I have some problems. Excuse me for my english.
我有一些问题。对不起我的英语。
I'ld display the datas but nothing !!
我显示数据但没有!
I dont know where is the probleme. I dont find it. Thnaks you for your help.
我不知道问题出在哪里。我找不到它。向你致以帮助。
code html
<div id="pie2" style="height:300px"></div>
code javascript
jQuery(document).ready(function ($){
var options = {
series: {
pie: {
show: true,
radius: 1,
label: {
show: true,
radius: 2 / 3,
tilt:0.5,
formatter: function(label, series)
{
return '<div style="font-size:8pt;text-align:center;padding:2px;color:white;">' + label + '<br/>' + Math.round(series.percent) + '% (' + series.data[0][1] + ')</div>';
},
background:
{
opacity: 0.8
}
}
}
},
legend: {
show: true
}
};
var dataset1 = <?php echo json_encode($pie);?>;
var data = [
{
"label": "Random Values",
"data": dataset1
}
];
var plotarea = $("#pie2");
$.plot( plotarea , data);
});
code PHP (source_pie.php)
代码PHP(source_pie.php)
$sql = "SELECT COUNT(rne) AS rne, dept FROM anuetab GROUP BY dept";
$result = mysql_query($sql);
while($row = mysql_fetch_assoc($sql)){
$pie[] = array(
'label'=>$row['dept'],
'data'=>$row['rne']
);
echo '<pre>';
print_r($pie);
echo '</pre>';
}
echo json_encode($pie);
1 个解决方案
#1
0
try with mysql_fetch_row instead of mysql_fetch_assoc
尝试使用mysql_fetch_row而不是mysql_fetch_assoc
#1
0
try with mysql_fetch_row instead of mysql_fetch_assoc
尝试使用mysql_fetch_row而不是mysql_fetch_assoc