PHP将数据提交到同一页面上的表

Question

试图找出如何将新帖子提交到已经使用 mysql 数据库中的数据条目创建的预先存在的表。我希望它首先测试请求是否来自 POST，如果是，则将新行插入数据库并在表中显示新数据。这是我到目前为止想出的，但在提交时，似乎什么也没发生，我的 table 消失了。非常感谢任何帮助。

这是我目前所拥有的:

      $result = mysqli_query($dbconnect, $query);
            $num_rows = mysqli_num_rows($result);
        }
        if ($num_rows > 0) // build a table to show results
        {
        echo "<table border='1'>";
        echo "<tr>";
        echo "<th>Post ID </th>"; echo "<th>Author</th>";
        echo "<th>Title</th>"; echo "<th>Post</th>";
        echo "</tr>";

     while($row = mysqli_fetch_array($result))
        {
        echo "<tr>";
        echo "<td>" . $row['pid'] . "</td>";
        echo "<td>" . $row['author'] . "</td>";
        echo "<td>" . $row['title'] . "</td>";
        echo "<td>" . $row['post'] . "</td>";
        echo "</tr>";
        }   
        echo "</table>";
    } else{
        echo "No rows returned.";
    }
    ?>

 <form name ="myForm" action ="second.php<?php echo htmlspecialchars($_SERVER['PHP_SELF']);?>"   
       method = "POST"> <br><br><br>

     <h3> Create a Post </h3>

     Author  <input type ="text" size ="40" name ="author"/><br>
     Title <input type ="text" size ="30" name ="title"/><br><br>
     Post <br><textarea rows ="15" cols ="10" name ="post"></textarea><br>
     <input type ="submit" name = "submitpost" value ="Submit Post"/>
    </form>
    <?php


  // $sql = "INSERT INTO blog_posts (pid, author, title, post) 
        VALUES (NULL, '$_POST[author]', '$_POST[title]', '$_POST[post]')";

  //if($_SERVER['REQUEST_METHOD'] === 'POST'){
   //if(isset($_POST['submitpost'])){

  //post the $sql back into the exisiting table somehow
   ?>

Answer 1

• 请注意 future 的读者。这个答案是基于 OP 的原始帖子。 See the revisions .

将您的 INSERT在条件语句中查询:

if($_SERVER['REQUEST_METHOD'] === 'POST'){

  if(isset($_POST['submitpost'])){

  $sql = mysqli_query($dbconnect, "INSERT INTO blog_posts (pid, author, title, post) 
        VALUES (NULL, '$_POST[author]', '$_POST[title]', '$_POST[post]')") 

     or die(mysqli_error($dbconnect));

    }

}

并更改action ="second.php<?php echo htmlspecialchars($_SERVER['PHP_SELF']);?>"只是action=""

使用条件 !empty()使用您的 POST 数组以确保您没有得到任何空数据并可能引发错误。

---

旁注:

您现在的密码是开放给SQL injection .使用 mysqli with prepared statements , 或 PDO with prepared statements ，它们更安全。

---

根据您的编辑，您缺少两个右大括号 }哪个error reporting如果使用它会发出通知。

 <form name ="myForm" action =""  method = "POST"> <br><br><br>

     <h3> Create a Post </h3>

     Author  <input type ="text" size ="40" name ="author"/><br>
     Title <input type ="text" size ="30" name ="title"/><br><br>
     Post <br><textarea rows ="15" cols ="10" name ="post"></textarea><br>
     <input type ="submit" name = "submitpost" value ="Submit Post"/>
    </form>
    <?php

   if($_SERVER['REQUEST_METHOD'] === 'POST'){

     if(isset($_POST['submitpost'])){

   $sql = "INSERT INTO blog_posts (pid, author, title, post) 
        VALUES (NULL, '$_POST[author]', '$_POST[title]', '$_POST[post]')";

   $data = mysqli_query($dbconnect, $sql)

        or die(mysqli_error($dbconnect));

    }

  }