NYCPHP Meetup

[Kristina Anderson]

What were we saying about how sometimes when we post to the list, we 
see what the problem is??????

In this case the problem is that I'm a MORON!!! I was calling $query1 
twice instead of calling $query2 in the second case....!!!

$result2 = mysql_query($query1)

{this is not the first time it took me hours to figure out that I was 
doing this....!!!  Hopefully the last time.)

> 
> 114   jkljlk tyuti ttyut tyutu 0 NULL 0 [0 NULL ]
> 113   jkljlk tyuti ttyut tyutu 0 NULL 0 0 NULL 
> 
> Last two fields in row 114 s/b "113"....but.
> 
> Query as follows looks OK to me...
> 
> INSERT INTO who_we_are SET eid=113,eid2='113', empname ='jkljlk', 
> title='tyuti', bodytext='ttyut tyutu', picture='' 
> 
> > Hi Kristina:
> > 
> > On Tue, Apr 01, 2008 at 07:57:48PM -0700, Kristina Anderson wrote:
> > > I'm pulling out the ID of the previously inserted row and then 
> > > inserting that as a lookup value in a duplicate row (two rows one 
> for 
> > > edit mode one for published mode).
> > > 
> > > Various other places in the app this works fine and there really 
> isn't 
> > > any reason this should be happening.
> > > 
> > > The query runs fine if I do it from within phpMyAdmin -- but from 
> the 
> > > PHP page the query does not error out but the value in the lookup 
> field 
> > > remains the default value
> > 
> > You say you're using "mysql_insert_row."  I assume you mean PHP's 
> > mysql_insert_id() function.  If so, there are two possible bugs 
that 
> come 
> > to mind.
> > 
> > 1) The table in question does not have auto_increment set for the 
> primary 
> > key on that table.  If that's not it...
> > 
> > 2) The PHP code has a bug...
> > 
> > You say this same logic works on other parts of the site.  So, do 
the 
> > various parts of the site use the _same_ _exact_ 
> files/lines/functions or 
> > do you have separate function/include/whatever for each section of 
> the 
> > application?
> > 
> > If you're using separate code for different sections, obviously the 
> PHP 
> > code you're using for the problematic insert is where the bug is.  
> > Perhaps the variable you assign the id to is different than the 
> variable 
> > you're using as the lookup value in the second query.
> > 
> > Again, if you're using separate code, you should refactor the 
system 
> to 
> > allow you to use the same code for the same purpose throughout the 
> > system.
> > 
> > --Dan
> > 
> > -- 
> >  T H E   A N A L Y S I S   A N D   S O L U T I O N S   C O M P A N Y
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