regarding the error handling , add this to the op of your script.

ini_set("display_errors","true");
error_reporting(E_STRICT|E_ALL);

and post the output of your error message.

On Thu, Feb 26, 2009 at 1:40 PM, Ashley Sheridan
<ash@stripped> wrote:
> On Thu, 2009-02-26 at 13:34 -0500, Darryle Steplight wrote:
>> Hi PJ,
>>    $db_host = 'biggie';
>> $db_user = 'root';
>> $db_pass = 'gugus@#$';
>> $db_name = 'biblane';
>>
>>
>>
>> Everyone here is trying to help you and that's cool, but EVERYONE on
>> this list may not be so nice. The above credentials is definitely the
>> type of information you want to keep private, unless you don't mind
>> people potentially accessing your database tables and doing whatever
>> they like with them.
>>
>> I suggest doing something like
>> $db_host = 'localhost;
>> $db_user = 'foo';
>> $db_pass= ''bar;
>> $db_name =''xxxxxx;
>>
>> if you are going to post it on the list.
>>
>> On Thu, Feb 26, 2009 at 1:22 PM, PJ <af.gourmet@stripped> wrote:
>> > Ricardo Dias Marques wrote:
>> >> Hi PJ,
>> >>
>> >> On Thu, Feb 26, 2009 at 17:28, PJ <af.gourmet@stripped>
> 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@stripped
>> >   http://www.ptahhotep.com
>> >   http://www.chiccantine.com
>> >
>> >
>> > --
>> > MySQL General Mailing List
>> > For list archives: http://lists.mysql.com/mysql
>> > To unsubscribe:    http://lists.mysql.com/mysql?unsub=1
>> >
>> >
>>
> I agree. I wouldn't trust me at all! ;)
>
>
> Ash
> www.ashleysheridan.co.uk
>
>
Thread
Re: [PHP] Re: catch the errorDarryle Steplight26 Feb
Re: [PHP] Re: catch the errorPJ26 Feb
Re: [PHP] Re: catch the errorPJ26 Feb
Re: [PHP] Re: catch the errorPJ26 Feb
Re: [PHP] Re: catch the errorPJ27 Feb