使用MySQL的PHP​​更新表

Question

我在我的数据库中有学生表，我需要在学生页面中创建额外的表单，以便他可以更新详细信息。然而即时得到这个错误（警告：mysqli_query（）预计至少2个参数，1 C中给出：\ WAMP \ WWW \ WebInterfaceLogIn \ studentEdit.php上线81）

<?php 
ini_set('display_errors', true); 
error_reporting(E_ALL); 
include ('includes/connection.php'); 

// Create connection 

if (isset($_POST["submit"])) 
{ //Determine if a variable is set and is not NULL. 
    if (!empty($_POST['ID']) && !empty($_POST['user']) && !empty($_POST['surr']) && !empty($_POST['course']) && !empty($_POST['mail']) && !empty($_POST['pass'])) 
    { //Determine if user enters both user name and password. 
     $ID = $_POST['ID']; // enters user ID in database 
     $user = $_POST['user']; // enters user name in database 
     $surr = $_POST['surr']; // enters user surname in database 
     $course = $_POST['course']; // enters user course in database 
     $pass = $_POST['pass']; // enters password in database 
     $mail = $_POST['mail']; 

     // $query = mysqli_query($con,"SELECT * FROM students WHERE Student_ID='".$ID."'"); // change to update 

     $query = mysqli_query("UPDATE students SET `course` = " . $_POST['course'] . ", `email` = " . $_POST['mail'] . " Student_ID='" . $ID . "'"); 
     $numrows = mysqli_num_rows($query); 
     if ($numrows == 0) 
     { 
      $sql = "INSERT INTO students(Student_ID,Name,Surname,Course,email,password) VALUES('$ID', '$user','$surr','$course','$mail','$pass')"; // insert user name and password to database 
      $result = mysqli_query($con, $sql); 

      // Checks does user enters the details 

      if ($result) 
      { 
       echo '<script language="javascript">; 
        alert("Account Successfully Updated"); 
         document.location.href="index.php"; 
        </script>'; 
      } 
      else 
      { 
       echo mysqli_error($con); 
      } 
     } 
    } 
    else 
    { 
     echo '<script language="javascript"> 
           alert("All fields required") 
          </script>'; 
    } 
} 

任何人都可以帮助解决这个问题？

Answer 1

你需要通过连接对象功能mysqli_query，是这样的：

mysqli_query($con,"UPDATE students SET `course` = " .$_POST['course']. ", `email` = " .$_POST['mail']. " Student_ID='".$ID."'"); 

查看更多详细信息： http://www.w3schools.com/php/func_mysqli_query.asp

Answer 2

的错误是不够明确。

附加$con作为第一个参数，您mysqli_query功能在这条线：

$query = mysqli_query("UPDATE students SET `course` = " . $_POST['course'] . ", `email` = " . $_POST['mail'] . " Student_ID='" . $ID . "'");