我收到以下 3 个变量的错误消息：
消息 1(未定义的变量：行 消息 2( 尝试获取非对象的属性
编辑：(终于找到了编辑按钮(，所以 3 个变量是：
$anatomic_code = $rows->NerveCode;
$ProxAccess = $rows->ProxAccess;
$DistAccess = $rows->DistAccess;
在我看来，正是上述3行给了我两个错误。
我没有包含该模型，因为我认为问题不存在，更重要的是因为它的代码相当多，因为开发人员在我之前使用了许多不同的表来存储数据......但是，如果需要，我准备发布它。
控制器 ->
function se_enkel_test()
{
$this->freakauth_light->check('submitter');
$tech_num = $this->uri->segment(6);
$this->load->model('view_single_test','',true);
$data['query'] = $this->view_single_test->query();
$data['query_tech_type'] = $this->view_single_test->query_tech_type();
$data['query_parameters'] = $this->view_single_test->query_parameters();
$data['query_tech_name'] = $this->view_single_test->query_tech_name();
$data['query_test'] = $this->view_single_test->query_test();
$data['query_temperat'] = $this->view_single_test->query_temperat();
$data['query_semiq'] = $this->view_single_test->query_semiq();
$data['query_conclusion'] = $this->view_single_test->query_conclusion();
$data['query_side'] = $this->view_single_test->query_side();
$this->load->model('mysql_create','',true);
$data['dropdown_conclusion_diagnosis'] = $this->mysql_create->get_dropdown_conclusion($tech_num);
$data['dropdown_conclusion_duration'] = $this->mysql_create->get_dropdown_duration_diagnosis();
$data['dropdown_conclusion_severity'] = $this->mysql_create->get_dropdown_severity_diagnosis();
$this->load->view('auh/se_enkel_test', $data);
}
视图 ->
elseif ($struct_id == "3") {
foreach($query->result() as $rows):
if ($rows->NerveCode > 10000 ){
echo "<h3>N." . $rows->NerveName . ") (" . $side->SideText.")</h3>";
}
else {
echo "<h3>N." . $rows->NerveName . " (" . $rows->ProxName . " - ";
if ($rows->Code < 3000) {
echo "m.";
}
echo $rows->DistName . ") (" . $side->SideText.")</h3>";
}
endforeach;
$anatomic_code = $rows->NerveCode;
$ProxAccess = $rows->ProxAccess;
$DistAccess = $rows->DistAccess;
}
这就是我的页面现在面临错误的样子，人们应该意识到这个页面上方有几个错误，如帖子前面所述。 但是所有 3 个变量都有相同的两个错误。
这是原始页面过去的外观。
从两个屏幕截图中，您可以看到我页面上的问题是我没有获得"段长"值..和肌肉名称： "腓蓿(脚踝 - 短指伸肌((罪。
具有函数"查询"的模型 ->
function query()
{
$exam = $this->uri->segment(3);
$type_id = $this->uri->segment(4);
$struct_id = $this->uri->segment(5);
$technique_id = $this->uri->segment(6);
$test_id = $this->uri->segment(7);
switch($type_id)
{
case 1:
$query_input = "SELECT EMUSCLE.*, NERVMUSCACCP.NAME AS MuscleName, NERVMUSCACCP.ShortName AS ShortMuscleName FROM EMUSCLE,NERVMUSCACCP WHERE Exam = ".$exam." AND Struct = ".$struct_id." AND EMUSCLE.MuscleCode = NERVMUSCACCP.Code;";
break;
case 2:
$query_input = "SELECT ENERVE.*, NERVMUSCACCP.NAME AS NerveName, NERVMUSCACCP.ShortName AS ShortNerveName FROM ENERVE,NERVMUSCACCP WHERE Exam = ".$exam." AND Struct = ".$struct_id." AND ENERVE.NerveCode = NERVMUSCACCP.Code;";
break;
case 3:
$query_input = "SELECT NewTable2.*, prox2.Code, prox2.Name AS DistName, prox2.ShortAccPName AS ShortDistName FROM (SELECT NewTable1.*, prox.Name AS ProxName, prox.ShortAccPName AS ShortProxName FROM (SELECT es.*, NERVMUSCACCP.NAME AS NerveName, NERVMUSCACCP.ShortName AS ShortNerveName FROM ESEGMENT es, NERVMUSCACCP WHERE Exam = ".$exam." AND Struct = ".$struct_id." AND es.NerveCode = NERVMUSCACCP.Code) AS NewTable1 LEFT OUTER JOIN PROXDISTINFO prox ON NewTable1.ProxAccess = prox.Code) AS NewTable2 LEFT OUTER JOIN PROXDISTINFO prox2 ON NewTable2.DistAccess = prox2.Code;";
break;
case 4:
$query_input = "SELECT NewTable1.*, NERVMUSCACCP.NAME AS MuscleName, NERVMUSCACCP.ShortName AS ShortMuscleName FROM (SELECT EJNCTION.*, NERVMUSCACCP.NAME AS NerveName, NERVMUSCACCP.ShortName AS ShortNerveName FROM EJNCTION LEFT OUTER JOIN NERVMUSCACCP ON EJNCTION.NerveCode = NERVMUSCACCP.Code WHERE EJNCTION.Exam = ".$exam." AND Struct = ".$struct_id.") AS NewTable1 LEFT OUTER JOIN NERVMUSCACCP ON NewTable1.MuscleCode = NERVMUSCACCP.Code";
break;
}
$query = $this->db->query($query_input);
return $query;
}
您似乎实际上并没有将查询结果传递给视图：
$data['query'] = $this->view_single_test->query();
如果不看到模型，就很难做出有根据的猜测。分配给 $data['query'] 的值应该是结果对象。 不是最干净的，但请尝试从模型中返回此$query->result();，并将其分配给控制器中的变量，然后将其传递给视图。