Question

我刚开始学习php，我的小项目有些问题。我创建一个DB，其中包含“members”表，其中包含“member_id”和“member_name”calumn。我创建了一个填充了成员名称的选择框。当我选择一个名称并单击提交按钮时，我想回显member_id。如果没有javascript / ajax，我怎么能这样做。

PS。我想在提交后留在当前页面。

非常感谢。

<html>
<body>
<h2>Select</h2>
<table>
  <tbody>
    <tr>
      <td>
        <form method="post" action="<?=($_SERVER['PHP_SELF'])?>">

          <table>
          <thead>
            <tr>
                <th>Nume</th>
                <th>Id</th>
            </tr>
          </thead>
          <tr>
            <td><select name="numeselect">
              <?php
                $numeselect=$_POST ["numeselect"];
                $host="localhost";
                $user="***";
                $password="***";
                $db_name="***";

                $con=mysqli_connect($host,$user,$password,$db_name);
                if(mysqli_connect_errno($con)){
                    echo "Eroare la coenxiune:" . mysqli_error();   
                }

                $selectare_nume = mysqli_query($con,"SELECT nume_membru, id_membru FROM membri");

                while($nume_selectat = mysqli_fetch_array($selectare_nume)){
                    echo '<option value="'.$nume_selectat['nume_membru'].'">'.$nume_selectat['nume_membru'].'</option>';
                }
                echo '</select></td>';
                while($id_selectat = mysqli_fetch_array($selectare_nume)){
                echo '<td>'.$id_selectat['$id_membru'].'</td>';
                }
                echo '</tr></table>';

                ?>
              <input type="submit" value="show id">
        </form>
      </td>
    </tr>
  </tbody>
</table>
</body>
</html>

Answer 1

您可以获得此类表单提交值

if(isset($_POST['numeselect'])){
    echo $_POST['numeselect'];
}

将选项更改为此

echo '<option value="'.$nume_selectat['id_membru'].'">'.$nume_selectat['nume_membru'].'</option>';

所以完整的脚本

<html>
<body>
<h2>Select</h2>
<table>
  <tbody>
    <tr>
      <td>
        <form method="post" action="<?=($_SERVER['PHP_SELF'])?>">

          <table>
          <thead>
            <tr>
                <th>Nume</th>
                <th>Id</th>
            </tr>
          </thead>
          <tr>
            <td><select name="numeselect">
              <?php
                $numeselect=$_POST ["numeselect"];
                $host="localhost";
                $user="***";
                $password="***";
                $db_name="***";

                $con=mysqli_connect($host,$user,$password,$db_name);
                if(mysqli_connect_errno($con)){
                    echo "Eroare la coenxiune:" . mysqli_error();   
                }

                $selectare_nume = mysqli_query($con,"SELECT nume_membru, id_membru FROM membri");

                while($nume_selectat = mysqli_fetch_array($selectare_nume)){
                    echo '<option value="'.$nume_selectat['id_membru'].'">'.$nume_selectat['nume_membru'].'</option>';
                }
                echo '</select></td>';
                while($id_selectat = mysqli_fetch_array($selectare_nume)){
                echo '<td>'.$id_selectat['$id_membru'].'</td>';
                }
                echo '</tr></table>';

                ?>
              <input type="submit" value="show id">
        </form>
      </td>
    </tr>
  </tbody>
</table>
<?php 
if(isset($_POST['numeselect'])){
    echo "<p>You selected member id: {$_POST['numeselect']}</p>";
}
?>
</body>
</html>

Answer 2

<html>
<body>
<h2>Select</h2>
<table>
  <tbody>
    <tr>
      <td>
        <form method="post" action="<?=($_SERVER['PHP_SELF'])?>">

          <table>
          <thead>
            <tr>
                <th>Nume</th>
                <th>Id</th>
            </tr>
          </thead>
          <tr>
            <td>
              <?php
                if(isset($_POST)) {
                  $numeselect = $_POST["numeselect"];
                  echo $numeselect;
                }
               ?>
              <select name="numeselect">
              <?php
                $host="localhost";
                $user="***";
                $password="***";
                $db_name="***";

                $con=mysqli_connect($host,$user,$password,$db_name);
                if(mysqli_connect_errno($con)){
                    echo "Eroare la coenxiune:" . mysqli_error();   
                }

                $selectare_nume = mysqli_query($con,"SELECT nume_membru, id_membru FROM membri");

                while($nume_selectat = mysqli_fetch_array($selectare_nume)){
                    echo '<option value="'.$nume_selectat['nume_membru'].'">'.$nume_selectat['nume_membru'].'</option>';
                }
                echo '</select></td>';
                while($id_selectat = mysqli_fetch_array($selectare_nume)){
                echo '<td>'.$id_selectat['$id_membru'].'</td>';
                }
                echo '</tr></table>';

                ?>
              <input type="submit" value="show id">
        </form>
      </td>
    </tr>
  </tbody>
</table>
</body>
</html>

我添加了什么？

                if(isset($_POST)) {
                  $numeselect = $_POST["numeselect"];
                  echo $numeselect;
                }

Answer 3

如果没有javascript / ajax，我怎么能这样做。

PS。我想在提交后留在当前页面。

如果留在当前页面上，则表示没有页面刷新，并且您不想使用javascript，这是不可能的。

如果您只是想要POST到同一页面，那么您将echo $_POST数组中的变量。

Answer 4

你可以使用switch case和2个函数，第一个用于表单，第二个用于显示值，代码为Joni Salmi，例如：

<?php

// action
$action = $_POST['action'];

if($action == "undefined" || $action == "") $action = "form_member";

switch($action){
  // Member form
  case 'form_member': 
        form_member();
        break;

  // Show Member id
  case 'show_member':
     show_member();
     break;

}

function form_member(){
?>

<html>
<body>
<h2>Select</h2>    
<table>
  <tbody>
    <tr>
  <td>
    <form method="post" action="<?=($_SERVER['PHP_SELF'])?>">
    <input type="hidden" name="action" value="show_member" />
      <table>
      <thead>
        <tr>
            <th>Nume</th>
            <th>Id</th>
        </tr>
      </thead>
      <tr>
        <td><select name="numeselect">
          <?php
            $numeselect=$_POST ["numeselect"];
            $host="localhost";
            $user="***";
            $password="***";
            $db_name="***";


            $con=mysqli_connect($host,$user,$password,$db_name);
            if(mysqli_connect_errno($con)){
                echo "Eroare la coenxiune:" . mysqli_error();   
            }

            $selectare_nume = mysqli_query($con,"SELECT nume_membru, id_membru FROM membri");

            while($nume_selectat = mysqli_fetch_array($selectare_nume)){
                echo '<option value="'.$nume_selectat['nume_membru'].'">'.$nume_selectat['nume_membru'].'</option>';
            }
                            echo '</select></td>';
            while($id_selectat = mysqli_fetch_array($selectare_nume)){
            echo '<td>'.$id_selectat['$id_membru'].'</td>';
            }
            echo '</tr></table>';

            ?>
          <input type="submit" value="show id">
        </form>
      </td>
    </tr>
  </tbody>
</table>
</body>
</html>



<?php
}


function show_member(){

  if(isset($_POST['numeselect'])){
    echo $_POST['numeselect'];
  }

}

?>

选择选项时的回显值 - 仅限PHP

4 个答案: