mysqli_query（）預計參數1是mysqli的，對象

Question

function isValidUser($dbc,$email) 
    { 

     $sql=sprintf("select id from user where email = '%s' and password ='%s' ", $_POST['email'],$_POST['password']); 
     $qresult = mysqli_query($dbc,$sql);//line 18 
     $test = mysqli_fetch_array($qresult);//line 19 
     echo $test[0]; 
     if(mysqli_num_rows($qresult)>0)//line21 
     { 
      echo "User Exists"; 
      return true; 
     } 
     else 
     { 
      echo "Not a User"; 
      return false; 
     } 
    } 

給予我想，也許我有一些範圍有關的問題（按類似問題的其他計算器答覆），但我不似乎understand.I是新來的PHP任何人都可以告訴我如何糾正這一點。

mysqli_query() expects parameter 1 to be mysqli, object given in... line 18 
mysqli_fetch_array() expects parameter 1 to be mysqli, null given in line 19 
mysqli_num_rows() expects parameter 1 to be mysqli, null given in line 21 

Answer 1

試試這個：

function isValidUser($dbc,$email) 
    { 
     $email = $_POST['email']; 
     $host = ""; 
     $username = ""; 
     $password_mysql = ""; 
     $database =""; 

     $password = $_POST['password']; 
     $link = mysqli_connect($host, $username, $password_mysql, $database); /* hard code the connection variables in to test if something is wrong here. */ 
     $sql="select id from user where email = '".$email."' and password ='".$password."'"; 
     $qresult = mysqli_query($link,$sql);//line 18 
     $test = mysqli_fetch_array($qresult, MYSQLI_BOTH);//line 19 
       while($test = mysqli_fetch_array($qresult, MYSQLI_BOTH)){  
       echo $test['id']."<br>"; 
       }; 
     if(mysqli_num_rows($qresult)>0)//line21 
     { 
      echo "User Exists"; 
      return true; 
     } 
     else 
     { 
      echo "Not a User"; 
      return false; 
     } 
    }