Re: Re: catch the error

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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


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