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using an array with an if statement

Hi, i am using this code below to generate my drop down box (all the select statements and everything are fine)

[code=php]
<select name=”Type” id=”Type”>
<option>Please Select</option>
<?PHP
//fill in the categories
while ($row = mysql_fetch_array($type_res)) {
extract($row);
echo “<option value=’$Type’>$Type</option>n”;
}
?>
</select>
[/code]

This works fine it populates the drop down box and everything its great.

what i have done below is put in an if script where if $Sub_Type is equal to $Type then it will echo selected so that, it will show what is in the database

so u know $Type is pulling from a different database to $Sub_Type and $Type is only to populate the drop down. what is happenen is i have a query to retrieve $sub_type from a product database and this is for an edit screen. so i need them to be able to change it but only to things populated from the $Type

[code=php]
<option>Please Select</option>
<?PHP
//fill in the categories
while ($row = mysql_fetch_array($type_res)) {
extract($row);
echo “<option value=’$Type’ <?PHP if($Sub_Type==$Type) echo”selected”;>?>$Type</option>n”;
}
?>
</select>
[/code]

This code works when i am not doin an arrary but the array messes it up i beleive becuase of the echo coz i usualy do this

[code=php]
<select name=”sex”>
<option value=”Male” <?php if($sex==”Male”) echo “selected”; ?>>Male</option>
<option value=”Female” <?php if($sex==”Female”) echo “selected”; ?>>Female</option>
</select>
[/code]

Any ideas y it aint working ?

Thanks Adam

to post a comment
PHP

2 Comments(s)

Copy linkTweet thisAlerts:
@NogDogSep 09.2004 — You have embedded php tags within a php block. Try something like this instead:
[code=php]
<option>Please Select</option>
<?PHP
//fill in the categories
while ($row = mysql_fetch_array($type_res))
{
extract($row);
echo "<option value='$Type'";
if($Sub_Type==$Type)
{
echo "selected";
}
echo ">$Type</option>n";
}
?>
</select>
[/code]
Copy linkTweet thisAlerts:
@Paul_JrSep 10.2004 — Alternately, so you don't have that if statement breaking up your output (I bet I'm the only one who isn't okay with that ?):
[code=php]
<option>Please Select</option>
<?PHP
//fill in the categories
while ($row = mysql_fetch_array($type_res)) {
extract($row);
$selected = $Sub_Type == $Type ? ' selected' : '';
echo "<option value='$Type' $selected>$Type</option>n";
}
?>
</select>
[/code]

That way, if $Sub_Type doesn't equal $Type, the string is just empty, otherwise the value is ' selected'.
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