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我創造，我想在MySQL數據庫中檢索數據，並以droplist顯示它的每個表中的搜索過程包含兩個字段一個是ID等是名稱PHP與MySQL數據庫檢索數據
的代碼或國家的功能工作得很好，但專業化的其他功能不顯示任何內容。如果有人能幫助PLZ
function.php
<?php
require_once('db.inc.php');
function connect(){
mysql_connect(DB_Host, DB_User ,DB_Pass)or die("could not connect to the database" .mysql_error());
mysql_select_db(DB_Name)or die("could not select database");
}
function close(){
mysql_close();
}
function countryQuery(){
$countryData = mysql_query("SELECT * FROM country");
while($record = mysql_fetch_array($countryData)){
echo'<option value="' . $record['country_name'] . '">' . $record['country_name'] . '</option>';
}
}
function specializtionQuery(){
$specData = mysql_query("SELECT * FROM specialization");
while($recordJob = mysql_fetch_array($specData)){
echo'<option value="' . $recordJob['specialization'] . '">' . $recordJob['specialization'] . '</option>';
}
}
?>
的index.php
<?php
require_once('func.inc.php');
connect();
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>testDroplistdown</title>
</head>
<body>
<p align="center">
<select name="dropdown">
<?php countryQuery(); ?>
</select>
<?php close(); ?>
</p>
<p align="left">
<select name="dropdown2">
<?php specializationQuery(); ?>
</select>
<?php close(); ?>
</p>
</body>
</html>
感謝您的答案，但仍然沒有得到任何數據顯示 – 2013-02-27 21:03:08