Codeigniter Form結果作爲另一個表單輸入值。第二種形式沒有做它應該做的事

Question

我製作了一個頁面，用戶可以從一個表格（表格數據）中搜索數據，然後將搜索結果輸入到另一個表格（表格XY）中。我試圖實現的是當搜索結果出現時，它出現在另一個表單中，並且當第二個表單被提交時，數據被插入到表XY中。但到目前爲止，我所做的只能實現第一個目標，即從表數據中搜索數據。當驗證運行False時，錯誤似乎完成了表單，但是當驗證運行爲True時，稍後檢查時數據不在XY表格中。沒有錯誤通知。我不知道我在哪裏做錯了。這裏是我的代碼： 控制器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(); 
    } 
} 

型號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(); 
} 

型號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的：

<?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() ?> 

This view printsc穎櫃面你不明白我的意思：

當我點擊「保存到XY」我希望爲我檢查表XY要插入的價值..

是不是因爲形式是一張桌子？

Answer 1

嘗試......

控制器： -

公共功能指數（）{

$data = array(); 

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

if($id!=''){ 

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

} 

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

公共職能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）;搜索表單