1

我有一些用于显示图形数据的 JS 代码：

series: [{
                name: 'Year 1800',
                data: [107, 31, 635, 203, 2]
            }, {
                name: 'Year 1900',
                data: [133, 156, 947, 408, 6]
            }, {
                name: 'Year 2008',
                data: [973, 914, 4054, 732, 34]
            }]

我需要让数据在 PHP 的 while 循环中显示。我试过这个：

<?php
                $sql="SELECT *, COUNT(category) AS my_groupcount from tickets where deleted = '' and DAY(datetime) = '".$day."' and MONTH(datetime) = '".$month."' and YEAR(datetime) = '".$year."' group by category order by datetime ASC ";
                $rs=mysql_query($sql,$conn);
                while($result=mysql_fetch_array($rs))
                {
                    echo "name: 'Cat ".$result["category"]."',";
                    echo "data: [".$result["my_groupcount"]."]";
                    echo "}, {";
                }
                ?>

我需要对门票表中的类别列进行分组并显示每个类别的图表但它不起作用 - 我认为这是因为在 while 循环中它以}, {但我需要它以结束}]

我该如何解决这个问题 - 门票表中的类别项目数量一直在变化，因为用户能够添加/删除类别。

4

4 回答 4

4

为什么不这样做：

<?php
    $sql = "[.. SQL Statement ..]";
    $rs = mysql_query($sql, $conn);
    $json = array();

    while($result = mysql_fetch_array($rs)) {
        $json[] = array( 
            'name' => 'Cat '. $result['category'],

            // This does assume that my_groupcount is an array with numbers
            // i.e. array(1, 34, 54, 345)
            // If not, you'll have to make it an array by doing:
            // explode(', ', $result['my_groupcount'])
            // This however does assume that the numbers are in 
            // the "12, 23" format
            'data' => $result['my_groupcount'],
        );
    }             

    echo json_encode($json);
于 2013-09-24T13:57:48.177 回答
1 
<?php
            $sql="SELECT *, COUNT(category) AS my_groupcount from tickets where deleted = '' and DAY(datetime) = '".$day."' and MONTH(datetime) = '".$month."' and YEAR(datetime) = '".$year."' group by category order by datetime ASC ";
            $rs=mysql_query($sql,$conn);
            $first = true;
            echo 'series: [{';
            while($result=mysql_fetch_array($rs))
            {
                if(!$first) {
                    echo "}, {";
                } else {
                    $first = false;
                }
                echo "name: 'Cat ".$result["category"]."',";
                echo "data: [".$result["my_groupcount"]."]";
            }
            echo '}]';
?>
于 2013-09-24T14:12:59.273 回答
0

而是将括号括起来，尽管最好先构建它然后回显它，这样您就可以摆脱最后一个逗号。

$string = '';
while($result=mysql_fetch_array($rs))
{
    string.= "{";
    string.= "name: 'Cat ".$result["category"]."',";
    string.= "data: [".$result["my_groupcount"]."]";
    string.= "},";
}
$string = trim($string,','); // gets rid of that last comma

echo "[$string]";
于 2013-09-24T13:57:30.983 回答
0

尝试

$sql="SELECT *, COUNT(category) AS my_groupcount from tickets where deleted = '' and DAY(datetime) = '".$day."' and MONTH(datetime) = '".$month."' and YEAR(datetime) = '".$year."' group by category order by datetime ASC ";
$rs=mysql_query($sql,$conn);
$output = '[';
while($result=mysql_fetch_array($rs))
{
    $output .= "name: 'Cat ".$result["category"]."',";
    $output .= "data: [".$result["my_groupcount"]."]";
    $output .= "}, {";
}

$output = substr($output, 0, -3) . ']';

但正如 Rich 所说，您真的不应该手动编写 JSON。

于 2013-09-24T13:59:05.200 回答