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Php Dropbox不返回值_Php_Html_Sql - Fatal编程技术网

Php Dropbox不返回值

Php Dropbox不返回值,php,html,sql,Php,Html,Sql,在这个程序中,get data方法似乎不正确,您能看到当用户与下拉框交互时,是什么导致下拉框中没有填充内容吗 更新get_数据类确实有效,感谢您的反馈 <?php require("db.php"); include("functions.php"); if(isset($_POST['submit'])) { $Date = date('Y-m-d H:i:s'); $FirstName = $mysqli->real_es

在这个程序中,get data方法似乎不正确,您能看到当用户与下拉框交互时,是什么导致下拉框中没有填充内容吗

更新get_数据类确实有效,感谢您的反馈

 <?php
    require("db.php");
    include("functions.php");
    if(isset($_POST['submit']))
    {
      $Date = date('Y-m-d H:i:s');
      $FirstName = $mysqli->real_escape_string($_POST['FirstName']);
      $LastName = $mysqli->real_escape_string($_POST['LastName']);
      $Rating = $mysqli->real_escape_string($_POST['Rating']);
      $Comment = $mysqli->real_escape_string($_POST['Comment']);

      $sql = "INSERT INTO guest(GuestID, FirstName, LastName, PostcodeFK, Email, Date, Rating,Comment)"
       ." VALUES ('', '$FirstName', '$LastName', '$Postcode', '$Email', '$Date', '$Rating', '$Comment')";

    if($mysqli->query($sql)==TRUE)
    {
      echo "<script>alert('Record Added.'); location.href='customers.html'</script>";
    }else{
     echo "<script>alert('Error'); location.href='#'</script>"; 
    }
    function get_data($mysqli)
    {

       $sql="SELECT `MealOption` FROM menu";
       $result=$mysqli->query($sql);
           while ($row=$result->fetch_assoc())
          {
            echo "<option value='". $row['MealOption'] . "'>". $row['MealOption'] ."</option>";
          }

    }
    }

    ?>
    <html>

    <body>
    <form action="" method="post">
    First Name: <input type="text" name="FirstName" require/><br>
    Last Name: <input type="text" name="LastName" require/><br>
    Postcode: <input type="number" name="Postcode" require/><br>
    Email: <input type="email" name="Email" require/><br>
    <button class="btn" onclick="get_data(Meal);">Add</button>
    <label for="Meal">Meal:</label>
    <select name="Meal">
    <?php
     //echo get_data($mysqli);   
    ?>
    </select><br>
    Rating: <input type="number" name ="Rating" min="1" max="5" require/><br>
    Comment: <input type="text" name ="Comment"><br>
    <input type="submit" name="submit">
    </form>
    </body>
    </html>
使用函数get_datamean的Add不能调用php函数

 <?php get_data($mysqli); ?>
因此,您必须创建一个javascript函数get_data,例如通过ajax请求信息


你应该从那个家伙那里拿回你的钱。

你需要将get_data函数移出if语句。我可以帮你

if(isset($_POST['submit']))
{
  $Date = date('Y-m-d H:i:s');
  $FirstName = $mysqli->real_escape_string($_POST['FirstName']);
  $LastName = $mysqli->real_escape_string($_POST['LastName']);
  $Rating = $mysqli->real_escape_string($_POST['Rating']);
  $Comment = $mysqli->real_escape_string($_POST['Comment']);

  $sql = "INSERT INTO guest(GuestID, FirstName, LastName, PostcodeFK, Email, Date, Rating,Comment)"
   ." VALUES ('', '$FirstName', '$LastName', '$Postcode', '$Email', '$Date', '$Rating', '$Comment')";

   if($mysqli->query($sql)==TRUE)
   {
       echo "<script>alert('Record Added.');location.href='customers.html'</script>";
   } else {
       echo "<script>alert('Error'); location.href='#'</script>"; 
   }
}

function get_data($mysqli)
{

   $sql="SELECT `MealOption` FROM menu";
   $result=$mysqli->query($sql);
       while ($row=$result->fetch_assoc())
      {
        echo "<option value='". $row['MealOption'] . "'>". $row['MealOption'] ."</option>";
      }

}
对于html使用,您需要在使用php函数之前添加php标记

 <?php get_data($mysqli); ?>

您的代码对于XSS攻击不安全-这可能是导致问题的原因,例如,如果您的用餐选项包含撇号

尝试更改此行:

echo "<option value='". $row['MealOption'] . "'>". $row['MealOption'] ."</option>";
echo "<option value='". htmlspecialchars($row['MealOption'],ENT_QUOTES) . "'>". htmlspecialchars($row['MealOption']) ."</option>";
关于这一行:

echo "<option value='". $row['MealOption'] . "'>". $row['MealOption'] ."</option>";
echo "<option value='". htmlspecialchars($row['MealOption'],ENT_QUOTES) . "'>". htmlspecialchars($row['MealOption']) ."</option>";

什么是不工作?当代码运行时会发生什么?@Cristik有一个下拉框,应该通过get_data函数填充它。这就是它的样子,其他一切都很好,只是菜单框付费让它检查,他说它应该工作,所以你问其他人,而不是检查它?你从别人那里得到钱让别人检查?你是说你付钱让别人检查?包含函数get_dataobj的Javascript在哪里?最后一句话中出现了单词“scammed”+1。一般来说,如果这个家伙不得不检查代码,并且不得不说什么是坏习惯,告诉OP他应该做什么不同的事情等等,他肯定不知道自己在做什么,他确实是个骗子。