javascript - javascript的对象对象结果

Question

我的 javascript 上有 [Object object] 结果。

[img] https://i.imgur.com/SQdIpYy.png[/img]

: "> ">

 $(document).ready(function(){
        if(!isAndroid) {
            $('.chosen-select').select2({
                theme: "classic"
            });
        }

        <?php $distance = (isset($data['distance']))? $data['distance'] != '' ?  $data['distance']  :  $this->config->item('default_distance') : $this->config->item('default_distance');  ?>

        var distance = parseInt('<?php echo $distance; ?>');

        var distance_unit = '<?php echo lang_key(get_settings("business_settings", "show_distance_in", "miles")); ?>';

        $("#distance-slider").ionRangeSlider({

			min: <?php echo $this->config->item('min_distance');?>,
			max: <?php echo $this->config->item('max_distance');?>,
			from: <?php echo $this->config->item('default_distance');?>,
			to: <?php echo $this->config->item('max_distance');?>,
			postfix: " <?php echo lang_key(get_settings("business_settings", "show_distance_in", "miles")); ?>",
			max_postfix: "+",
      		value: distance,

            slide: function (event, ui) {

                $("#location-distance-slider").val(ui.value);
                $("#amount").html( ui.value + ' ' + distance_unit );

            }

        });
		
        $("#location-distance-slider").val(distance);
        $("#amount").html($( "#distance-slider" ).ionRangeSlider( "value") + ' ' + distance_unit);


    });

<?php echo lang_key('distance_around_my_position'); ?>: <span class="text-danger" id="amount"></span>
										<input id="distance-slider"/>
										<input type="hidden" id="location-distance-slider" name="distance" value="">
										<input type="hidden" id="geo_lat" name="geo_lat" value="<?php echo (isset($data['geo_lat']))?$data['geo_lat']:''; ?>">
										<input type="hidden" id="geo_lng" name="geo_lng" value="<?php echo (isset($data['geo_lng']))?$data['geo_lng']:''; ?>">

滑块数据应显示在我的位置周围的距离上：[object Object] kms

标签： javascript

Answer 1

您可以使用 json 编码和 json 解码。

获取对象对象结果是因为如果我们在从数据库中获取数据时使用精确选择。它发生在我的案例中。

您尝试将该变量放入 =>

 $value['code'] = json_decode(json_encode($value['group_code'], true));

打印那个$result。

例子：

$data['obli'] = execSelect(" 

                         SELECT 
                         CONCAT(bp_brokerage_history_id,'-',DATE_FORMAT(created_at, '%d/%m/%Y')) AS bp_brokerage_history_id
                         FROM
                         brokerage_payout
                         WHERE
                         status = 2
                         ORDER BY bp_brokerage_history_id DESC;",[]);

我的结果$data['obli']给出了对象对象。那时使用“收集”，您可以使用json_decode(json_encode($value['group_code'], true));或收集。

 $data['date'] = collect($data['obli'])-> 
                  pluck('bp_brokerage_history_id','bp_brokerage_history_id');