我的 php 腳本有問(wèn)題。對(duì) php 很陌生，所以我知道它不是最好的。提交表單時(shí)出現(xiàn)以下錯(cuò)誤。注意：未定義索引：第 6 行 C:\xampp\htdocs\AppX1\signin.php 中的用戶名 注意：第 7 行 C:\xampp\htdocs\AppX1\signin.php 中未定義索引：密碼 致命錯(cuò)誤：未捕獲錯(cuò)誤：調(diào)用 C:\xampp\htdocs\AppX1\signin.php:13 中未定義的函數(shù) mysql_real_escape_string() 堆棧跟蹤：第 13 行 C:\xampp\htdocs\AppX1\signin.php 中拋出 #0 {main}我讀過(guò)一些文章說(shuō)要使用一種isset方法，盡管我不確定該將其放在哪里或者它是否適用于我的情況。<form action="signin.php" id="signInForm">    <fieldset>        <label class="fieldLabel" for="uname">Email Address</label>        <input class="fieldInput" type="text" id="uname" name="username">        <label class="fieldLabel" for="pword">Password</label>        <input class="fieldInput" type="password" id="pword" name="password">        <input class="submitBtn" type="submit" value="Submit">    </fieldset></form><?php  $username = $_POST['username'];  $password = $_POST['password'];  // sql injection prevention  $username = stripcslashes($username);  $password = stripcslashes($password);  $username = mysql_real_escape_string($username);  $password = mysql_real_escape_string($password);  //connect to server and select SQLiteDatabase  mysql_connect("localhost", "root", "");  mysql_select_db("user");  //query db for user  $result = mysql_query("SELECT * FROM user where email = '$username' and password = '$password'")            or die("Failed to query database ".mysql_error());  $row = mysql_fetch_array($result);  if ($row['email'] == $username && $row['password'] == $password) {    echo "Login Success!".$row['email'];  }  else {    echo "Login Failed";  }?>