警告:mysqli_error()只需要1个参数,给定0

时间:2017-03-22 12:49:40

标签: php html mysql post

我在这里搜索了同样的问题被问到并发现许多答案根本没有解决我的错误,通常只是创造了更多:P 这是完整的错误

Warning: mysqli_error() expects exactly 1 parameter, 0 given in F:\Jacob's Project rehash\register.php on line 54.

以下是它所居住的整个PHP页面:

 <!DOCTYPE html>
<html>

<head>

<title>Membership Confirmation</title>
    <script type='text/javascript' src='gen_validatorv31.js'></script>
</head>

<body>
<h1></h1>
<p>Thank for your membership support!</p>


<br>


<?php

//assign connection details to PHP variables
$servername="localhost";                                             
$username="root";                                                    
$password="";                                                        
$database="bryanbook"; 

//connect to the database server and select Bryanbook database
$link=mysqli_connect($servername,$username,$password,$database);
//check for successful connection
if (mysqli_connect_error())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }


//assign values submitted by HTML form to PHP variables
$email = $_POST['email'];                         
$firstname = $_POST['firstname'];                         
$surname = $_POST['surname'];                             
$addressline1 = $_POST['addressline1'];            
$password =   $_POST['password'];                     
$towncity = $_POST['towncity'];                            
$postcode = $_POST['postcode'];                            
$favouritecolour = $_POST['favouritecolour'];                         
$likes = $_POST['likes'];                          
$dislikes = $_POST['dislikes'];                           




//create query to add member details to database
$query = "INSERT INTO users (email,password,firstname,surname,`address 
line 1`,town/city,postcode,`fave colour`,likes,dislikes) VALUES ('$email','$password','$firstname','$surname','$addressline1','$towncity','$postcode','$favouritecolour','$likes','$dislikes')";
//execute SQL query to add details to the member table
$data = mysqli_query($link, $query)or die(mysqli_error());   <------- Line 54


//check that query has been successful
if($data)
  {
  //display message to notify user that details have been added
  echo "Your registration has been successful";
  }


//close server connection
mysqli_close($link);

//exit PHP
?>


<br>
<p><a href="Home Page.html">Click here to return to the Home Page</a></p>
</body>

</html>

1 个答案:

答案 0 :(得分:2)

更改$data = mysqli_query($link, $query)or die(mysqli_error()); <------- Line 54

to(简单地将$ link连接传递给mysqli_error()函数)

$data = mysqli_query($link, $query)or die(mysqli_error($link)); <------- Line 54