Question

错误视图中的“提交”按钮不会传递值。

单击提交按钮后，将值从视图传递到控制器，然后传递给模型。数据库结构由表（预订）和列名（scheduled_id，名称和移动）组成。但是该值不会传递给控制器也不会传递给数据库。

查看

<form method="post" action="">
    <input type="text" value="<?php echo isset($post)?$post->name:''; ?>" name="manakamana[name]"/>
    <input type="text" value="<?php echo isset($post)?$post->mobile:''; ?>" name="manakamana[mobile]"/>
    <button type="submit" class="btn btn-success">POST</button>
</form>

控制器

public function bookManakamana(){

    if ($data = $this->input->post('manakamana')) {
        $data['created_on'] = date('Y-m-d H:i:s');
        $this->booking->add($data);
        redirect('booking');
    } else {
        $this->load->view('index');
    }

}

模型

function add($data)
{
    $this->db->insert('booking', $data);
}

Answer 1

未给出表格行动

<form method="post" action="<?php echo base_url().'controller name/function name' ?>">
<input type="text" value="<?php echo isset($post)?$post->name:''; ?>" name="name"/>
<input type="text" value="<?php echo isset($post)?$post->mobile:''; ?>" name="mobile"/>
<button type="submit" class="btn btn-success">POST</button>

控制器中的

public function bookManakamana(){
if ($this->input->post('name')) {
    $data['created_on'] = date('Y-m-d H:i:s');
    $data['name'] = $this->input->post('name');
    $data['mobile'] = $this->input->post('mobile');
    $this->booking_models->add($data); //booking_model is the name of model
    redirect('booking');
} else {
    $this->load->view('index');
}}

Codeigniter提交不起作用

1 个答案: