使用php交叉制表查询

Question

我试图想出一种有效的方式来显示一些数据，但没有成功。

该表是动态地从几桌建，以显示与填充显示结果通过网站如列标题的行：

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); 

查询

所以问题是，有没有办法从上述查询（或其修改后的版本）中获取数据来动态构建表格，并使所有结果正确排列？到目前为止，我只能让它垂直构建。

对不起，编辑时间（已与下面的答案的工作，我意识到我没有措辞得到了很好的问题）

我想要得到的是从查询列标题的行（ tblriskcategories.risksection AS risksection）这些水平填充列名称。

然后下方，网站名称，显示与对应于列标题以上即

<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> 

下相关的表结构的结果：
tblresultsnew resultsid，auditid，risksectid，总
tblriskcategories risksectid，risksection
tblself selfauditid，SITEID，auditid
tblsite SITEID，站点名

因此，tblself拥有我们需要的数据和相关auditid的站点列表，
tblresultsnew保存结果 - 总列 - 每个risksectid和每个auditid，例如，一个auditid可以具有大约8个risksectid，每个具有相应的总数
tblriskcategories拥有列标题
tblsite拥有网站数据，使其意味着
我希望这可以进一步解释这个问题。

再次感谢您的帮助。
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"; 

给出：

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

是那个样的解决方案，你是后？