我一直在尝试各种各样的事情,出于某种原因,无论我做什么,我似乎无法用一个可用的查询...到目前为止,这是我的代码,它不是全部,但我的针指向我的地方问题是......
<?php
$hostname = 'localhost'; // Your MySQL hostname. Usualy named as 'localhost', so you're NOT necessary to change this even this script has already online on the internet.
$dbname = 'ServiceHistoryDB'; // Your database name.
$username = 'root'; // Your database username.
$password = ''; // Your database password. If your database has no password, leave it empty.
// Let's connect to host
mysql_connect($hostname, $username, $password) or DIE('Connection to host is failed, perhaps the service is down!');
// Select the database
mysql_select_db($dbname) or DIE('Database name is not available!');
$custnum = '1';
function connect(){
mysql_connect($hostname, $username, $password) or DIE('Connection to host is failed, perhaps the service is down!');
mysql_select_db($dbname) or DIE('Database name is not available!');
}
function close(){
mysql_close();
}
function query(){
$mydata = mysql_query("SELECT * FROM vehicles WHERE CustNum=1");
while($record = mysql_fetch_array($mydata)){
echo '<option value="' . $record["Model"] . '">' . $record["Model"] . '</option>';
}
}
?>
<!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>testdropdown</title>
</head>
<body>
<select name="dropdown">
<?php query() ?>
</select>
<?php close() ?>
</body>
</html>
答案 0 :(得分:0)
好吧,你可以简单地写下面的变量:
$variable = 'someData';
"Select someColumn from yourTable where someColumn = $variable";
但是,如果您是通过函数执行此操作,则可能需要声明global
变量以检索值 如果 ,则将值设置为外部功能:
function someFunc(){
global $variable; // set above
"Select someColumn from yourTable where someColumn = $variable";
}
您应该切换到mysqli prepared statements
,因为它们更安全:
if($stmt = $mysqli->prepare($yourSQL)){
$stmt->bind_param("s", $variable);
$stmt->execute();
$stmt->bind_result($thisHoldsThe_QueryResult);
$stmt->fetch();
$stmt->close();
}