Question

您好我有一个插入一些数据的查询，然后检索插入的ID。然后，inert_id用于另一个连接2个表的查询中。数据正在插入但未显示。

欢迎任何帮助，

麦克

<?php
public  function saveRecord(){
    $this->load->helper('date');

    $data = array(
    'user' => $this->input->post('user'),
    'qId' => $this->input->post('id'),
        'date' => now()
    );

    $sql = $this->db->insert('videos',$data);
    $answer_id = $this->db->insert_id();

    if($this->db->affected_rows() != 1)
    {
    echo "failed to add to table";
    } else {
        $this->db->where('id', $answer_id);
        $this->db->select('*');
        $this->db->from('videos');
        $this->db->join('questions', 'questions.id = videos.qId','inner');

        $sql2 = $this->db->get();

        if($sql2->num_rows() > 0){
            foreach($sql2->result() as $r){
?> 

<script>
    $('#answers_wrapper').append("<div class='answer_recorded_container'></div>");
</script>

<?php 
            }
        } else {
?>

<script>
    $('#answers_wrapper').html("<div class='answer_recorded_container'>nothing found!</div>");
</script>

<?php 
        }
    }
}
?>

选择Join不使用insert_id（）codeIgniter

0 个答案: