MrFahad
asked on
How to create a dynamic drop down menu from a feild in my database
Hello, i have a database for real estate website. i want to want to make a dynamic drop-down menu that will show the records from a feild in my database's table.
the feild name is PropertyName.
Thank you
the feild name is PropertyName.
Thank you
SOLUTION
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striker46
Penguin, I think you forgot closing the </option> in your code...
ASKER
Penguin it is only display 1st letter of the PropertyName
ASKER
What's the problem with Penguin's code? Apart that there's a missing >?
echo '<SELECT>';
$result = mysql_query('SELECT PropertyName FROM tablename');
foreach (mysql_fetch_array($result) as $property)
{
echo '<option value="'.$property['PropertyName'].'">'.$property['PropertyName'].'</option>';
}
echo '</SELECT>';
@striker46 : Thanks for the missing >
@MrFahad : This piece of code should display exacly what is in the field of the database. Just to make sure we are talking about a field not a row?
@MrFahad : This piece of code should display exacly what is in the field of the database. Just to make sure we are talking about a field not a row?
ASKER
no it's the row ( record )
here is the screen shot of my DB i pointed with any arrow what i want to disply
ss2.jpg
here is the screen shot of my DB i pointed with any arrow what i want to disply
ss2.jpg
Could you please post the outputted HTML code of the dropdown menu? From the first screenshot, with the fetched results.
ASKER
<label for="PropertyName"><em>Ple
<SELECT><option value="9">9</option><optio
<SELECT><option value="9">9</option><optio
That's weird, it retrievied twice a same row? I see the code has two options "9" but in the table you showed in the 2nd screenshot there's just 1 row, isn't there?
Try the code below
Try the code below
<?php
$sql = "SELECT PropertyName
FROM properties;
$result = mysql_query($sql);
while ($row = mysql_fetch_assoc($result)) {
echo '<option value="'.$row["PropertyName"].'">'.$row["PropertyName"].'</option>';
}
?>
Sorry, forgot the <SELECT>
echo "<SELECT>";
$sql = "SELECT PropertyName
FROM properties;
$result = mysql_query($sql);
while ($row = mysql_fetch_assoc($result)) {
echo '<option value="'.$row["PropertyName"].'">'.$row["PropertyName"].'</option>';
}
echo "</SELECT>";
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ASKER
Worked Great Striker46, I'm also giving points for penguin Thanks guys great work.