php中的sql在運行時不會返回任何東西

Question

每次嘗試登錄時，頁面都會刷新並且什麼都不會發生。我可以打破這個PHP並讓它返回一個錯誤，但是現在寫出來，沒有錯誤發生。這是我第一次使用SQL，我渴望學習，所以任何和所有的幫助是巨大的讚賞。

<?php 

session_start(); 

$link=mysqli_connect("localhost", "cl44-thediary", "password", "cl44-thediary"); 

if ($_POST["submit"]=="Sign Up") { 

    if (!$_POST["email"]) $error.="<br />Please enter your email."; 
     else if (!filter_var($_POST["email"], FILTER_VALIDATE_EMAIL)) $error.="<br/>Please enter a valid email address."; 

    if (!$_POST["password"]) $error.="<br />Please enter your password."; 
     else { 

      if (strlen($_POST["password"])<8) $error.="<br />Please enter a password with at least 8 characters."; 
      if (strlen($_POST["password"])>55) $error.="<br />Please enter a password with less than 55 characters."; 
      if (!preg_match('`[A-Z]`', $_POST["password"])) $error.="<br />Please include at least one capital letter in your password."; 
      if (!preg_match('`[0-9]`', $_POST["password"])) $error.="<br />Please include at least one number in your password."; 
      if (!preg_match('`[a-z]`', $_POST["password"])) $error.="<br />Please include at least one lower-case letter in your password."; 

     }; 

    if ($error) echo "There were error(s) in your signup details".$error; 
     else { 

      $query="SELECT * FROM `login` WHERE email='".mysqli_real_escape_string($link, $_POST['email'])."'"; 

      $result=mysqli_query($link, $query); 

      $results=mysqli_num_rows($result); 

      if ($results) echo "That email address is already registered. Do you want to login?"; 
       else { 

        $query="INSERT INTO `login` (`email`, `password`) VALUES('".mysqli_real_escape_string($link, $_POST['email'])."','".md5(md5($_POST['email']).$_POST['password'])."')"; 

        mysqli_query($link, $query); 

        echo "You've signed up!"; 

        $_SESSION["id"]=mysqli_insert_id($link); 

        print_r($_SESSION); 

        //Redirect to logged in page 

       } 


     } 

    } 

    if ($_POST["submit"]=="Login") { 

     $query="SELECT * FROM `login` WHERE email='".mysqli_real_escape_string($link, $_POST['LoginEmail'])."' AND password='".md5(md5($_POST['LoginEmail']).$_POST['loginPassword'])."' LIMIT 1"; 

     $result=mysqli_query($link, $query); 

     $row=mysqli_fetch_array($result); 

     print_r($row); 
    }; 
?> 

<form method="post"> 

    <input type="email" name="email" id="email" value="<?php echo addslashes($_POST['email']);?>"/> 

    <input type="password" name="password" id="password" value="<?php echo addslashes($_POST['password']);?>"/> 

    <input type="submit" name="submit" value="Sign Up"/> 
</form> 

<form method="post"> 

    <input type="email" name="loginEmail" id="loginEmail" value="<?php echo addslashes($_POST['loginEmail']);?>"/> 

    <input type="password" name="loginPassword" id="loginPassword" value="<?php echo addslashes($_POST['loginPassword']);?>"/> 

    <input type="submit" name="submit" value="Login"/> 
</form> 

Answer 1

嘗試在mysql中捕獲錯誤。使用mysqli_error()

$query="SELECT * FROM `login` WHERE email='".mysqli_real_escape_string($link, $_POST['email'])."'"; 
$result=mysqli_query($link, $query); 
if($result == false) { 
    echo mysqli_ee($link)." ".mysqli_error($link); 
}