2017-10-07 81 views
1

我是编程新手,我在使用爆炸函数时遇到了问题,即在将值爆炸并将这些值传递给模型以从db中检索记录之后,只有爆炸数据的最后一个值显示。 ex我有爆炸数据(test,test1,test2)之后的数据就像(testtest1test2)在这里我传递这些值到模型从一个字段名称等于爆炸值的表中取得记录。我不'不知道如何正确解释,请原谅我。codeigniter中的爆炸函数

public function location($id) { 
    $query = $this->gt_pav_model->select($id); 
    $data['selectdata'] = $query->result(); 

    foreach($data['selectdata'] as $s) { 
     $explode = $s->package_id; // this is to get imploded data from db 
    } 

    $t = explode(",", $explode); 

    foreach($t as $tt) { 
     $query = $this->gt_pav_model->select_pa($tt); 
     $data['dataa'] = $query->result(); 
     $this->load->view('pav/pavdetails',$data); 
    } 
} 
+0

Sooraj仍然面临的问题?使用我的答案代码后? –

回答

1
public function location($id) { 
$final_result = array(); 
$result= $this->gt_pav_model->select($id)->result(); 
foreach($result as $key=> $re) { 
     $explode    = $re->package_id; 
     $t      = explode(",", $explode); 
     $param     = $t[1]; 
     $datas = $this->gt_pav_model->select_pa($param)->result(); 
     $final_result[$key] => $datas; 
    }  
    $this->load->view('pav/pavdetails',array('datas'=>$final_result)); 
} 

可以打印导致您的视图页面<?php print_r($datas); ?>