PHP将数据提交到同一页面上的表

试图弄清楚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.我希望它测试请求是否首先来自POST,如果是,则将新行插入数据库并在表中显示新数据.这是我到目前为止所提出的,但在提交时,似乎没有任何事情发生,我的桌子就消失了.任何帮助是极大的赞赏.

这是我到目前为止所拥有的：

      $result = MysqLi_query($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
   ?>

解决方法:

>请注意未来的读者.这个答案是基于OP的原始帖子. See the revisions.

将INSERT查询放在条件语句中：

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']);？>”只是动作=“”

对POST数组使用条件！empty()以确保不会获得任何空数据并可能抛出错误.

边注：

您现在的代码是SQL injection开放.使用mysqli with prepared statements或PDO with prepared statements,它们更安全.

根据你的编辑,你错过了两个结束括号},如果使用它,error reporting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));

    }

  }