php - 将 json 数据解析为变量并在每次迭代时提交给函数

Question

我创建了一个生成 json 响应的 php 脚本

这是输出的示例:

[[],{"idgps_unit":"2","lat":"40","lon":"40","name":"ML350","notes":"Andrew","dt":"2012-10-29 19:43:09","serial":"3602152","speed":"44","odometer":"208.49"},{"idgps_unit":"1","lat":"42","lon":"39","name":"unit1","notes":"fake unit 1","dt":"2012-10-18 18:16:37","serial":"12345","speed":"0","odometer":"0.16"}]

这就是我在 PHP 中形成响应的方式:

   $data[] = array();
    foreach ($list->arrayList as $key => $value) {

        $unit = new Unit();
        $unit = $value;
        //create array for json output
        $data[] = array('idgps_unit' => $unit->idgps_unit, 'lat' => $unit->lat,
            'lon' => $unit->lon, 'name' => $unit->name, 'notes' => $unit->notes,
            'dt' => $unit->dt, 'serial' => $unit->serial, 'speed' => $unit->speed,
            'odometer' => $unit->odometer);
    }
    echo json_encode($data);

现在，在 JS 中我这样做了:

function getCheckedUnits() {
    jQuery(function($) {    
        $.ajax( {           
            url : "json.php?action=get",
            type : "GET",
            success : function(data) {
                var jsonData = JSON.parse(data);

                ///PARSE VALUES AND SUBMIT TO A FUNCTION :: START
                var C_longitude = 0;
                var C_name = 0;
                var C_idgps_unit = 0;
                var C_serial = 0;
                var C_speed= 0;
                var C_notes= 0;
                var C_dt = 0;
                var C_time = 0;
                var C_odometer = 0;

                initialize(C_longitude,C_name,C_idgps_unit, C_serial,C_speed, C_notes, C_dt, C_time, C_odometer);
                ///PARSE VALUES AND SUBMIT TO A FUNCTION :: END
            }
        });
    });
}

我需要将 json reponce 解析为值

Answer 1

假设 JSON.parse(data) 仅获取 JSON 响应中的关联数组，您应该能够获取 JSON 数据中的值，如下所示:

var i = 1;
var C_longitude = jsonData[i]["lon"];
var C_name = jsonData[i]["name"];

假设第一个空数组没有被 JSON.parse() 删除，i = 1 将获取第一批数据，i = 2 将获取第二批数据。

解析后的 JSON 的行为方式与 JavaScript 中定义的行为方式相同