Php 代码点火器中的空结果

我想从我的模型中获取查询结果并在视图中使用它。我不知道为什么，但我的结果是空的。你能看看这个，告诉我我做错了什么吗

控制器：

    $peugeot = $this->input->post('peugeot');
    $citroen = $this->input->post('Citroen C-Elysee');
    $nissan= $this->input->post('Nissan Evalia');
    $renault = $this->input->post('Renault Trafic 9-os');
    $this->load->model('Edit_model');
    $this->Edit_model->peugeot($peugeot);
    if($peugeot)
    {
        $result = $this->Edit_model->peugeot($peugeot);
        $data['result'] = $result;
    }

型号：

    public function peugeot($peugeot)
    {
        $this->db->like('model', $peugeot, 'after'); 
        $query = $this->db->get('cars');
        $result = $query->result();
        return $result;
    }

视图：

---

有点奇怪，但它是有效的

我把控制器换成了这个

        $data['peugeot'] = $this->input->post('peugeot');
    $data['citroen'] = $this->input->post('citroen');
    $data['nissan'] = $this->input->post('nissan');
    $data['renault'] = $this->input->post('renault');
    $this->load->model('Edit_model');
    if($data['peugeot'])
    {
        $this->Edit_model->peugeot($data['peugeot']);
        $result = $this->Edit_model->peugeot($data['peugeot']);
        $data['result'] = $result;
        $this->load->view('content/editprocess',$data);
    }

---

在返回结果之前是否尝试过打印SQL？通过这种方式，您可以确认查询是否返回结果。ofc，在您的模型代码中，尝试执行var_dump（$this->db->last_query（））；这将显示它是否与您想要的完全匹配。string（61）“选择*来自

汽车

WHERE

model

像'Peugeot%'ESCAPE'！”string（61）“选择*来自

cars

WHERE

model

像'Peugeot%'ESCAPE'！”我想我的if-in控制器有问题，但我不知道是什么。

        $data['peugeot'] = $this->input->post('peugeot');
    $data['citroen'] = $this->input->post('citroen');
    $data['nissan'] = $this->input->post('nissan');
    $data['renault'] = $this->input->post('renault');
    $this->load->model('Edit_model');
    if($data['peugeot'])
    {
        $this->Edit_model->peugeot($data['peugeot']);
        $result = $this->Edit_model->peugeot($data['peugeot']);
        $data['result'] = $result;
        $this->load->view('content/editprocess',$data);
    }