-1

大家好，我在模型中给出的功能如下

$query = $this->db->get();
            $result = $query->result();
            $data[] = $result;

但由于遇到 PHP 错误而出现错误

严重性：通知

消息：未定义变量：数据

文件名：models/survey_model.php

行号：845

无效的

这是为了什么。有人能帮助我吗

function get_actual_details_model($fin_year,$state){
        $this->db->select('*');
        $this->db->from('survey_respondent_info');
        $this->db->where('state', $state);
        $queryYr = $this->db->get();
        $resYr = $queryYr->result();
        foreach ($resYr as  $surveyId) {
            $this->db->select('budgets.*, budget_funding.*, marketing_budget.*, personnel_budget.*, grants_budget.*');
            $this->db->from('budgets');
            $this->db->join('budget_funding', 'budget_funding.budget_id = budgets.budget_id', 'left');
            $this->db->join('marketing_budget', 'marketing_budget.budget_id = budgets.budget_id', 'left');
            $this->db->join('personnel_budget', 'personnel_budget.budget_id = budgets.budget_id', 'left');
            $this->db->join('grants_budget', 'grants_budget.budget_id = budgets.budget_id', 'left');
            $this->db->where('budgets.financial_year',$fin_year);
            $this->db->where('budget_option_id', 1);
            $this->db->where('survey_id', $surveyId->survey_id);
          $data = array();
$query = $this->db->get();
$result = $query->result();
$data[] = $result;
        }
        return $data;
      }
4

4 回答 4

2

您应该使用的代码示例。这只是一个例子！

$data = array();
$query = $this->db->get( 'table' );
$result = $query->result();
$data[] = $result;

return $data;

您的代码几乎没有更新

function get_actual_details_model($fin_year,$state){
    $data = array(); // Data need to be in start
    $this->db->select('*');
    $this->db->from('survey_respondent_info');
    $this->db->where('state', $state);
    $queryYr = $this->db->get();
    $resYr = $queryYr->result();
    foreach ($resYr as  $surveyId) {
        $this->db->select('budgets.*, budget_funding.*, marketing_budget.*, personnel_budget.*, grants_budget.*');
        $this->db->from('budgets');
        $this->db->join('budget_funding', 'budget_funding.budget_id = budgets.budget_id', 'left');
        $this->db->join('marketing_budget', 'marketing_budget.budget_id = budgets.budget_id', 'left');
        $this->db->join('personnel_budget', 'personnel_budget.budget_id = budgets.budget_id', 'left');
        $this->db->join('grants_budget', 'grants_budget.budget_id = budgets.budget_id', 'left');
        $this->db->where('budgets.financial_year',$fin_year);
        $this->db->where('budget_option_id', 1);
        $this->db->where('survey_id', $surveyId->survey_id);
        $query = $this->db->get();
        $result = $query->result();
        $data[] = $result;
    }
    return $data;
}
于 2013-08-23T09:21:49.227 回答
2

你必须先定义你$data。

$data = array();

改变

$data[] = $result;

至

$data = $result;
于 2013-08-23T09:22:33.320 回答
1

在将任何内容放入 $data 之前，请将其定义为空数组，如下所示：

$data = array();
于 2013-08-23T09:22:16.237 回答
1

您的$data变量未定义。它附有$result

以下是更新后的代码：

$data = array();
$query = $this->db->get();
$result = $query->result();
$data[] = $result;
于 2013-08-23T09:22:31.343 回答