2017-03-31 57 views
-3

我想把我的SQL數據放到HTML數據中,這是我迄今爲止的代碼,但它現在還沒有工作。我試圖把HTML到我的PHP代碼,我想知道我是否應該這樣做的另一種方式圓(把PHP轉換成HTML),任何答覆不勝感激如何把sql數據放到html表中

while ($rowObj = $queryResult->fetch_object()) { 
echo "<tr>"; 
echo"<td>" . $AE_events ['eventTitle'] . "</td>; 
echo"<td>" . $AE_events ['eventDescription'] . "</td>; 
echo"<td;>" . $AE_events ['eventStartDate'] . "</td>; 
echo"<td>" . $AE_events ['eventEndDate'] . "</td>; 
echo"<td>" . $AE_events ['eventPrice'] . "</td>; 

    $AE_eventTitle = $rowObj->eventTitle; 
    $AE_eventDescription = $rowObj->eventDescription; 
    $AE_eventStartDate = $rowObj->eventStartDate; 
    $AE_eventEndDate = $rowObj->eventEndDate; 
    $AE_eventPrice = $rowObj->eventPrice; 
    echo "<div> $AE_eventTitle<br> $AE_eventDescription<br> 
$AE_eventStartDate<br> 
$AE_eventEndDate<br> $AE_eventPrice<br> </div>"; 

} 
+1

語法錯誤,解析 –

+0

你缺少一個' 「'每一個'' – Jerodev

+0

'回聲後」​​「$ AE_events [ 'EVENTTITLE']錯誤。 「;'First error –

回答

0

如果你想畫一個表出來的數據在PHP中,你可以使用嵌套循環。第一個循環管理行並且嵌套管理列。例如:

$data = [ 
    [1, 3, 5], 
    [2, 4, 6] 
]; 
print "<table>"; 
for ($i = 0; $i < count($data); $i++) { 
    $row = $data[$i]; 
    print "<tr>"; 
    for ($k = 0; $k < count($row); $k++) { 
     print "<td>{$row[$k]}</td>"; 
    } 
    print "</tr>"; 
} 
print "</table>"; 

輸出:

<table> 
    <tr> 
     <td>1</td> 
     <td>3</td> 
     <td>5</td> 
    </tr> 
    <tr> 
     <td>2</td> 
     <td>4</td> 
     <td>6</td> 
    </tr> 
</table>