Re: Re: catch the error

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Ashley Sheridan wrote:
> 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@xxxxxxxxxxxx> wrote:
>>>>>>     
>>>>>>         
>>>>>>             
>>>>>>> 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
>>>>>>>
>>>>>>>
>>>>>>> --
>>>>>>> MySQL General Mailing List
>>>>>>> For list archives: http://lists.mysql.com/mysql
>>>>>>> To unsubscribe:    http://lists.mysql.com/mysql?unsub=dsteplight@xxxxxxxxx
>>>>>>>
>>>>>>>
>>>>>>>       
>>>>>>>           
>>>>>>>               
>>>>> 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@xxxxxxxxxxxxx
>>>>    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@xxxxxxxxxxxxx
>>    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.
>   
I type too fast and am too speedy... :-)

I'll have to look up about the variables.
Thanks & good night. 'Til the morrow.


-- 

Phil Jourdan --- pj@xxxxxxxxxxxxx
   http://www.ptahhotep.com
   http://www.chiccantine.com


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