2014-04-22 34 views
0

所以我有這個奇怪的問題,我有一堆值和列的表,當我在sql studio中運行我的查詢它的工作原理,但當我得到顯示我的表時,到日期列它不顯示它,它只是插入一個記錄,即使有很多現在我不知道爲什麼它這樣做的查詢工作。它爲此數據庫中任何表的任何日期列執行此操作。我使用MSSQL 2008與iis和php這裏是我的代碼。任何幫助非常感謝PHP mssql顯示列不工作

<?php 
echo "Door 1 Current State= Closed<br />"; 
echo "Door 2 Current State= Closed<br />"; 
echo "Door 3 Current State= Closed<br />"; 
echo "Door 4 Current State= Closed<br />"; 
echo "state1= Green State0 = Yellow state84=red stateA0 = Red state3=green<br />"; 
require('connection.php'); 


$query = "SELECT * FROM dbo.v_records"; 

$result = sqlsrv_query($conn, $query) or die (sqlsrv_errors()); 
if(sqlsrv_has_rows($result)) 
{ 
echo "<table border='1'> 
<tr> 
<th>ID</th> 
<th>Card</th> 
<th>Date</th> 
<th>Dorr</th> 
<th>Reader</th> 
<th>Name</th> 
<th>State</th> 
</tr>"; 
while($row = sqlsrv_fetch_array($result, SQLSRV_FETCH_ASSOC)) 
    { 
    echo "<tr>"; 
    echo "<td>" . $row['ID'] . "</td>"; 
    echo "<td>" . $row['Card'] . "</td>"; 
    echo "<td>" . $row['Date'] . "</td>"; 
    echo "<td>" . $row['ReaderN'] . "</td>"; 
    echo "<td>" . $row['ReaderNo'] . "</td>"; 
    echo "<td>" . $row['Name'] . "</td>"; 
    echo "<td>" . $row['Warn'] . "</td>"; 
    echo "</tr>"; 
    } 
echo "</table>"; 
} 

else 
{echo "No results were found.<br />";} 


/* Close the connection. */ 
sqlsrv_close($conn); 
?> 

我將此添加到循環SQLSRV_FETCH_ASSOC,但它沒有解決問題。我試圖創建一個視圖,並給列的別名,但也不起作用。

回答

0

改變你的循環是這樣的:

while($row = sqlsrv_fetch_array($result, SQLSRV_FETCH_ASSOC)) { 
//...