dtmooir3395 2017-10-23 10:48 采纳率: 0%
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可捕获的致命错误:类mysqli的对象无法在第6行的/home/cabox/workspace/writing/login.php中转换为字符串

This is my code:

<?php
$conn = mysqli_connect("localhost", "root", '060419');
mysqli_select_db($conn, "lol");
$login ="SELECT * FROM username";
$sql = mysqli_query($conn,$login);
$pw = "mysqli_fetch_assoc($sql)";
$username = $pw["username"];
$id = $pw["id"]
?>
<!DOCTYPE html>
<html>
  <head>
    <meta charset="utf-8">
  </head>
  <body>
  <?php
  if($_POST["username"] == $username){
    echo '<a href="http://php-
 noamboy2006504805.codeanyapp.com/writing/realindex.php?user='.$id.'">로그인 
성공
</a>';
  }else{
    echo '로그인 실패<a href="http://php-
noamboy2006504805.codeanyapp.com/writing/index.php">다시 시도</a>';
  }
  ?>
  </body>
</html>

I keep getting this error:

Catchable fatal error: Object of class mysqli could not be converted to string in /home/cabox/workspace/writing/login.php on line 6

Can anyone help?

  • 写回答

2条回答 默认 最新

  • dsjhejw3232 2017-10-23 10:50
    关注
    $pw = "mysqli_fetch_assoc($sql)";
    $username = $pw["username"];     
    

    Problem is here, maybe you meant to do:

     $pw = mysqli_fetch_assoc($sql);
     $username = $pw["username"];    
    

    Because right now, $pw is a string that contains "mysqli_fetch_assoc($sql)", not an array

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论
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