mysql - Laravel 的 View 的 ajax 命名路由

Question

我尝试检索数据以查看但失败。有人会帮我吗？这是我的代码:路线

Route::post('/test',array('as'=>'ajaxTest','uses'=>'TestController@getTestData'));    
Route::get('/test',function(){ return View::make('test3');});    

测试 Controller

<?php

Class TestController extends BaseController {

    public function getTestData()
    {
        //$companies = People::all();
        //http://trirand.com/blog/jqgrid/jqgrid.html
        $page = $_GET['page'];  // get the requested page 
        $limit = $_GET['rows']; // get how many rows we want to have into the grid 
        $sidx = $_GET['sidx'];  // get index row - i.e. user click to sort 
        $sord = $_GET['sord'];  // get the direction if(!$sidx) $sidx =1;
        if(!$sidx) $sidx =1;    
        // connect to the database 
        $sql = DB::table('people')->get();
        $db = mysql_connect('localhost', 'root', 'mysql') 
        or die("Connection Error: " . mysql_error()); 
        mysql_select_db('people') or die("Error conecting to db.");

        $result = mysql_query("SELECT COUNT(*) AS count FROM invheader a, 
            clients b WHERE a.client_id=b.client_id"); 
        $row = mysql_fetch_array($result,MYSQL_ASSOC); 
        $count = $row['count']; 
        if( $count >0 ) { $total_pages = ceil($count/$limit); } 
            else { $total_pages = 0; } 
        if ($page > $total_pages) $page=$total_pages; 
        $start = $limit*$page - $limit; // do not put $limit*($page - 1) 
        $SQL = "SELECT a.id, a.invdate, b.name, a.amount,a.tax,a.total,a.note FROM invheader a, clients b WHERE a.client_id=b.client_id ORDER BY $sidx $sord LIMIT $start , $limit"; 
        $result = mysql_query( $SQL ) or die("Couldn t execute query.".mysql_error()); $responce->page = $page; $responce->total = $total_pages; $responce->records = $count; $i=0; while($row = mysql_fetch_array($result,MYSQL_ASSOC)) { $responce->rows[$i]['id']=$row[id]; $responce->rows[$i]['cell']=array($row[id],$row[invdate],$row[name],$row[amount],$row[tax],$row[total],$row[note]); $i++; } 
        dd($result);
        return Response::json($result);
    }
}    

查看

<script>    
$(document).ready(function () {
    jQuery("#list").jqGrid({ 
    //url: '{{ URL::route("ajaxTest") }}' ,
    //url: 'ajaxTest', 
    url: <?php echo URL::to("ajaxTest");?>,
    datatype: "json", 
    colNames:['ID','Name', 'Age'], 
    colModel:[ 
        {name:'id',index:'id', width:55}, 
        {name:'Name',index:'Name', width:90}, 
        {name:'Age',index:'Age', width:50} 
        ], 
       rowNum:10, 
       rowList:[10,20,30], 
       pager: '#pager2', 
       sortname: 'id', 
    viewrecords: true, 
    sortorder: "desc", 
    //autowidth: true, 
    height:280, 
    viewrecords: true, 
    multiselect: true, 
    multiselectWidth: 25, 
    sortable:true,      
    caption:"JSON Example"
    });
jQuery("#list").jqGrid('navGrid','#pager',{edit:false,add:false,del:false});
});

问题:
1. 如何访问命名路由ajax.test和TestController？
将调用方式更改为以下后，路线似乎现在可以工作:

url: <?php echo URL::to("ajaxTest");?>,

但仍然给我带来错误:

SyntaxError: missing } after property list


url: http://ecom1/ajaxTest,

2 我怎样才能使 TestController 的 getTestData 以 Laravel 方式工作？非常感谢

Answer 1

您在 JavaScript 中指定的 URL 为:url: 'ajaxTest'，而不是指定路由的 URL ajax.test。

为什么你要做那些mysql_connect和mysql_query的事情？您不需要使用 Laravel 执行任何这些操作。