Php 如何使我的JSON动态地看起来与下面谷歌图表的示例相同？

首先，检查此示例：

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其次，我需要知道如何通过知道下面的代码是针对数据库的，从而使我的JSON数据看起来与此相同：

<?php

$con=mysql_connect("localhost","root","") or die("Problem while connecting to the database!");

mysql_select_db("charts", $con); 

$sth = mysql_query("SELECT item_code,serial_no,date FROM googlecharts group by item_code");
$rows = array();
$flag = true;

$table = array();

$table['cols'] = array(
//i guess i need a loop through the first column to get all items and make them labels
    array('label' => 'item code', 'type' => 'string'),
    array('label' => 'sum of serial numbers', 'type' => 'number'),
    array('label' => 'sold month', 'type' => 'number')
);

$rows = array();
while($r = mysql_fetch_assoc($sth)) {
    $temp = array();

$temp[] = array('v' => (string) $r['item_code']); 
$temp[] = array('v' => (int) $r['serial_no']);
$temp[] = array('v' => (int) $r['sold_date']);

$rows[] = array('c' => $temp);
}

$table['rows'] = $rows;

$jsonTable = json_encode($table);

echo $jsonTable;
?>

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通过在jsfiddle.net中看到这个例子，我想你会对我的想法有所了解，如果可以的话，请帮助我。。我是stackoverflow的新手，这是我第一个订购的问题XD，谢谢大家。

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竖起大拇指^ ^

亲爱的，我通过一种叫做“pivot”的方式在互联网上搜索数据透视表或数据透视视图，实现了这一点。。。如果有人想这么做，告诉我我会向你解释的。 谢谢大家

<?php

$con=mysql_connect("localhost","root","") or die("Problem while connecting to the database!");

mysql_select_db("charts", $con); 

$sth = mysql_query("SELECT item_code,serial_no,date FROM googlecharts group by item_code");
$rows = array();
$flag = true;

$table = array();

$table['cols'] = array(
//i guess i need a loop through the first column to get all items and make them labels
    array('label' => 'item code', 'type' => 'string'),
    array('label' => 'sum of serial numbers', 'type' => 'number'),
    array('label' => 'sold month', 'type' => 'number')
);

$rows = array();
while($r = mysql_fetch_assoc($sth)) {
    $temp = array();

$temp[] = array('v' => (string) $r['item_code']); 
$temp[] = array('v' => (int) $r['serial_no']);
$temp[] = array('v' => (int) $r['sold_date']);

$rows[] = array('c' => $temp);
}

$table['rows'] = $rows;

$jsonTable = json_encode($table);

echo $jsonTable;
?>