old*_*ete 2 php mysql mysqli function
我花了几个小时阅读并试图弄清楚为什么这不起作用,但下拉菜单没有填充.我认为这很简单,但我看不到它.任何人？
dbconn.php
<?php
define('DB_NAME' , 'artprints');
define('DB_USER' , 'root');
define('DB_PASS' , '');
define('DB_HOST' , 'localhost');
func.php
<?php
include_once 'dbconn.php';
function connect(){
$connection = mysqli_connect(DB_HOST, DB_USER, DB_PASS) or die ('Could not connect to the database' . mysl_error());
mysqli_select_db($connection, DB_NAME);
}
function close(){
mysql_close();
}
function query(){
$myData = mysql_query("SELECT * FROM artists");
while ($record = mysql_fetch_array($myData)){
echo '<option value="' . $record['artistID'] . '">' . $record['artistID'] . '</option>';
}
}
test.php的
<?php
include_once 'func.php';
connect();
?>
<html>
<head>
<title>Drop down testing</title>
</head>
<body>
<select name='artist'>
<?php query() ?>
</select>
<?php close() ?>
</body>
</html>
你正在混合mysqli_*和mysql_*
mysqli_connect 和 mysqli_select_db
反对
mysql_query 和 mysql_fetch_array
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