如何将我的sql结果作为变量发送到Codeigniter中的控制器 [英] How to send the my sql result to controller in codeigniter as a variable

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本文介绍了如何将我的sql结果作为变量发送到Codeigniter中的控制器的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

在我的模式下,我选择的字段为

In my mode I am selecting a field as

$query1 = $this->db->query("SELECT dPassword 
                    FROM tbl_login
                    WHERE  dEmailID='a@a.in'");

如何将 dpassword 作为变量返回给我的控制器我这样尝试过 return dpassword;

How to return dpassword as a variable to my controller I tried this way return dpassword;

推荐答案

以下内容也不错:

if($query1->num_rows() > 0){
    $row = $query1->row();
}
return $row->dPassword;

然后,如果您的查询要返回多个行,则可以对结果进行如下操作:

Then if your query was to return more than a single row you could operate on the results like so:

foreach($query1->result() as $row){
echo $row->field1;
echo $row->field2;
echo $row->etcetera;
}

对于单行结果,我通常像这样直接从模型返回行:

For single row results i usually return the row directly from the model like so:

return $query1->row();

这是一个例子:

function select_provider_details($provider_id)
    {
        $this->db->select('*');
        $this->db->from('providers');
        $this->db->where('provider_id', $provider_id);

    $query = $this->db->get();

    if($query->num_rows() > 0)
    {
        $result['success'] = TRUE;
        $result['query'] = $query->row();
    }
    else
    {
        $result['success'] = FALSE;
        $result['error'] = "Provider not found in database";
        $result['errorcode'] = "E003";
        $result['query'] = $query->row();
    }
        return $result;
}

或者对于预期返回多个结果的查询,我将返回整个结果对象:

Or for a query expected to return multiple results i return the entire results object:

return $query1;

这篇关于如何将我的sql结果作为变量发送到Codeigniter中的控制器的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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