Question

我使用PHP和MySQL制作HTML / CSS搜索表单。 IT不显示我想要的数据。不确定我做错了什么。我已经有其他选项的PHP工作输出，只是这种搜索方法没有找到任何结果。被困了一段时间后决定问。

    if (isset($_POST["searchcrews"]))
  {
    $searchparam = "%{$_POST['search']}%";
    $stmt=$conn->prepare("SELECT * FROM crewdata WHERE name LIKE ? AND is_approved =? ORDER BY (vote_up - vote_down) DESC LIMIT ?,?");
    $stmt->bind_param('s',$searchparam);
    $stmt->execute();

    $result=$stmt->get_result();

  }

HTML / CSS

    <form id="searchcrews">          
<div class="box">
  <div class="container-4">
    <input type="text" id="search" placeholder="Search crew.." />
   <button type="submit" class="icon"><i class="fa fa-search"></i></button>

  </div>
</div>
          </form>

Answer 1

将“name”属性添加到输入标记，将方法添加到表单标记。

<form id="searchcrews" method="POST">          
  <div class="box">
    <div class="container-4">
      <input type="text" id="search" placeholder="Search crew.." />
      <button type="submit" class="icon"><i class="fa fa-search"></i></button>
    </div>
  </div>
</form>

Answer 2

试试这个希望它有助于在表单和按钮中进行一些修改

 <form id="searchcrews" method="post">          
<div class="box">
  <div class="container-4">
    <input type="text" id="search" name="search" placeholder="Search crew.." />
   <input type="submit" class="icon" name="searchcrews">

  </div>
</div>
          </form>

搜索表单未显示任何结果

2 个答案: