小编IBM*_*dig的帖子

当容器在小型设备中具有显示属性none时我的饼图有问题,并且我有以下错误:

未捕获的IndexSizeError:无法在'CanvasRenderingContext2D'上执行'arc':提供的半径(-0.5)为负数.

我的HTML代码是

<div class="container-fluid charts hidden-xs">
    <div class="row">
        <div class="col-md-6 col-xs-12">
            <div class="panel panel-default">
                <div class="panel-heading">Annual sales</div>
                <div class="panel-body">
                    <canvas class="annualChart" width="400" height="400"></canvas>
                </div>
            </div>
        </div>
        <div class="col-md-6 col-xs-12">
            <div class="panel panel-default">
                <div class="panel-heading">Visitors per shop</div>
                <div class="panel-body">
                    <canvas class="visitorsChart" width="400" height="400"></canvas>
                    <div id="legend"></div>
                </div>
            </div>
        </div>
        <div class="col-md-6 col-xs-12">
            <div class="panel panel-default">
                <div class="panel-heading">Fans</div>
                <div class="panel-body">
                    <canvas class="annualfanschart" width="400" height="400" ></canvas>
                </div>
            </div>
        </div>

        <div class="col-md-6 col-xs-12">
            <div class="panel panel-default">
                <div class="panel-heading">General sales</div>
                <div class="panel-body">
                    <canvas class="generalSalesChart" …

我有两个recipes_sa带列的表：

recipes_id   recipes_name   recipes_chef
----------   ------------   ------------

并chefs_sa带有列：

chefs_id   chefs_name
--------   ----------

我想获得有限数量的食谱，其中包含他们的厨师详细信息，使用INNER JOIN和LIMIT

我做了以下功能：

function getLimitJoinData($data, $tbls, $ids, $abr, $type, $limit) {

            $dataToSelect = implode($data, ',');

            $q = "SELECT $dataToSelect";

            $q.= " FROM (SELECT * FROM $tbls[0] LIMIT $limit) $abr";


            for ($i=1; $i < count($tbls); $i++) { 
                $q .= " ".$type." JOIN ". $tbls[$i] ." ON " . $abr.'.recipes_chef' .' = '. $ids[$i-1][0];   
            }
        }

查询是这样的

SELECT chefs_sa.chefs_name,
       recipes_sa.recipes_name 
FROM (SELECT * …