NYCPHP Meetup

[Brian O'Connor]

Hmm, have you tried outputting the results of $_POST['editjob']?  It might
not be set to something that evaluates to true, which would cause none of
the information you're looking for to be processed.

On 8/14/07, Anthony Wlodarski <aw at sap8.com> wrote:
>
>  I have been staring at this for about a week and it is turning into a
> mind bender.  Here is the code excerpt:
>
>
>
> // pre: none
>
> // post: displays our controls for …
>
> function displayControls()
>
> {
>
>         if($_POST['editjob'])
>
>         {
>
>                 $query = "SELECT * FROM `jobsdb` WHERE `id`
> =".$_POST['id']."";
>
>                 $result = mysql_query($query) or die("Error editing
> data.");
>
>                 $row = mysql_fetch_assoc($result, MYSQL_ASSOC);
>
>
>
>                 echo "Debug: ".$_POST['id']." - ".$row['id']."\n";
>
>
>
>                 // display all our form data
>
>                 echo "<fieldset style='color: #ffffff;'><legend>Edit a
> job</legend>\n".
>
>                                 "<form action='".$_SERVER['PHP_SELF']."'
> method='post'>\n".
>
>                                 "Recruiter: <input name='recruiter'
> type='text' value='".$row['recruiter']."'></br>".
>
>                                 " Job ID: <input name='jobid' type='text'
> value='".$row['jobid']."'></br>".
>
>                                 " Position: <input name='position'
> type='text' value='".$row['position']."'></br>".
>
>                                 " Position Link: <input name='link'
> type='text' value='".$row['link']."'></br>".
>
>                                 " Additional Info: <input
> name='additionalinfo' type='text' value='".$row['additionalinfo']."'></br>".
>
>                                 " Salary: <input name='salary' type='text'
> value='".$row['salary']."'></br>".
>
>                                 " Hiring Manager: <input
> name='hiringmanager' type='text' value='".$row['hiringmanager']."''></br>".
>
>                                 " Company: <input name='company'
> type='text' value='".$row['company']."'></br>".
>
>                                 " Status: <input name='status' type='text'
> value='".$row['status']."'></br>".
>
>                                 " Postings: <input name='postings'
> type='text' value='".$row['postings']."'></br>".
>
>                                 " Job Color: \n".
>
>                                 "<select name='color'>\n".
>
>                                 "<option value='#ff0000'>Red</option>\n".
>
>                                 "<option
> value='#ffa500'>Orange</option>\n".
>
>                                 "<option
> value='#ffff00'>Yellow</option>\n".
>
>                                 "<option value='#00ff7f'>Spring
> Green</option>\n".
>
>                                 "<option value='#32cd32'>Lime
> Green</option>\n".
>
>                                 "<option
> value='#c0c0c0'>Silver</option>\n".
>
>                                 "<option value='#0000cd'>Medium
> Blue</option>\n".
>
>                                 "<option value='#191970'>Midnight
> Blue</option>\n".
>
>                                 "<option value='#4169e1'>Royal
> Blue</option>\n".
>
>                                 "<option value='#c71585'>Medium Violet
> Red</option>\n".
>
>                                 "<option
> value='#ff00ff'>Fuchsia</option>\n".
>
>                                 "</select>\n".
>
>                                 "<input name='saveedit' type='submit'
> value='Save Edit'>".
>
>                                 "<input name='cancel' type='submit'
> value='Cancel'>".
>
>                                 "</form>".
>
>                         "</fieldset>\n";
>
>         }
>
>
>
> ……. Rest of application.
>
>
>
> Now the big problem are the three PHP statements after the "if"
> statement.  When I click edit job nothing appears even in the debug
> statement, I am not returning a valid id from $row['id'] nor does any of the
> data appear in the form(which is just an integer).  I have scrutinized the
> SQL query a million times, even tried to qualify it with the database name
> using `foobar`.`jobsdb` and that still not do anything.  I know the
> $_POST['id'] variable is passing the right id based on the echo command.  Is
> this a SQL issue or a PHP issue?  I have looked at my databases in
> phpMyAdmin and even checked them using the command line on our box and they
> have the right permissions for the users that can access them.  If this is
> a SQL issue can you guys point me in the right direction?  I was thinking of
> pulling my data base handle code and the SQL queries and just throwing them
> into a bare script to see what happens.  If anyone needs to see the HTML
> output I can get that as well.
>
>
>
> Any help is appreciated.
>
>
>
> Thanks,
>
>
>
> *Anthony*
>
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-- 
Brian O'Connor
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