Question

该函数从数据库中获取数据，但问题是，在打印第一行后，它不会获取下一行。

请检查并告诉我我做错了什么？

Public function Return_Config($P_name, $WW, $KPI) {
    log_message('debug', 'hd');
    log_message('debug', $P_name . ' ' . $WW . ' ' . $KPI);
    $P_name .= '_one_voice_perf_measured';
    $select_query = "SELECT config FROM $P_name WHERE KPI = '$KPI' AND Config = '$WW'";
    //echo $select_query;

    $query = $this->db->query($select_query);
    if ($query->row() > '0') {
        return $query->row()->config;
    } else {
        return 'Not Measured';
    }
}

它直接进入其他部分。请帮忙

Answer 1

这是因为您使用->row()而不是->result()，例如：

Public function Return_Config($P_name, $WW, $KPI)
{
    log_message('debug', 'hd');
    log_message('debug', $P_name . ' ' . $WW . ' ' . $KPI);
    $P_name.= '_one_voice_perf_measured';
    $select_query = "SELECT config FROM $P_name WHERE KPI = '$KPI' AND Config = '$WW'";

    // echo $select_query;

    $query = $this->db->query($select_query);
    if (!empty($query->result()))
    {
        return $query->result();
    }
    else
    {
        return 'Not Measured';
    }
}

https://www.codeigniter.com/userguide3/database/results.html

的更多详情

Answer 2

试试这个

 Public function Return_Config($P_name,$WW,$KPI){
      log_message('debug','hd');
      log_message('debug', $P_name.' '.$WW.' '.$KPI);
            $P_name .= '_one_voice_perf_measured';
        $select_query = "SELECT config FROM $P_name WHERE KPI = '$KPI' AND Config = '$WW'";
        //echo $select_query;

        $query = $this->db->query($select_query)->row_array();
        if (!empty($query)){
          return query['config'];
           } 
      else{
                return 'Not Measured' ;
               }
    }

功能不正常，codeigniter

2 个答案: