Re: Re: catch the error

On Thu, 2009-02-26 at 14:15 -0500, PJ wrote: > Ashley Sheridan wrote: > > On Thu, 2009-02-26 at 13:56 -0500, PJ wrote: > > > >> Ashley Sheridan 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@videotron.ca> wrote: > >>>> > >>>> > >>>>> Ricardo Dias Marques wrote: > >>>>> > >>>>> > >>>>>> Hi PJ, > >>>>>> > >>>>>> On Thu, Feb 26, 2009 at 17:28, PJ <af.gourmet@videotron.ca> > >>>>>> 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@ptahhotep.com > >>>>> http://www.ptahhotep.com > >>>>> http://www.chiccantine.com > >>>>> > >>>>> > >>>>> -- > >>>>> MySQL General Mailing List > >>>>> For list archives: > >>>>> http://lists.mysql.com/mysql > >>>>> To unsubscribe: � > >>>>> K�\�$qwRl�g�http://lists.mysql.com/mysql?unsub=dsteplight@gmail.com > >>>>> > >>>>> > >>>>> > >>>>> > >>> I agree. I wouldn't trust me at all! ;) > >>> > >>> > >>> Ash > >>> www.ashleysheridan.co.uk > >>> > >>> > >>> > >>> > >> Yeah.... very stupid of me...but I found the error: see if you can catch it: > >> <? > >> //include ("../lib/db1.php"); // Connect to database > >> > >> $db_host = 'xxx'; > >> $db_user = 'xxx; > >> $db_pass = 'xxx'; > >> $db_name = 'xxx'; > >> > >> > >> $db = mysql_connect($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(); > >> ?> > >> > >> -- > >> > >> Phil Jourdan --- pj@ptahhotep.com > >> http://www.ptahhotep.com > >> http://www.chiccantine.com > >> > >> > >> > > $db_user has not had the string terminated. pray tell was that the > > answer you were looking for?! > > > > > > Ash > > www.ashleysheridan.co.uk > > > > > > > No. Damn those typos! > > What seems to have made it work is just > > $db = mysql_connect($db_host, $db_user, $db_pass); > mysql_select_db($db_name,$db); > > not using mysql_select in a string.... > but would you use it in a string? how & why? > > -- > > Phil Jourdan --- pj@ptahhotep.com > http://www.ptahhotep.com > http://www.chiccantine.com > > Yeah, you'd typo'd on the variable name. Also, the $ sign doesn't actually denote a string, but a scaler variable, which can be any type, complex or simple. Ash www.ashleysheridan.co.uk