php dropdown list box

[ilikephp]

Hi,

 

I am trying to call the data from Mysql but I am getting an empty drop down list, this is the code:

 

 

mysql:

 

create table years
(  yearID integer auto_increment,
   year varchar(30),
   primary key (yearID)
);

insert into years (yearID, year) values ('1', '2007-2008');
insert into years (yearID, year) values ('2', '2008-2009');
insert into years (yearID, year) values ('3', '2009-2010');
insert into years (yearID, year) values ('4', '2010-2011');
insert into years (yearID, year) values ('5', '2011-2012');
insert into years (yearID, year) values ('6', '2012-2013');

 

PHP:

 

<?php require_once('../Connections/connection.php'); ?>
<?php

$result = @mysql_query( "select yearID, year, from sss.years");

print "<p>Select a year:\n";
print "<select name=\"yearID\">\n";
while ($row = mysql_fetch_assoc($result)){
$yearID = $row[ 'yearID' ];
$year = $row[ 'year' ];
print "<option value=$yearID>$year\n";
}
print "</select>\n";
print "</p>\n";
?>

 

 

Thank you!

[trq]

Remove the error suppression and you might actually see an error message. Better, yet, check for one.

 

<?php require_once('../Connections/connection.php'); ?>
<?php

if ($result = mysql_query( "select yearID, year, from sss.years")) {
  if (mysql_num_rpws($result)) {
    print "<p>Select a year:\n";
    print "<select name=\"yearID\">\n";
    while ($row = mysql_fetch_assoc($result)) {
      $yearID = $row[ 'yearID' ];
      $year = $row[ 'year' ];
      print "<option value=$yearID>$year\n";
    }
    print "</select>\n";
    print "</p>\n";
  } else {
    // no record found.
  }
} else {
  trigger_error(mysql_error());
}

[ilikephp]

thx for your help!

 

I am getting this error:

 

Notice: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'from sss.years' at line 1

[eevan79]

$result = mysql_query( "select yearID, year from sss.years")

[ilikephp]

I am using the code below, it is working fine, but how can I add to the first line of the list an item without a value like: "Select please"

 

<?php
$query="SELECT yearID,year FROM years";

$result = mysql_query ($query);
echo "<select name=education_level value=''>Education Level</option>";

while($nt=mysql_fetch_array($result)){
echo "<option value=$nt[yearID]>$nt[year]</option>";
}
echo "</select>";
?>

[trq]

So, you have chosen to simply ignore the handling of errors like I showed you?

[ilikephp]

For sure not but I did not understand it :S

[trq]

Its simply checking to make sure the called functions actually work before using there results in another function.

 

As for you question. Just change....

 

echo "<select name=education_level value=''>Education Level</option>";

 

to....

 

echo "<select name=education_level value=''>Education Level</option>";
echo "<option value=''>Select PLease</option>";

[ilikephp]

THANKS A LOT (Y)

 

still have one issue:

 

when the form is submitted successfully the success.php page will open

 

$insertGoTo = "success.php";

 

 

if someone tries to put this link www.mywebsite.php/success.php it will be opened, how can I restricted from being viewed unless the form is submitted?

 

 

Thanks,

[trq]

If you are redirecting them to the success page there isn't much you can do. You could store a flag in the $_SESSION array but it seems like a bit of overkill to me.

[ilikephp]

is there a tutorial about this?

[trq]

I wouldn't know.