Re: Problem with INSERT Query

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<irinchiang@xxxxxxxxxxxxxxxxx> wrote in message
news:1072768171.3ff124abe14e8@xxxxxxxxxxxxxxxxxxxxxxxx
>
>
> Hi all:
>
> Well, when i bring out the page with the drop down list it was able to
display
> all tutors' names from "tutor_name" column. Anyway here's a review of my
code
> (snip) again before i continue:
> --------------------------------------------------------------------------
-----
> <snip>
> $sql = "INSERT INTO class (class_code, tutor_name, edu_level,
timetable_day,
> timetable_time)
>         VALUES
>
('$class_code','$tutor_name','$edu_level','$timetable_day','$timetable_time'
)";
>
>
> <?//retrieve data from DB & display in dynamic drop down ?>
>
> <SELECT class="textarea" name="tutor_name" />
> <?
>
>
 $sqltutor = mysql_query("SELECT DISTINCT tutor_name FROM tutor ");
 while ($row = mysql_fetch_array($sqltutor))
 {
  print "<OPTION VALUE=\"$row ["tutor_name"]. \" SELECTED> " .$row
 ["tutor_name"]. "</option> ";
>  }
>  $result = $db-> query($sql);
>
> ?>
> </select>
>
> <?
>
> while($selected_tutor_name == $tutor_name)
> echo $_POST["tutor_name"];
>
How about
if ($tutor_name ){
echo $_POST["tutor_name"];
$sql = "INSERT INTO class (class_code, tutor_name, edu_level, timetable_day,
 timetable_time)
         VALUES

('$class_code','$tutor_name','$edu_level','$timetable_day','$timetable_time'
)";}

is it working?


Hadi

> ?>
>
> </snip>
>
> --------------------------------------------------------------------------
-----
>
> so when i submit the form, i am suppose to echo the values i have entered
into
> the field and then INSERT the values into DB (Queries stated above).
However i
> was able to echo all other values eg. class_code, edu_level, etc...but
> not "tutor_name"....same thing happen when i do an INSERT, all other
values
> are inserted into DB but not $tutor_name....why is this so???Really need
some
> help here...Anyway i have already specify a name to be reference :
>
> <SELECT class="textarea" name="tutor_name" >
>
> and then I also did an echo of "tutor_name" being selected:
>
> while($selected_tutor_name == $tutor_name)
> echo $_POST["tutor_name"];
>
> All help are greatly appreciated =)
>
> Irin.

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