1

我有下表:

在此处输入图像描述

在查询中,我使用 group_concat 获取错误 ID。我想为每个错误 ID 提供单独的链接。例如:要给出的链接是“show_bug.cgi?id =3743200”

以下是我使用过的代码:

 $query = "select count(cbm.bug_id) as count,(select     concat(round((count(cbm.bug_id)/(select count(*) from techzilla.category_bug_map cbm,techzilla.bugs b where b.assigned_to =$userId  and cbm.bug_id=b.bug_id) * 100 ),2),'%')) as Percentage ,group_concat(b.bug_id separator ',') as BugIds from techzilla.bugs b left join techzilla.category_bug_map cbm on cbm.bug_id = b.bug_id where b.assigned_to = $userId and b.creation_ts >= '$fromDate 00:00:00' and b.creation_ts <= '$toDate 00:00:00' and cbm.os IN ('$opess')";
   $result = mysql_query($query) or die ("Bad query: " . mysql_error() );
   while($row = mysql_fetch_row($result)) {
   echo "<tr><td>$opess</td>
            <td>$row[0]</td>
            <td>$row[1]</td>
            <td>$row[2]</td></tr>";

   }

我怎样才能做到这一点?

4

3 回答 3

2
$bugIdString="3743121,3743125,3743126,3743193";

$bugIdArray=explode(",",$bugIdString);

foreach($bugIdArray as $bug_id){
    echo "<a href='show_bug.cgi?id=$bug_id'>$bug_id</a>, ";
}
于 2013-06-06T07:37:30.383 回答
1

您可以使用爆炸功能。

$ids = explode(',', $bugs);
foreach($bugs as $b) {
    echo '<a href="...">'.$b.'</a>';  
}
于 2013-06-06T07:38:53.313 回答
1

尝试使用此示例中的代码:

// here gets //selects the row with BUG IDs
$bug_id = '3743200,3743234,3743212';
$arrids = explode(',', $bug_id);
$links = '';
for($i=0; $i<count($arrids); $i++) {
  $links .= '<br/><a href="show_bug.cgi?id='.trim($arrids[$i]).'">'.$arrids[$i].'</a>';
}
echo $links;
于 2013-06-06T07:39:43.660 回答