Question

我已填充sql php结果中的下拉列表，现在我尝试将所选值设置为同一页面中的php变量，但它无法正常工作。你能帮我吗。以下是代码。

<?php
     $mid="mario";
     $sql = "SELECT * FROM tbl_prdy" ;
     $result = mysqli_query($conn,$sql);
     echo "<select name='list'>";
     while ($row = mysqli_fetch_array($result,MYSQLI_ASSOC)) {
     echo "<option value='" . $row['col_of_fac'] . "'>" . $row['col_of_fac'] . "
     </option>";
     }
     echo "</select>";
     $varsel = $_POST['list'];
     echo "hai";
     echo $varsel;

 ?>

$varsel = $_POST['list'];无效。

Answer 1

你应该

1-使用表格

2-使用方法“POST”将变量发送到同一个文件php

 <?php
     // display the errors
     error_reporting(E_ALL);
     ini_set("display_errors", 1);
 ?>

 <?php
     // when the form submitted
     if(!empty($_POST['list'])){
       echo $_POST['list'];
     }
 ?>

 <?php
     // connection test
     $conn=mysqli_connect("localhost","user","pass","db"); //replace the (user, pass, db) with your parameters
     if (mysqli_connect_errno())
     {
       echo "Failed to connect to MySQL: " . mysqli_connect_error();
       die();
     }

     //if your connection succeded
     $sql = "SELECT * FROM tbl_prdy" ;
     $result = mysqli_query($conn,$sql);
 ?>

 <form method="post" action="<?= $_SERVER['PHP_SELF']; ?>">
  <select name="list">
    <?php while ($row = mysqli_fetch_array($result,MYSQLI_ASSOC)): ?>
      <option value="<?= $row['col_of_fac']; ?>"><?= $row['col_of_fac']; ?></option>
    <?php endwhile; ?>
  </select>
  <input type="submit" value="valider">
 </form>

如果您想打印所选的值而不是没有表单

然后你必须使用jquery

 <?php
      // display the errors
      error_reporting(E_ALL);
      ini_set("display_errors", 1);
  ?>

  <?php
      // when the form submitted
      if(!empty($_POST['list'])){
        echo $_POST['list'];
      }
  ?>

  <?php
      // connection test
      $conn=mysqli_connect("localhost","user","pass","db"); //replace the (user, pass, db) with your parameters
      if (mysqli_connect_errno())
      {
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
        die();
      }

      //if your connection succeded
      $sql = "SELECT * FROM tbl_prdy" ;
      $result = mysqli_query($conn,$sql);
  ?>

  <select name="list">
     <?php while ($row = mysqli_fetch_array($result,MYSQLI_ASSOC)): ?>
       <option value="<?= $row['col_of_fac']; ?>"><?= $row['col_of_fac']; ?></option>
     <?php endwhile; ?>
   </select>

   <p id="value-selected"></p>

  <script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>
  <script>
       $(function(){
           $("select[name=list]").on("change", function () {
                 var valueSelected = $(this).val();
                 $("#value-selected").html(valueSelected);
           });
       });
  </script>

Answer 2

$ varsel = $ _POST ['list'];不管用。：只有在通过post方法提交表单时才会起作用，

（打印当前代码选择框，您无需提交表单即可获取帖子）逻辑应该是这样的在表格选择框中使用您的以下代码和用户选择，然后单击提交：

 <?php
     $mid="mario";
     $sql = "SELECT * FROM tbl_prdy" ;
     $result = mysqli_query($conn,$sql);
     echo "<select name='list'>";
     while ($row = mysqli_fetch_array($result,MYSQLI_ASSOC)) {
     echo "<option value='" . $row['col_of_fac'] . "'>" . $row['col_of_fac'] . "
     </option>";
     }
     echo "</select>";
 ?>

在下面提交的表单应该执行后，在网络控制台中，在页眉中，您可以看到所需的数据是否已传输到请求的页面

从php html

2 个答案: