我不能登录系统

Question

这里是我的login.php

<? 
    session_start(); 
    if (isset($_POST['Login'])) { 
    include_once("connect.php"); 
    $user_ok = false; 
    $usernamer=$_POST['Usernamer']; 
    $passaword=$_POST['passsword']; 
    $id = "SELECT id FROM websiteusers WHERE userName='$Usernamer' AND pass='$passaword' LIMIT 1"; 
    function evalLoggedUser($link,$id,$usernamer,$passaword){ 
     $sql = "SELECT ip FROM websiteusers WHERE userName='$Usernamer' AND pass='$passaword' LIMIT 1"; 
     $query = mysqli_query($link, $sql); 
     $numrows = mysqli_num_rows($query); 
     if($numrows > 0){ 
      return true; 
     } 
    } 
     $user_ok = evalLoggedUser($link,$usernamer,$passaword,$id); 
     if(isset($_POST['Login'])){ 
     if($user_ok == true){ 
     header("location: user.php?u=".$_POST["Usernamer"]); 
     }else{ 
     echo "username was not found"; 
     } 
     } 
    } 
    ?> 

它给像

mysqli_num_rows误差修改（）预计参数1被mysqli_result，空给出

即使当我输入正确的用户名时，它也会回显username was not found

Answer 1

I thi NK应该

if($numrows < 0){ 
    return true; 
    } 

和您的查询应该是：

$id = "SELECT id FROM websiteusers WHERE userName='$Usernamer' AND pass='$passaword' LIMIT 1"; 
//change $Usernamer to $usernamer 

Answer 2

呀！参数一必须是结果集。这意味着你必须通过$查询不是你的SQL查询

Answer 3

原来的一些变化与适当的意见

<?php 
    session_start(); 
    if (isset($_POST['Login'])) { 

     include_once("connect.php"); 
     $user_ok = false; 

     $usernamer=$_POST['Usernamer']; 
     $passaword=$_POST['passsword']; 
     /* to debug */ 
     echo "POSTED username:".$usernamer." POSTED password: ".$passaword; 

     /* 
      this gets passed as a parameter to the function but never used... 
      $id = "SELECT id FROM websiteusers WHERE userName='$Usernamer' AND pass='$passaword' LIMIT 1"; 
     */ 
     function evalLoggedUser($conx, $u, $p){ 
      /* To use the variables above ($usernamer & $passaword) you would need to declare them as global within the function */ 
      /* I assumed, perhaps wrongly, that this ought to have been id rather than ip */ 

      $sql = "SELECT `id` FROM `websiteusers` WHERE `userName`='$u' AND `pass`='$p' LIMIT 1"; 

      /* to debug */ 
      exit("PARAM Username:".$u ."<br />PARAM Password:". $p ."<br />sql:" .$sql); 

      $query = mysqli_query($link, $sql); 
      $numrows = mysqli_num_rows($query); 

      /* previously the function only returned a value if there was a recordset */ 
      return ($numrows > 0) ? true : false; 
     } 


     /* 
      The sequence of parameters was incorrect 
     */ 
     $user_ok = evalLoggedUser($link, $usernamer, $passaword); 

     if(isset($_POST['Login'])){ 
      if($user_ok === true){ 
       header("location: user.php?u=".$usernamer); 
      }else{ 
       echo "username was not found"; 
      } 
     } 
    } 
?>