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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。