drci47425 2014-10-23 06:13 采纳率: 100%
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尝试从codeigniter中的不同表中获取部门名称

Model: Here is my Model i want to get record from department against department id.

<?php
class Functions extends CI_Model{
    function getName($table,$column,$id){
        $this->db->select($column)->from($table)->where('id',$id);
        $query = $this->db->get();
        return $query->result_array();
    }
}
?>

View:

  33) <td><kbd><?php echo $rows['id'];?></kbd></td>
  34) <td><?php echo $rows['doctor_name'];?></td>
  35) <td><?php echo $this->functions->getName('mansoora_department','department_name',$rows['doctor_department_id']);?></td>
  36) <td><?php echo $rows['doctor_nic'];?></td>

Error:

A PHP Error was encountered

Severity: Notice

Message: Array to string conversion

Filename: doctor/index.php

Line Number: 35

  • 写回答

1条回答 默认 最新

  • dsstlsqt696435 2014-10-23 06:19
    关注

    Since your method returns an Array. As clearly stated in your codes. Treat it as such:

    return $query->result_array(); // an array. duh! result array :)
    

    Either put it in another loop

    <td>
        <?php 
        $names = $this->functions->getName('mansoora_department','department_name',$rows['doctor_department_id']);
        foreach($names as $name) {
            echo $name . ' ';
        }
        ?>
    </td>
    

    Or if you're expecting just one row:

    return $query->row()->department_name;
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

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