我遇到問題從我的數據庫使用where子句與for循環變量檢索數據。此代碼當前可以從數據庫中檢索數據,但是從數據庫中檢索到的值存儲。在for循環中的數字1仍然很好。但是當它進入下一個。收集的數據是1和2的組合。當它最終到達循環結尾時,顯示所有數據。 哪個代碼的一部分,必須進行編輯,以數據的顯示改變到非累積PHP SQL其中聲明與循環變量
<?php
$sectorcount = $row_SectorCount['COUNT(*)'];
//number of rows in database
$spendingname= array();
$spendingpercent= array();
$spendingid= array();
?>
// Add some data to the details pie chart
options_detail_1.series.push({ name: 'Tax Spending Detail', data: [] });
$(document).ready(function() {
chart_main = new Highcharts.Chart(options_main);
chart_detail_1 = new Highcharts.Chart(options_detail_1);
})
function push_pie_detail(sectorid) {
switch(sectorid)
{
<?php
for ($i=1; $i<=$sectorcount; $i++)
{
echo "case 'sector".$i."':" ,
"setTimeout", "(" , '"chart_detail_1.series[0].remove(false)", 1000);' ,
"chart_detail_1.addSeries(options_detail_1, false);" ,
"chart_detail_1.series[1].setData([";
mysql_select_db($database_conn2, $conn2);
$query_Spending = "SELECT CONCAT(spending.SectorID, spending.ExpenditureID) AS 'SpendingID',
expenditure.ExpenditureName, spending.SpendingPercent, spending.SectorID
FROM spending
INNER JOIN expenditure ON spending.ExpenditureID = expenditure.ExpenditureID
WHERE spending.SectorID = '.$i.'";
$Spending = mysql_query($query_Spending, $conn2) or die(mysql_error());
$totalRows_Spending = mysql_num_rows($Spending);
while($row_Spending = mysql_fetch_assoc($Spending))
{
$spendingname[] = $row_Spending['ExpenditureName'];
$spendingpercent[] = $row_Spending['SpendingPercent'];
$spendingid[]= $row_Spending['SpendingID'];
}
mysql_free_result($Spending);
$a = 0;
foreach ($spendingid as $sgi => $sgid)
{
if ($a > 0) echo ', '; // add the comma
echo "{
name: '$spendingname[$sgi]',
y: ". $spendingpercent[$sgi] * $tax .",
id: ' $sgid',
}";
$a++;
}
echo "], false);" ,
"chart_detail_1.redraw();",
"break;";
}
?>
default:
break;
}
}
你如何設法解決它?即時通訊真的很新的PHP請指導我:D – 2012-01-04 02:39:22
其不工作:(餅圖即時嘗試顯示仍然顯示幾乎所有累計 – 2012-01-04 02:53:02
然後我會指出一個問題與餅圖,而不是SQL或循環,我還沒有餅圖功能的知識 – 2012-01-04 03:03:34