Question

我刚接触php并且还在学习。我尝试自定义bootstrap模板，并尝试将表单连接到数据库。

insert.php

$con=mysqli_connect("localhost","root","","finalproject");
// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }

$sql="INSERT INTO stokbarang (Merek, Tipe, Harga)
VALUES
('$_POST[merek]','$_POST[tipe]','$_POST[harga]')";

    if (!mysqli_query($con,$sql))
      {
      die('Error: ' . mysqli_error($con));
      }

header("Location: stock.php");

以及表格为stock.php的页面，表格如下：

<form name="sentMessage" class="well" id="contactForm" novalidate action = "insert.php" method = "post">
<legend>Masukkan Data</legend>
     <div class="control-group">
     <div class="controls">
                     <label>Merek Mobil</label>
               <input type="text" class="form-control" placeholder="Merek mobil" id="merek" required
                         data-validation-required-message="Masukkan Merek mobil" name = "merek"/>
                         <p class="help-block"></p>
     </div>
         </div>         
   <div class="control-group">
  <div class="controls">
     <label>Tipe Mobil</label>
     <input type="text" class="form-control" placeholder="Tipe Mobil" id="email" required
                             data-validation-required-message="Masukkan tipe mobil" name = "tipe"/>
                </div>
         </div>         
                <div class="control-group">
                  <div class="controls">
                   <label>Harga</label>
                        <input type="number" class="form-control" placeholder="Harga"
                            id="email" required
                            data-validation-required-message="Masukkan harga mobil" name = "harga"/>
                </div>      

        <br>
         <div id="success"> </span></div> <!-- For success/fail messages -->
         <br>
         <button type="submit" class="btn btn-primary pull-right">Insert</button><br/><br>
          </form>

表单是可行的，我可以通过单击“插入”按钮来插入数据。现在我想在将表单提交到stock.php

中的数据库后添加模式作为警报

我修改了“插入”按钮，如下所示

<button type="submit" class="btn btn-primary pull-right" data-toggle="modal" data-target="#myModal">Insert</button><br/><br>

这是模态：

<!-- Modal -->
<div class="modal fade" id="myModal" tabindex="-1" role="dialog" aria-labelledby="myModalLabel" aria-hidden="true">
  <div class="modal-dialog">
    <div class="modal-content">
      <div class="modal-header">
        <button type="button" class="close" data-dismiss="modal" aria-hidden="true">&times;</button>
        <h4 class="modal-title" id="myModalLabel">SUCCESS</h4>
      </div>
       <div class="modal-body">
        <p>Data Inserted!!</p>
      </div>
      <div class="modal-footer">
        <button type="button" class="btn btn-default" data-dismiss="modal">Close</button>

      </div>
    </div><!-- /.modal-content -->
  </div><!-- /.modal-dialog -->
</div><!-- /.modal -->

但似乎按钮只触发模态而不将表单提交给数据库。在成功将数据插入数据库后（重定向到stock.php后），是否有任何建议使模式出现？或者可能有更好的方法在重定向后发出警报？谢谢你的时间和帮助：）

Answer 1

更改insert.php上的标头位置，如下所示：

header("Location: stock.php?sucsess=true");

然后stock.php在头上：

<script type="text/javascript">
<?php
if ($_GET['sucsess'] =='true'){
echo '$(function() {
$( "#myModal" ).dialog();
});'
}
?>
</script>

这是仅针对提醒demo

的演示

将数据插入数据库后，将bootstrap模式加载为警报

1 个答案: