Question

我正在构建一个搜索列表，会列出一些用户，当我尝试加载更多页面以显示更多用户时，我收到此错误：

SyntaxError：JSON.parse：JSON数据第1行第1列的意外字符

var result = JSON.parse（res）;

这是我的index.volt中的ajax函数

function(cb){
        $.ajax({
            url: '/search/search?q=' + mapObject.q + '&sort=<?php echo $sort;?>' +  '&page=' + mapObject.page,
            data:{},
            success: function(res) {
                var result = JSON.parse(res);
                if (!result.status){
                    return cb(null, result.list);
                }else{
                    return cb(null, []);
                }
            },
            error: function(xhr, ajaxOptions, thrownError) {
                cb(null, []);
             }
    });

这是我的Controller中的searchAction（）：

public function searchAction()
    {
        $q = $this->request->get("q");
        $sort = $this->request->get("sort");
        $searchUserModel = new SearchUsers();
        $loginUser = $this->component->user->getSessionUser();
        $page = $this->request->get("page");
        $limit = 1;
        if (!$page){
            $page = 1;
        }
        $list = $searchUserModel->findTeachers($q, $loginUser->id, ($loginUser?true:false), $page, $limit, $sort);
        $list['status'] = true;
        echo json_encode($list);
    }

Console.log（res）返回：

<table class='xdebug-error xe-notice' dir='ltr' border='1' cellspacing='0' cellpadding='1'> <tr><th align='left' bgcolor='#f57900' colspan="5">( ! ) Notice: Trying to get property of non-object in C:\xampp\htdocs\tute\app\controllers\SearchController.php on line 62</th></tr> <tr><th align='left' bgcolor='#e9b96e' colspan='5'>Call Stack</th></tr> <tr><th align='center' bgcolor='#eeeeec'>#</th><th align='left' bgcolor='#eeeeec'>Time</th><th align='left' bgcolor='#eeeeec'>Memory</th><th align='left' bgcolor='#eeeeec'>Function</th><th align='left' bgcolor='#eeeeec'>Location</th></tr> <tr><td bgcolor='#eeeeec' align='center'>1</td><td bgcolor='#eeeeec' align='center'>0.0009</td><td bgcolor='#eeeeec' align='right'>192016</td><td bgcolor='#eeeeec'>{main}( )</td><td title='C:\xampp\htdocs\tute\public\index.php' bgcolor='#eeeeec'>...\index.php:0</td></tr> <tr><td bgcolor='#eeeeec' align='center'>2</td><td bgcolor='#eeeeec' align='center'>0.0019</td><td bgcolor='#eeeeec' align='right'>205824</td><td bgcolor='#eeeeec'><a href='http://www.php.net/Phalcon\Mvc\Application.handle' target='_new'>handle</a> ( )</td><td title='C:\xampp\htdocs\tute\public\index.php' bgcolor='#eeeeec'>...\index.php:239</td></tr> <tr><td bgcolor='#eeeeec' align='center'>3</td><td bgcolor='#eeeeec' align='center'>0.0032</td><td bgcolor='#eeeeec' align='right'>256968</td><td bgcolor='#eeeeec'><a href='http://www.php.net/Phalcon\Dispatcher.dispatch' target='_new'>dispatch</a> ( )</td><td title='C:\xampp\htdocs\tute\public\index.php' bgcolor='#eeeeec'>...\index.php:239</td></tr> <tr><td bgcolor='#eeeeec' align='center'>4</td><td bgcolor='#eeeeec' align='center'>0.0161</td><td bgcolor='#eeeeec' align='right'>452064</td><td bgcolor='#eeeeec'>SearchController->searchAction( )</td><td title='C:\xampp\htdocs\tute\public\index.php' bgcolor='#eeeeec'>...\index.php:239</td></tr> </table> {"total":"2","numPages":2,"page":"2","list":[{"id":"23","lesson_complete":"4","num_rating":"4","rating":"3.75","intro_video":"0","uploads":"0","experience_filled":"1","lat":"9999","lng":"9999","usr_id":"23","price":"30","distance":"0","rank":"4.33","search_id":"18","firstname":"Yolo","lastname":"yolo","avatar":null,"lan":"lang_zh-CN","usr_teach":1,"usr_rate":"30","skills":"Ice Hockey"}],"status":true}

SearchController.php中的第62行引用searchAction（）中的$list = $searchUserModel->findTeachers($q, $loginUser->id, ($loginUser?true:false), $page, $limit, $sort);

所以我的问题是如何在res中只返回那些没有那些html的JSON？

Answer 1

要发送JSON响应，您必须禁用该视图。

$this->view->disable();

我还会使用$this->response->setContentType('application/json')明确设置内容类型。

Answer 2

所有htmls都是在表格中列出的xdebug消息。产生错误是因为用户在搜索时未登录。通过设置$loginUser = new StdClass; $loginUser->id = '';，问题就会解决。

如何在res中只获得JSON？

2 个答案: