function search_review($input, $serviceName){
$ipJson = json_encode($input);
$this->db->select('*');
$this->db->from('reviews');
$this->db->where('reviewee_name', $input['reviewee_name']);
$query = $this->db->get();
$result = $query->result();
if (!empty($result)) {
foreach ($query->result() as $row)
{
$data['reviewer_id'] = $row->reviewer_id;
$data['reviewee_name'] = $row->reviewee_name;
$data['tournament_played_c_s'] = $row->tournament_played_c_s;
}
$data['message'] = 'Review details retrieved successfully.';
$status = $this->clamo_lib->return_status('success', $serviceName, $data, $ipJson);
} else {
$data['message'] = 'Unable to retrieve review details.';
$status = $this->clamo_lib->return_status('error', $serviceName, $data, $ipJson);
}
return $status;
}
这是我的代码,我可以检索具有相同名称的单行数据。例如reviewee_name =“sangeetha”意味着我需要显示名称为sangeetha的细节。但现在单行数据只有我得到我需要做的?在codeigniter的一个表中检索多个具有相同名称的数据
谢谢很多sundar其工作正常.. – sangee
一个疑问孙大信 – sangee
' <_0> 3 reviewer_id> <_1> deepi reviewee_name> <_2> 奈,TN tournament_played_c_s> <_3> 4 reviewer_id> <_4> deepi reviewee_name> <_5> 卡纳,成功检索班加罗尔 tournament_played_c_s> 评分的信息。 ' –
sangee