Question

我知道我的问题是我在Controller $ name和$ year中调用变量的地方，我不知道如何将该数组格式化为Controller中的变量。

型号：

function getId($id)
{
    $this->db->distinct();
    $this->db->select('*');
    $this->db->where('id', $id);
    $result = $this->db->get('HWC');

    return $result->result();
}

function getVariations($name, $year)
{
    $this->db->distinct();
    $this->db->select('*');
    $this->db->where(array('ModelName' => $name, 'Year' => $year));
    $variations = $this->db->get('HWC');

    if(mysql_num_rows($variations) != 0)
    {
        return $variations->result();
    }
    else
    {

    }
}

控制器：

$this->load->model('testingsearch');
$data['records'] = $this->testingsearch->getId($id);

$name = $records->ModelName;
$year = $records->Year;

$data['variations'] = $this->testingsearch->getVariations($name, $year);

查看：

 foreach($variations as $row){
    echo $row['name'];
    echo $row['color'];
    echo $row['Tampo'];
 }

Answer 1

function getId($id)
{
    $this->db->distinct();
    $this->db->select('*');
    $this->db->where('id', $id);
    $result = $this->db->get('HWC');
    return $result->result();
}

You are returning object with result() but displaying as array

foreach($variations as $row){
    echo $row['name'];
    echo $row['color'];
    echo $row['Tampo'];
}

像这样改变

return $result->result_array();

现在查看工作正常。在控制器中，你在这里犯了错误

$data['records'] = $this->testingsearch->getId($id);

将其更改为

$records    =   $this->testingsearch->getId($id);

现在这些说明应该可以正常工作但条件

if($records){
    $name = $records->ModelName;
    $year = $records->Year;

    $data['variations'] = $this->testingsearch->getVariations($name, $year);
}

2 SQL在CodeIgniter MVC环境中搜索一个id

1 个答案: