Question

如何将JSON字符串转换为javascript对象。我正在尝试将以下字符串转换为从服务器获取的JSON对象

JSON字符串：

["{"title":"Admin Dhaka","href":"#0","dataAttrs":[],"data":["{\"title\":\"BNS HAJI MOHSIN\",\"href\":\"#0\",\"dataAttrs\":[\"{\\\"title\\\":\\\"id\\\",\\\"data\\\":\\\"51\\\"}\"]}","{\"title\":\"BNS ISSA KHAN\",\"href\":\"#1\",\"dataAttrs\":[\"{\\\"title\\\":\\\"id\\\",\\\"data\\\":\\\"1\\\"}\"]}","{\"title\":\"BNT KHADEM\",\"href\":\"#2\",\"dataAttrs\":[\"{\\\"title\\\":\\\"id\\\",\\\"data\\\":\\\"6\\\"}\"]}","{\"title\":\"BN DOCKYARD\",\"href\":\"#3\",\"dataAttrs\":[\"{\\\"title\\\":\\\"id\\\",\\\"data\\\":\\\"13\\\"}\"]}","{\"title\":\"BNT SEBAK\",\"href\":\"#4\",\"dataAttrs\":[\"{\\\"title\\\":\\\"id\\\",\\\"data\\\":\\\"7\\\"}\"]}","{\"title\":\"Naval Aviation\",\"href\":\"#5\",\"dataAttrs\":[\"{\\\"title\\\":\\\"id\\\",\\\"data\\\":\\\"89\\\"}\"]}","{\"title\":\"BNS SAIKAT\",\"href\":\"#6\",\"dataAttrs\":[\"{\\\"title\\\":\\\"id\\\",\\\"data\\\":\\\"40\\\"}\"]}","{\"title\":\"BNS Novojatra\",\"href\":\"#9\",\"dataAttrs\":[\"{\\\"title\\\":\\\"id\\\",\\\"data\\\":\\\"119\\\"}\"]}","{\"title\":\"BNS SHAH AMANAT\",\"href\":\"#10\",\"dataAttrs\":[\"{\\\"title\\\":\\\"id\\\",\\\"data\\\":\\\"11\\\"}\"]}"]}"]

作为示例，我在上面的代码中给了一个对象，实际上Array是这样的对象列表

[obj1, obj2....]

我尝试如下：

var arr = '<?php echo !empty($treeView) ? $treeView : "[]"; ?>';
    arr = JSON.parse(arr);
    console.log(arr);

出现以下错误：

Uncaught SyntaxError: Unexpected token t in JSON at position 4

PHP代码：

function ship_by_area_zone(){

        $area_list = [];
        $ship_list = [];
        $zone = $this->utilities->findAllByAttributeWithOrderBy("bn_navyadminhierarchy", array("ADMIN_TYPE" => 1, "ACTIVE_STATUS" => 1), "CODE");
        $area = $this->utilities->findAllByAttributeWithOrderBy("bn_navyadminhierarchy", array("ADMIN_TYPE" => 2, "ACTIVE_STATUS" => 1), "CODE");


        // area wise ship
        foreach ($area as $key=>$value)
        {
            $row = $this->db->query("select * from bn_ship_establishment where AREA_ID = $value->ADMIN_ID and ACTIVE_STATUS = 1 order by CODE asc")->row();
            if($row)
            {
                $dataAttrs = array();
                $dataAttrs['title'] = 'id';
                $dataAttrs['data'] = $row->SHIP_ESTABLISHMENTID;
                $dataAttrs = json_encode($dataAttrs);

                $ship_row = array();
                $ship_row['title'] = $row->NAME;
                $ship_row['href'] = "#$key"; //"#1"
                $ship_row['dataAttrs'] = [$dataAttrs];
                $ship_list[] = json_encode($ship_row);
            }
        }

        // zone wise area
        foreach ($zone as $key=>$value)
        {

            $row = $this->db->query("select * from bn_navyadminhierarchy where ACTIVE_STATUS = 1 and PARENT_ID = $value->ADMIN_ID order by CODE asc")->row();
            if($row)
            {
                $area_row = array();
                $area_row['title'] = $row->NAME;
                $area_row['href'] = "#$key";
                $area_row['dataAttrs'] = [];
                $area_row['data'] = $ship_list;
                $area_list[] = json_encode($area_row);
            }
        }

        return json_encode($area_list);
    }

有人可以帮助我吗？

谢谢！

Answer 1

您需要在PHP中使用json_encode：

var arr = '<?php echo json_encode(!empty($treeView) ? $treeView : "[]"); ?>';

Answer 2

首先，您所拥有的json字符串无效。您可以在线免费检查以验证json。拥有有效的json后，您可以使用JSON.parse（）将其转换为JSONObject

Answer 3

// converting a simple javascript object to JSON object
 my_details =  
                {
                   "name"  : "SL",
                   "age "  : "30" ,
                   "photo" : "imgMe.jpg"
                }
my_details_in_json = JSON.stringify(my_details);

// "{"name":"SL","age ":"30","photo":"imgMe.jpg"}" --> my_details_in_json

如何从JSON制作Javascript对象

3 个答案: