Question

所以我已经使用post将数据从HTML表单插入到XAMPP上运行的MySQL数据库中，然后如何在表格中的另一个HTML页面上显示这些数据呢？当我尝试从localhost运行它时，它会出现一个空白页面，顶部有一行代码。我是新手，这是我的代码：我做得对吗？

d3.request("https://erudite-master-api-awsmaui.lab.expts.net/erudite/search")
  .header("Content-Type", "application/x-www-form-urlencoded")
  .mimeType("application/json")
  .post("intent=data-quality", function (error,data){...});

Answer 1

尝试以下代码并查看有关程序更改的注释

<html>
  <head>
  </head>
  <body>
    <?php
      // MySQL has been deprecated so use mysqli or pdo.
      $con = mysqli_connect('localhost', 'root', '', 'form_process') die("Can not connect: " . mysql_error());

      $sql = "SELECT * FROM `form_submissions`";
      $myData = mysqli_query($con, $sql);

      echo "<table border=1>
        <tr>
          <th>First Name</th>
          <th>Last Name</th>
          <th>Phone Number</th>
          <th>Class interested in</th>
        </tr>"; 

      // mysqli_fetch_array should have query result in parameters
      while($row = mysqli_fetch_array($myData)){
        echo "<tr>";
        // use proper array in this case its $row as in while condition
        echo "<td>" . $row['First'] . "</td>";
        echo "<td>" . $row['Last'] . "</td>";
        echo "<td>" . $row['Phone'] . "</td>";
        echo "<td>" . $row['Class'] . "</td>";
        echo "</tr>";
      }
      echo "</table>";

      mysqli_close();

    ?>

  </body>
</html>

Answer 2

你应该使用mysqli或pdo，不推荐使用mysql。 mysqli类似于mysql。

mysqli代码吼叫

  $con = mysqli_connect('localhost', 'root', '', 'form_process') die("Can not connect: " . mysql_error());

  $sql = "SELECT * FROM `form_submissions`";
  $myData = mysqli_query($con, $sql);

  while($row = mysqli_fetch_array($myData)){
  /// some code
  }
  mysqli_close();

如何将数据从SQL数据库显示到PHP / HTML表中

2 个答案: