Question

I am trying to join two tables. Lets say t1 and t2. t1 has fk t2_id. but when I run the code nothing is shown in my view.

<<>斯特凡>

 //..
 $data[ city ]= $this->state_model->name();
 $this->load->view( viewt , $data);

<<>Model>:

    function name(){
    $this->db->select( * );
    $this->db->from( state );
    $this->db->join( city ,  city.state_id = state.id );
    $sql = $this->db->get();

    if ($sql->num_rows () >0) {
        foreach($sql->result() as $row) {
        $this->db->where( state_id , state.id );
        $r = $this->db->get( city );
        }

    return $r->result();

     }
     else {
     return null;   
            }

View:

 <?php foreach($city as $row):?>
  <?php echo $row->cityname; ?></br></br></br>
  <?php endforeach;?></br></br>

预收

询问:

  SELECT `id`, `statename`
  FROM (`state`)
  ORDER BY `id` ASC

  SELECT *
FROM (`state`)
JOIN `city` ON `city`.`state_id` = `state`.`id`  

  SELECT *
  FROM (`city`)
  WHERE `state_id` =  state.id   

  SELECT *
  FROM (`city`)
  WHERE `state_id` =  state.id   

  SELECT *
  FROM (`city`)
  WHERE `state_id` =  state.id

Answer 1

你的美元变数只能是最后的变数,因此,这一功能只能回到最后的结局。

if ($sql->num_rows () >0) {
    foreach($sql->result() as $row) {
    $this->db->where( state_id ,$row->state_id);
    $r[] = $this->db->get( city )->row();
    }

return $r;

Answer 2

最新查询:$this->db->last_query()。

Answer 3

我不明白,为什么你需要在一间休息室内进行第二次询问。你可以这样做:

if ($sql->num_rows() > 0) {
    return $sql->result();
}

?

Answer 4

因此,你从你们的加入中重新获得三个记录,因此我们知道这项工作。但是,你再次提出其他3个问题,认为是不正确的。为此:

if ($sql->num_rows () >0) {
    $r = array();
    foreach($sql->result() as $row) {

        // here, you need to replace  state.id  with the actual
        // results from your join query ($row->id which is the id
        // from the state table
        $this->db->where( state_id ,$row->id);
        $r[] = $this->db->get( city )->result();
    }
    return $r;
 }
 else 
 {
     return null;   
 }

因此,现在,你应当取得一系列成果,并且能够与你的看法相呼应。

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