Question

Mates，我有这个下拉菜单，我需要在任何用户点击它时用一个单选按钮更新一个部分和数据库。我不知道如何实现它。
我的下拉列表看起来像这样

<form class="form">
    <div class="switch-field">
      <div class="switch-title">Make Resume Public?</div>
      <input type="radio" id="switch_left" name="switch_2" value="yes" checked/>
      <label for="switch_left">Yes</label>
      <input type="radio" id="switch_right" name="switch_2" value="no" />
      <label for="switch_right">No</label>
    </div>
</form>  

    <?php
    $switch_2 = $_POST['switch_2'];

    $qry = "UPDATE resumes SET public = :public WHERE memberID = :memberID";
    $stm = $db->prepare($qry);
    $stm->bindParam(':public', $switch_2, PDO::PARAM_STR);
    $stm->bindParam(':memberID', $uid, PDO::PARAM_STR);
    $stm->execute();
    ?>

我需要更新此部分，无论点击哪个都应该更新数据库并保持选中状态

{{1}}

感谢您的帮助和时间。

Answer 1

我假设你能够更新数据库。

以下是选择单选按钮的代码

        <?php
        $switch_2 = $_POST['switch_2'];

        $qry = "UPDATE resumes SET public = :public WHERE memberID = :memberID";
        $stm = $db->prepare($qry);
        $stm->bindParam(':public', $switch_2, PDO::PARAM_STR);
        $stm->bindParam(':memberID', $uid, PDO::PARAM_STR);
        $stm->execute();
        ?>

        <?php
        $stmt = $db->prepare("select * from jobs where jobposterID = '".$_SESSION['memberID']."'");
                $stmt->execute();
                $jobs = $stmt->rowCount();    
        ?>              
        <a href="<?php echo DIR; ?>my-jobs.php"><i class="fa fa-suitcase"></i> MY JOBS (<?php echo $jobs; ?>) </a>
        <a href="<?php echo DIR; ?>applied-jobs.php"><i class="fa fa-check-square-o"></i> APPLIED JOBS </a>     
        <a href="<?php echo DIR; ?>viewed-jobs.php"><i class="fa fa-eye"></i> VIEWED JOBS </a>
        <a href="<?php echo DIR; ?>logout.php"><i class="fa  fa-power-off"></i> LOGOUT </a>
      </div>
      </div>
    <form class="form" id="form_abc" action="">
        <div class="switch-field">
          <div class="switch-title">Make Resume Public?</div>

          <!-- Here you need to get the public field from database on load -->
          <input type="radio" id="switch" name="switch_2" value="yes" <?php echo ($public =='Yes') ? 'checked' : ''; ?>/>

          <label for="switch_left">Yes</label>

          <!-- Here you need to get the public field from database on load -->
          <input type="radio" id="switch" name="switch_2" value="no" <?php echo ($public =='No') ? 'checked' : ''; ?> />
          <label for="switch_right">No</label>
        </div>
    </form>  

<Script>
$("#switch").on('click',function(){
// see is this alert comes?
alert('Its working');
$("#form_abc").submit();
});
</Script>

需要从下拉列表部分更新查询

1 个答案: