PHP SQL数据库查询错误

Question

我在对 mysql 数据库进行看似正确的查询时遇到了麻烦。有问题的查询是:

"SELECT * FROM Users WHERE Email =".$email.";". The Query itself is executing fine but the $result that is returned back is false (if i replace "Email =".$email."" with "Id = 1" it works and returns a value).

  if($emailCheck = TRUE){
               echo "<script type='text/javascript'>alert('Email check true.');</script>";
               $sql = "SELECT * FROM Users WHERE Email =".$email.";";
               echo $sql;
               $result = $conn->query($sql);
               if ($result){
                  $row = mysqli_fetch_array($result) ;
                  echo "<script type='text/javascript'>alert('".(string)$row['FirstName']."');</script>"; 
               } else { echo "<script type='text/javascript'>alert('bad result');</script>";}
           }

一些信息:

• $emailCheck = TRUE 工作正常。

• 当使用“Id = 1”而不是“Email =".$email.””时，一切正常

• echo $sql;返回“SELECT * FROM Users WHERE Email =zxzx@hotmail.com;”

使用“Email =".$email."”时为什么 $result 返回 false 有任何帮助吗？

Answer 1

这是因为如果您使用 id 那么它是整数，所以不需要将其放在引号('')中但是如果您使用电子邮件那么它是字符串所以您需要将其写在引号('')中，如下所示

$sql = "SELECT * FROM Users WHERE Email ='" . $email . "'";