Question

我创建一个页面，用户可以从表中搜索数据（表格数据），然后将搜索结果输入另一个表格（表格XY）。我想要实现的是当搜索结果出现时，它出现在另一个表单中，并且当提交第二个表单时，数据将插入到表XY中。但到目前为止，我所做的只能实现第一个目标，即从表格数据中搜索数据。当验证运行False时，错误似乎完成表单，但是当验证运行为True时，我稍后检查时数据不在表XY中。没有错误通知。我不知道我哪里做错了。这是我的代码： Controller Data.php：

public function index(){
 $this->load->view('form');
}
public function search_xy(){
 $this->form_validation->set_rules('id', 'ID', 'required|numeric|exact_length[16]');
  if($this->form_validation->run() == false) {
   $this->index();
  } else {
   $this->m_data->search_xy();
   $this->index();
  }
}

public function insert_xy(){
 $this->form_validation->set_rules('id', 'ID', 'required');
 $this->form_validation->set_rules('xname', 'X Name', 'required|alphanumeric');
 $this->form_validation->set_rules('yname', 'Y Name', 'required|alphanumeric');
 $this->form_validation->set_rules('xage', 'X Age', 'required');
 $this->form_validation->set_rules('yage', 'Y Age', 'required');
 $this->form_validation->set_rules('children', 'Children', 'required|numeric');
  if($this->form_validation->run() == false) {
   $this->index();
  } else {
   $this->m_xy->insert_xy();
   $this->session->set_flashdata('message', '<div class="alert alert-success" data-dismiss="alert">Insert Data to XY Table Success!</div>');
   $this->index();
  }
}

模型m_data.php：

public function search_pus(){
 $id = $this->input->post('id');
 $this->db->empty_table('searched', TRUE);
 $this->db->query("INSERT INTO 'searched' SELECT a.id, a.name as xname, a.age as xage, b.name as yname, b.age as yage FROM searched a, searched b WHERE a.gender = 1 AND b.gender = 2 AND a.id = '$id' AND b.id = '$id'");
  return $this->db->query("SELECT * FROM searched")->result_array();
}

模型m_xy.php：

public function insert_xy(){
 $data = array(
  'id' => $this->input->post('id'),
  'xname' => $this->input->post('xname'),
  'yname' => $this->input->post('yname'),
  'xage' => $this->input->post('xage'),
  'yage' => $this->input->post('yage'),
  'children' => $this->input->post('children'));

  $this->db->insert('XY', $data);
}

查看form.php：

<?php $attributes = array("class" => "form-inline"); echo form_open('data/search_xy', $attributes);?>
 <fieldset>
  <div class="form-group">
   <label class="control-label" for="id">Enter ID :</label>
     <input type="text" name="id" class="form-control" id="id" value="<?php echo set_value('id'); ?>" />
     <button class="btn btn-info btn-md" type="submit"> Search </button><span class="text-danger"><?php echo form_error('id');?></span>
  </div>
 </fieldset>
<?php echo form_close(); ?>

<?php echo form_open('data/insert_xy'); ?>
 <fieldset>
  <div class="form-group">
   <h4 class="text-left text-primary">Search Result:</h4>
    <table style="font-size:11px;" class="table table-condense table-bordered table-hover display" cellspacing="0" width="100%">
     <thead>
      <tr class="primary">
       <th></th>
       <th>X Name</th>
       <th>Y Name</th>
       <th>X age</th>
       <th>Y age</th>
      </tr>
     </thead>
     <tbody>
      <?php foreach($search_result as $row): ?>
       <tr>
        <td><input type="checkbox"  class="form-control" name="no_kk" id="no_kk" value="<?php echo $row['no_kk'];?>" /></td>
        <td><input class="form-control" name="xname" value="<?php echo $row['xname'];?>" /></td>
        <td><input class="form-control" name="yname" value="<?php echo $row['yname'];?>" /></td>
        <td><input class="form-control" name="xage" value="<?php echo $row['xage'];?>" /></td>
        <td><input class="form-control" name="yage" value="<?php echo $row['yage'];?>" /></td>
       </tr>
      <?php endforeach ?>
     </tbody>
    </table>
    <div class="col-md-3">
     <select class="form-control" name="children" value="<?php echo set_value('children');?>">
      <option value="">-- How Many Children? --</option>
      <option value="0">0</option>
      <option value="1">1</option>
      <option value="2">2</option>
      <option value=">3">3 or more</option>
     </select>
     <span class="text-danger"><?php echo form_error('children'); ?></span>
    </div>
    <div class="col-md-3">
     <input class="btn btn-primary btn-md" type="submit" value="Save" />
    </div>
   </div>
  </fieldset>
 <?php echo form_close() ?>

这个视图打印屏幕，如果你没有得到我的意思：

当我点击“保存到XY”时，我希望将我检查的值插入表XY ..

是因为表格在表格中吗？

Answer 1

试试这个......

控制器： -

公共职能指数（）{

$data = array();

$id =  $this->input->get('id');

if($id!=''){

    $data = $this->m_data->search_xy();

}

$this->load->view('form',$data);  }

public function insert_xy（）{

$this->form_validation->set_rules('id', 'ID', 'required');

$this->form_validation->set_rules('xname', 'X Name', 'required|alphanumeric');

$this->form_validation->set_rules('yname', 'Y Name', 'required|alphanumeric');

$this->form_validation->set_rules('xage', 'X Age', 'required');

$this->form_validation->set_rules('yage', 'Y Age', 'required');

$this->form_validation->set_rules('children', 'Children', 'required|numeric');

if($this->form_validation->run() == false) {

    redirect('data');

} else {

    $this->m_xy->insert_xy();

    $this->session->set_flashdata('message', '<div class="alert alert-success" data-dismiss="alert">Insert Data to XY Table Success!</div>');

    redirect('data');
} }

查看： -

更改echo form_open（'data'，$ attributes）;搜索表单

Codeigniter表格结果作为另一种表格输入值。第二个表格没有做它应该做的事情

1 个答案: