doufangmu9087 2015-10-06 09:48
浏览 152
已采纳

警告:mysqli_error()期望参数1为mysqli,字符串为

I want to connect to a database (with xampp) called ranch and insert into this db some form data. The browser displays the error:

Warning: mysqli_error() expects parameter 1 to be mysqli, string given in C:\xampp\htdocs\project1egister.php on line 38

ERROR: Could not able to execute INSERT INTO child_parent('childname','childsurname','age','gender','name','surname','address','tk','city','telephone','mobile','email','parea','pass') VALUES('nikos','ads','34','Αγόρι','sds','sds','dsd','34','dsds','34','434','mail@hotmail.com','sds','34').

I have the .php file that is below. How can i fix it?

<?php

ini_set('display_errors', 'On');
session_start();
unset ($msg);
    echo "kajsj<br>";
    $conn= new mysqli("localhost","root","","ranch");
    
    if (mysqli_connect_errno())
    { printf("Connect failed: %s
",mysqli_connect_error());//error message
    }
    else
    {
    printf("Connect achieved<br>"); 
    echo $_GET['childname'];
    $childname=$_GET['childname'];
    $childsurname=$_GET['childsurname'];
    $age=$_GET['age'];
    $gender=$_GET['gender'];
    $name=$_GET['name'];
    $surname=$_GET['surname'];
    $address=$_GET['address'];
    $tk=$_GET['tk'];
    $city=$_GET['city'];
    $telephone=$_GET['telephone'];
    $mobile=$_GET['mobile'];
    $email=$_GET['email'];
    $parea=$_GET['parea'];
    $pass=$_GET['pass'];
    // Insert data into mysql 
    $query1="INSERT INTO child_parent('childname','childsurname','age','gender','name','surname','address','tk','city','telephone','mobile','email','parea','pass')
                                VALUES('$childname','$childsurname','$age','$gender','$name','$surname','$address','$tk','$city','$telephone','$mobile','$email','$parea','$pass')";
    
    if(mysqli_query($conn, $query1))
    {echo "Records added successfully.";
    } 
    else{echo "ERROR: Could not able to execute $query1. " . mysqli_error($query1);
    }
    }
?>

</div>
  • 写回答

1条回答 默认 最新

  • douyangqian5243 2015-10-06 09:53
    关注

    There are three main problems with your code:

    1. Wrong SQL. You have to learn basic SQL syntax prior trying to use SQL from PHP.
    2. Wrong mysqli syntax. You have to use prepared statements instead of adding variables right in the query.
    3. Wrong error reporting. You have to set mysqli in exception mode instead of checking every query result manually.

    To solve them one by one, read these three helpful answers:

    1. When to use single quotes, double quotes, and backticks in MySQL
    2. How can I prevent SQL injection in PHP?
    3. How to get mysqli error in different environments?
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

报告相同问题?