使用php交叉製表查詢

Question

我試圖想出一種有效的方式來顯示一些數據，但沒有成功。

該表是動態地從幾桌建，以顯示與填充顯示結果通過網站如列標題的行：

Site name | Column A | Column B | Column C => column headers continue from DB query 
Site A | Result A | Result B | Result C 
Site B | Result C | Result B | Result C 
Site C | Result C | Result B | Result C 

Results keep going vertically. 

下面是我使用

$risk_section=$row_risk_category['risksectid']; 
    mysql_select_db($database_auditing, $auditing); 
$qry_selfsites = sprintf("SELECT 
tblself.siteid AS selfsite, 
tblsite.sitename AS sitename, 
tblsite.address AS address, 
tblpct.pctname AS pctname, 
    tblresultsnew.total, 
    tblresultsnew.auditid AS auditid, 
    tblriskcategories.risksectid AS risksectid, 
    tblriskcategories.risksection AS risksection 
FROM tblself 
LEFT JOIN tblresultsnew ON tblself.auditid = tblresultsnew.auditid 
    LEFT JOIN tblsite ON tblself.siteid = tblsite.siteid 
    LEFT JOIN tblpct ON tblsite.pctid = tblpct.pctid 
    LEFT JOIN tblriskcategories ON 
tblresultsnew.risksectid=tblriskcategories.risksectid 
WHERE tblsite.pctid IN (SELECT pctid FROM tblreportpcts WHERE 
pctreportid='$pctreportid') 
AND tblsite.sitetypeid IN (SELECT sitetypeid FROM tblreportsites WHERE 
pctreportid='$pctreportid') 
    AND tblself.year = %s 
    ORDER BY tblsite.pctid,tblsite.sitename", 
GetSQLValueString($yearofrpt, "int")); 
$selfsites = mysql_query($qry_selfsites, $auditing) or die(mysql_error()); 
$totalRows_selfsites = mysql_num_rows($selfsites); 

查詢

所以問題是，有沒有辦法從上述查詢（或其修改後的版本）中獲取數據來動態構建表格，並使所有結果正確排列？到目前爲止，我只能讓它垂直構建。

對不起，編輯時間（已與下面的答案的工作，我意識到我沒有措辭得到了很好的問題）

我想要得到的是從查詢列標題的行（ tblriskcategories.risksection AS risksection）這些水平填充列名稱。

然後下方，網站名稱，顯示與對應於列標題以上即

<table> 
<tr> 
<th>Sitename</th> 
<?php while($row_selfsites=mysql_fetch_assoc($selfsites){ 
//loop through the section headers pulled from the DB (tblriskcategories) 
<th><?php echo $row_selfsites['risksection'];//show the section headers?></th> 
<?php } 
//end header loop then start another loop to show a row for each site pulled 
//out by the query and show the relevant results in the correct column 
while($row_selfsites=mysql_fetch_assoc($selfsites)) { 
//do the vertical drop matching the header rows with the sitenames from tblsite 
//and the results from tblresultsnew 
?> 
<tr> 
<td><?php echo $row_selfsites['sitename'];?></td> 
<td><?php echo $row_selfsites['total']; 
//these need to grow to fit the headers and each site?></td> 
<tr> 
<?php } //end displayed data loop?> 
</table> 

下相關的表結構的結果：
tblresultsnew resultsid，auditid，risksectid，總
tblriskcategories risksectid，risksection
tblself selfauditid，SITEID，auditid
tblsite SITEID，站點名

因此，tblself擁有我們需要的數據和相關auditid的站點列表，
tblresultsnew保存結果 - 總列 - 每個risksectid和每個auditid，例如，一個auditid可以具有大約8個risksectid，每個具有相應的總數
tblriskcategories擁有列標題
tblsite擁有網站數據，使其意味着
我希望這可以進一步解釋這個問題。

再次感謝您的幫助。
戴夫

Answer 1

$rows = array(
0 => array('headingA' => 'results1a', 'headingB' => 'results1b', ), 
1 => array('headingA' => 'results2a', 'headingB' => 'results2b', ), 
2 => array('headingA' => 'results3a', 'headingB' => 'results3b', ), 
); 
// $rows is spoofing a db result set as an array 
//var_dump($rows); 

$first = true ; 
$header_row = "START TABLE <br />" ; 
$data_rows = ""; 
foreach($rows as $row) { 
    foreach($row as $key=>$value){ 
    if($first === true){ 
    $header_row .= "$key | "; 
    } 
    $data_rows .= "$value |"; 
    } 
if($first) $header_row .="<br />"; 

$data_rows .= "<br />"; 
$first = false; 
} 
echo $header_row . $data_rows . "END TABLE"; 

給出：

START TABLE 
headingA | headingB | 
results1a |results1b | 
results2a |results2b | 
results3a |results3b | 
END TABLE 

是那個樣的解決方案，你是後？