我有一个查询，该查询在foreach循环内运行，但仅在以下代码中立即运行，如下所述：
public function functional_job($functional)
{
$functions = explode(",",$functional);
foreach($functions as $row)
{
$this->db->select('*');
$this->db->from('job');
$this->db->where('functional',$row);
$query = $this->db->get();if($query->num_rows()>0)
{
$result = $query->result_array();
return $result;
}
else
{
$this->session->set_flashdata('msg','<p style="color:red;font-weight: 600;">No Job Found.</p>');
}
}
}
在哪里$functional = IT,marketing,ecommerce。当我点击查询时，它只会运行一个类似的查询SELECT * FROM job WHERE functional = 'IT'。但我希望它运行三遍SELECT * FROM job WHERE functional = 'IT' SELECT * FROM job WHERE functional = 'marketing' SELECT * FROM job WHERE functional = 'ecommerce'
您的代码将返回一个结果的值并停止处理。
您可以更改代码以将结果存储在数组中，然后返回该代码？
public function functional_job($functional){
$functions = explode(",",$functional);
$results = array();
foreach($functions as $row)
{
$this->db->select('*');
$this->db->from('job');
$this->db->where('functional',$row);
$query = $this->db->get();
if($query->num_rows()>0)
{
$results[] = $query->result_array();
}
else
{
$this->session->set_flashdata('msg','<p style="color:red;font-weight: 600;">No Job Found.</p>');
}
}
return $results;
}
如前所述，您还可以更改查询以使用MySql的IN功能。
SELECT * FROM job WHERE functional IN ('IT', 'marketing', 'ecommerce')
如果不知道codeigniter，它可能看起来像：
$this->db->select('*');
$this->db->from('job');
$this->db->where_in('id', $functional);
$query = $this->db->get();
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