Need PHP Select records into dropdown box help

malibu65k · Feb 15, 2011

I have php code to select records from a table in mysql. I have a dropdown box that isn't filling in with the records. Not sure what I am doing wrong. Thank you in advance for any help you can provide.

Here is my code...

<html>
<body>
<head>
<title>Search for a Pattern</title>
</head>
<body bgcolor="#ffffff">

<h2>Search for a Pattern</h2>

<form action="action" method="post">
<select name="option">

<?php

$user = "Admin";
$host = "LocalHost"
$password = ""
$dbName = "mypatterns"

mysql_connect($host, $user, $password) OR DIE( "Unable to connect to database");

mysql_select_db($dbName);

$sql = "SELECT PatternNo FROM patterns";
$result = mysql_query($sql) or die(mysql_error());

while ($row = mysql_fetch_array($result))

{
echo '<option value="' . $row['PatternNo'] . '">' . $row['PatternNo'] . '</option>';
}

?>
</select>

</form>

jstreich · Feb 15, 2011

First, mysql_ functions are a bad idea; try switching to mysqli_ for better sanitation and being type safe.

The die(...) function should be lowercase in both instances.

besides that, what error(s) are you getting?

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jpadie · Feb 15, 2011

mysql_ functions are a bad idea

I disagree.

The die(...) function should be lowercase in both instances.

there is no need for this. functions in php are case insensitive.

http://fr2.php.net/manual/en/functions.user-defined.php

@OP
the errors that I can see are that your variable assignments are not terminated by a semicolon.

Code:

$user = "Admin";  
$host = "localhost" ;
$password = "" ;
$dbName = "mypatterns";

make sure that the case of your user name and password is correct.

it is a good idea when debugging to add the following to the start of your file

Code:

error_reporting(E_ALL);
ini_set('display_errors', true);

adding some debug to your mysql_* calls is a good idea

Code:

mysql_select_db($dbName) or die (mysql_error());

perhaps use mysql_fetch_assoc rather than _array as you are referencing the columns by their name.

Code:

while ($row = mysql_fetch_assoc($results)){

}

other things to be aware of:

1. might the patternno have any quotes in it? if so the value attribute might be breaking.
2. you might consider running htmlspecialchars on the value of PatternNo when in the text part of the <option> tag
3. the <select> element must be terminated

Code:

echo '</select>';

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Need PHP Select records into dropdown box help

malibu65k

Programmer

jstreich

Programmer

jpadie

Technical User

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