嘿,对不起,如果这个问题是在我之前问的。我是CodeIgniter的新手,想知道这里将使用什么条件,以便它显示特定的emp_id
数据。在我的代码中,它使用user_id
显示数据,我想在url上显示user_id
,但根据emp_id
显示表单,它显示user_id
的所有emp_id
数据
需要帮助
$q="Select emp_id ,month from employee where user_id= $emp_id and year =".date("Y");
$details = $this->data['details'] = $this->evaluation_model->q($q);
//echo"<pre>";print_r($details);die;
$q ='select distinct section_permissions.assigned_for ,section_id , employee.emp_id ,employee.user_id, employee.month from employee join section_permissions on (section_permissions.assigned_for = employee.user_id) where section_permissions.user_id = '.$user_id;
$assigned_name = $this->data['assigned_name'] = $this->evaluation_model->q($q);
视图代码<?php foreach($assigned_name as $assigned){ ?>
<?php foreach($details as $detail){?>
<?php if($assigned['emp_id'] == $detail['emp_id']){?>
我认为你在查询中犯了错误,这就是为什么你得到表的所有行,而不是只有一个用户。
使用以下代码:
$dat = date("Y");
$q = "Select emp_id, month from employee where user_id = '$emp_id' and year = '$dat=date' ";