如何從php的mysql查詢中排除列？

Question

我有一個簡單的查詢（在mySQL視圖中），php使用它來創建表：

select `swtickets`.`ownerstaffid` AS `ID`,
`swtickets`.`ownerstaffname` AS `Owner`,
sum(if((`swtickets`.`ticketstatusid` = 2),1,0)) AS `In Progress`,
sum(if((`swtickets`.`ticketstatusid` = 4),1,0)) AS `Pending Customer`,
sum(if((`swtickets`.`ticketstatusid` = 5),1,0)) AS `Customer Replied`,
sum(if((`swtickets`.`ticketstatusid` = 9),1,0)) AS `Suggested Resolution`,
sum(if((`swtickets`.`priorityid` in (3,6) and `swtickets`.`ticketstatusid` in     (2,4,5,9)),1,0)) AS `High/Critical`,
sum(if((`swtickets`.`ticketstatusid` in (2,4,5,9)),1,0)) AS `Total Workable` 
from `swtickets` where ((`swtickets`.`departmentid` = 14) 
and ownerstaffid in (select staffid from swstaff where staffgroupid=4 and isenabled =  1)
and (`swtickets`.`ownerstaffname` <> '')) group by `swtickets`.`ownerstaffname`

我在while循環中調用數據，但似乎無法擺脫“ ID”列，我需要調用它，因為它用作“所有者”列的超鏈接的目標。 理想的標頭是：

業主| 進行中| 待定客戶| 客戶回復| 建議的分辨率| 總可行

的PHP：

 $ticketloadquery = mysql_query("SELECT * from open_tickets;")
or die(mysql_error()); 

$ticketloadfield_num = mysql_num_fields($ticketloadquery);

echo "<div><h1>Work Load</h1>";
if(mysql_num_rows($ticketloadquery)==0) {
echo "<i>There are no currently unassigned tickets!</i>";
}
else{
echo "<table border='1'><tr>";
for($i=0; $i<$ticketloadfield_num; $i++)
{
$ticketloadfield = mysql_fetch_field($ticketloadquery);
echo "<td>{$ticketloadfield->name}</td>";
}
echo "</tr>\n";

while($ticketloadinfo = mysql_fetch_array( $ticketloadquery )) 
{ 

echo "<tr>"; 
echo "<td><a href='https://support.mysite.com/staff/dashboard.php?   id=".$ticketloadinfo['ID']."', target='_blank'>".$ticketloadinfo['Owner']."</a></td> "; 
echo "<td>".$ticketloadinfo['Owner'] . "</td> "; 
echo "<td>".$ticketloadinfo['In Progress'] . "</td> "; 
echo "<td>".$ticketloadinfo['Pending Customer'] . " </td>"; 
echo "<td>".$ticketloadinfo['Customer Replied'] . " </td>";
echo "<td>".$ticketloadinfo['Suggested Resolution'] . " </td>"; 

echo "<td>".$ticketloadinfo['Total Workable'] . " </td></tr>"; 
} 
echo "</div>";
echo "</table>"; 
}

有什么建議么？

Answer 1

刪除任一

echo "<td><a href='https://support.mysite.com/staff/dashboard.php?   id=".$ticketloadinfo['ID']."', target='_blank'>".$ticketloadinfo['Owner']."</a></td> "; 

要么

echo "<td>".$ticketloadinfo['Owner'] . "</td> "; 

Answer 2

擺脫只有所有者的單元格，因為它與先前的單元格相同，只有超鏈接。

Answer 3

即使您現在設法對ID列進行例外處理，這種方法也會在將來導致問題，因為您是根據標題的查詢自動生成列，然后手動添加所需信息的列顯示。

有兩種選擇：

1. 您會自動生成所有信息，以便標題和內容始終保持同步。 不建議您這樣做，因為您在顯示信息方面沒有任何靈活性
2. 也手動添加標題行。 這樣，您可以顯示所需的任何內容，並且可以使用簡單的html塊來實現。

作為一個旁注，我還建議在標題行中使用th元素而不是td元素。

因此，您將替換為：

for($i=0; $i<$ticketloadfield_num; $i++)
{
$ticketloadfield = mysql_fetch_field($ticketloadquery);
echo "<td>{$ticketloadfield->name}</td>";
}

與：

?>
<th>Owner</th>
<th>In Progress</th>
<th>Pending Customer</th>
<th>Customer Replied</th>
<th>Suggested Resolution</th>
<th>Total Workable</th>
<?php