I need some help

[bluwe]

I want to retrieve a record from my database and display the record as the default choice in a html drop down menu - is this possible?

Cheers...

[gluck]

It is possible. You just need to add details of what specifics you want

[bluwe]

Thanks for replying. I've got a title column in my customer details table with the following options: Mr, Mrs, Miss, Ms, Dr.

When a user edits the customer details record it is obviously retireved from the db using a SELECT statement and displayed in a html table. I want to populate the first entry in the title drop down menu in the table with the value returned from the db, but still have the other options available. If you see what I mean!

Cheers!

[keeB]

[code=php:0]
<select name="zomg">
<option name="ms">Ms.</option>
<option name="dr" selected>Dr.</option>
<option name="mr">Mr.</option>
</select>
[/code]

Notice the "selected" attribute in <option>

[craygo]

this will do it

[code]<select name=title>
<?php
$title = array("Mr", "Mrs", "Miss", "Ms", "Dr.");
foreach($title as $tname){
  if($tname == $dbvariable){
  echo "<option value=$title selected>$title</option>";
  } else {
  echo "<option value=$title>$title</option>";
  }
}
?>
</select>[/code]

Ray

[bluwe]

That's excellent! Thanks you all very much!!!

[bluwe]

Thanks for the start Ray, needed a bit of debugging though, but finally got it working like so...

[code]<select name=title>
<?php
$dbvariable = "Miss";
$title = array ("Mr", "Mrs", "Miss", "Ms", "Dr");
foreach($title as $tname){
  if($tname == $dbvariable){
  echo "<option value=$dbvariable selected>$dbvariable</option>";
  } else {
  echo "<option value=$title>$tname</option>";
  }
}
echo "</select>";
?>[/code]

[bluwe]

Actually the above doesn't work, hmm....

[bluwe]

This deffo works though...

[code]<select name=title>
<?php
$dbvariable = "Dr";
$title = array ("Mr", "Mrs", "Miss", "Ms", "Dr");
foreach($title as $tname){
  if($tname == $dbvariable){
  echo "<option value=$tname selected>$dbvariable</option>";
  }
}
foreach($title as $tname){
if($tname !=$dbvariable){
  echo "<option value=$title>$tname</option>";
}
}
echo "</select>";
?>[/code]

[kenrbnsn]

Actually your were closer with this:
[code]<?php
<select name=title>
<?php
$dbvariable = "Miss";
$title = array ("Mr", "Mrs", "Miss", "Ms", "Dr");
foreach($title as $tname){
  if($tname == $dbvariable){
  echo "<option value=$dbvariable selected>$dbvariable</option>";
  } else {
  echo "<option value=$title>$tname</option>";
  }
}
echo "</select>";
?>[/code]

The problem is that you can't use the variable "$title" as a value, since all that will give you is the string "Array".

Try this instead:
[code]<?php
<select name=title>
<?php
$dbvariable = "Miss";
$title = array ("Mr", "Mrs", "Miss", "Ms", "Dr");
foreach($title as $tname){
  $sel = ($tname == $dbvariable)?' selected':''; // sets $sel to either selected or nothing
  echo '<option value="' . $tmame . '"' . $sel . 'selected>' . $tname. "</option><br>"; // make sure all values of attributes are quoted
  }
echo "</select>";
?>[/code]

Ken

[craygo]

SHIT good catch ken. I buzzed though it so quick I put the wrong one in

[code]<select name=title>
<?php
$title = array("Mr", "Mrs", "Miss", "Ms", "Dr.");
foreach($title as $tname){
  if($tname == $dbvariable){
  echo "<option value=$tname selected>$tname</option>";
  } else {
  echo "<option value=$tname>$tname</option>";
  }
}
?>
</select>[/code]

Ray

[bluwe]

Thnaks for all your help, much appreciated!!