[Resolved] Populating SELECT

jlpierce

Hello all,

Thanks to craig, in another post, I have a piece of code that will dynamically re-populate a select field with the selected option if the form has to be submitted again for validation.

<TD>Password Hint:</TD>
<TD>
<SELECT name=hint>
<OPTION VALUE=maiden<? if ($hint == 'maiden') print " selected"; ?>>Mother's maiden name?
<OPTION VALUE=color<? if ($hint == 'color') print " selected"; ?>>Your favorite color?
<OPTION VALUE=dog<? if ($hint == 'dog') print " selected"; ?>>Your dogs name?

<OPTION VALUE=eatery<? if ($hint == 'eatery') print " selected"; ?>>Your favorite restaurant?
</SELECT>
</TD>

This code works well for the static nature of this select statement.

However, I am dynamically generating the options for a select list options from a database and I cannot figure out how to redisplay the selected option if the form must be resubmitted to the user for correction.

This is the code that I am using to generate the option statements.

function populateselectoptions();
{

// Set up table query
$thisconnect = user_account_db_connect();
$local_qry = "select state_prov from stat_prov";
$result = mysql_query( $local_qry ) or die ( mysql_error() );

// Construct HTML code

if ( $result )
{
while ( $row = mysql_fetch_array( $result, MYSQL_ASSOC ) )
{
while( list ($key, $value) = each($row) )
{
echo "<option";
echo " value=" . ">";
echo "$value \n";
}
}
}
} // End populateselectoptions

No matter how I try to incorporate the following code into the option statement above parts of the php code will end up displaying in the field and will not print the selected option on the second draw of the form after a failed validation.

I hope someone can assist me, I have already lost enough hair, and the rest is going fast.

John

mharju

Well, as to the syntax of HTML, I've always thought that the HTML also needs the closing tags, so the syntax would be:

<SELECT>
<OPTION></OPTION> <-- notice the closing tag
...
</SELECT

Secondly, why on earth are you doing 3 echos? and I think the HTML parser doesn't like OPTION tags like you print:
<OPTION value=>something and they are also without <SELECT></SELECT>-tags ;-)

Try:
#before the while-loop put:
echo "<SELECT NAME=\"something\">\n";

#to the loop, use this instead of the three
echo "<OPTION
VALUE=\"$value\">$value</OPTION>\n";

#and finally
echo "</SELECT>\n";

There are also a few more things in your code that I noticed, but they shouldn't matter.

Hope this helps you!

🙂Mikko

Anon

What about the code for calling data from a table which are going to be displayed in the <option>?

jlpierce

Ok, I have the opening and closing pair for SELECT in the html page with the call to the function in the middle. ie:

If I statically populate an <option> tag like the following then when the form has to be redisplayed it will print the value that had been previously selected. ie:

<OPTION VALUE=test1<? if ($something == 'test1') print " selected"; ?>>you selected test1!</OPTION>

So when I try to use a function to populate the options and at the same time insert the php code above to print the selected value, I see parts of the php code also displayed.

while ($row=mysql_fetch_array($result,MYSQL_ASSOC ) )

{
while( list ($key, $value) = each($row) )
{
echo "<OPTION
VALUE=\"$value\"<? print \" selected\";?>>$value</OPTIONS>\n\";
}
}

Anon

make your query and the populate the select... like this:
<?
<select name="something">
?>
$query = mysql_query("slect * from table");
while ($row = mysql_fetch_array($query))
{
echo "<option>" . $row["field"] . "</option>"
}
<?
</select>
?>

jlpierce

First, are the opening and closing PHP tags in the wrong spot.

<?
<SELECT NAME="something">
?>

Doesn't this tell the parser that what's in the middle should be treated as PHP.

Secondly, I can populate the OPTIONS correctly already. I am developing a new user registration form and if the form is submitted and data in the form is not valid ( as in they did not supply an email address or username ) then I want to redisplay the form with all of the pre-entered data to save the user typing, and then indicate what needs to be corrected.

However, by dynamically populating the OPTIONS when the form is re-displayed, it will always show the first OPTION, in this case it displays ALABAMA, which is the first record in the table.

I have successfully re-displayed the selected option if the OPTIONS are populated statically, that is hardcoded into the SELECT statement. See previous posting.

John