php - codeigniter mysql查询输出数据显示特定条件

Question

以下代码:

Controller .php

function getchart() {
$prac = $this->input->post('prac_name');
$datee = $this->input->post('datee');
$this->load->model('appoint');
$results['appoint'] = $this->appoint->getappoint($prac , $datee);
$this->load->view('ajax/getappchart' , $results);
}

模型.php

function getappoint($prac , $datee) {
        $this->db->select('rdv.id as rdvid, startTime, endTime, day, firstname, lastname');
        $this->db->from('rdv');
        $this->db->join('contact', 'contact.id = rdv.contact_id');
        $this->db->where('people_id',$practicien);
        $this->db->where('DATE(day)', $datee);
        $this->db->order_by('TIME(startTime)', 'ASC'); 
        $query = $this->db->get();
        //print_r($this->db->last_query());
        return $query;
    }

查看.php

if ($appoint->num_rows() > 0) {
        foreach($appoint->result() as $sub_row)
        {
// display output.
}
    } else {
        echo 'No Appointments on Above Date.';
    }
    ?>

我需要的是，有相同时间和日期的约会(大部分是2个相同的)。如果超过2个，我需要为两个约会设置两种不同的类(class)风格。我怎样才能实现这个目标？

谢谢。

Answer 1

最终答案:在@minhaz-ahmed的帮助下完成

if ($appoint->num_rows() > 0) {
    $appoint_counter = array();
    foreach ($appoint->result() as $sub_row) {
        //i am assuming your startTime is H:M:S, and  day Y-M-D format 
        $key = strtotime($sub_row['day'] . ' ' . $sub_row['startTime']);
        if (!isset($appoint_counter[$key])) {
            $appoint_counter[$key] = 0;
        }
        $appoint_counter[$key] ++;
        $style_class = 'YOUR_1ST_CLASS';
        if ($appoint_counter[$key] > 2) {
            $style_class = 'YOUR_2ND_CLASS';
        }

       //REST VIEW CODE
   }
} 
else {
    echo 'No Appointments on Above Date.';
}