我正在尝试使用PHP代码从MySQL数据库中选择数据,但总是出错。
<?php
$dbhost = "localhost";
$dbuser = "";
$dbpass = "";
$db = "test";
$connect = mysql_connect($dbhost, $dbuser, $dbpass, $db)
or die ("connexion impossible");
mysql_select_db($db) or die ("selection de la base échoué");
$username = $_POST['username'];
$password = $_POST['password'];
query = mysql_query("SELECT * FROM table2 WHERE username= '$username' AND password='.$password'");
$num = mysql_num_rows($query);
if($num == 1) {
while($list = mysql_fetch_assoc($query)){
$output = $list;
echo json_encode($output);
}
mysql_close();
}
?>
错误:
Notice: Undefined variable: username in C:'wamp'www'projet'connect.php on line 11
Notice: Undefined variable: password in C:'wamp'www'projet'connect.php on line 11
替换该行
query = mysql_query("SELECT * FROM table2 WHERE username= '$username' AND password='.$password'");
跟
$query = mysql_query("SELECT * FROM table2 WHERE username= '".$username."' AND password='".$password."'");
您错过了变量查询之前的 $,并且存在一些字符串连接问题
请使用 PDO 而不是已弃用的mysql_*
在
mysql查询中$password
之前有额外的点,query
应该$query
,因为$query
是可变的。
query = mysql_query("SELECT * FROM table2 WHERE username= '$username' AND password='.$password'");
应该是
$query = mysql_query("SELECT * FROM table2 WHERE username= '$username' AND password='$password'");