我一直試圖根據他們的empid
輸出我的數據庫字段,但我不知道怎麼辦。它給了我這個錯誤..根據id添加數據庫字段php
警告:mysql_fetch_array()預計參數1是資源,布爾在C中給出:_webhost \ Apache24 \ htdocs中\ EIS \上線usercp.inc.php 16
<?php
$firstname = getuserfield('txtFname');
$lastname = getuserfield('txtLname');
echo 'Hello '.$firstname.' '.$lastname.'.';
$empid = getuserfield('empid');
$query = "SELECT type_of_leave,specific_reason,date_from,date_to,num_of_days FROM `hrf_leave` WHERE `empid` = '$empid' AND `formStatus` = 0";
$query_run = mysql_query($query);
echo "<table border=1>
<tr>
<th>Type of Leave</th>
<th>Specific Reason</th>
<th>Date From</th>
<th>Date To</th>
<th>Number of Days</th>
</tr>";
while($record = mysql_fetch_array($query_run)){ // line 16
echo "<tr>";
echo "<td>" . $record['type_of_leave'] . "</td>";
echo "<td>" . $record['specific_reason'] . "</td>";
echo "<td>" . $record['date_from'] . "</td>";
echo "<td>" . $record['date_to'] . "</td>";
echo "<td>" . $record['num_of_days'] . "</td>";
echo "</tr>";
}
echo "</table>";
?>
在mysql_query之後放置或'die(mysql_error())'; – artragis
我試過了。我發現這個問題,我做了一些修改後忘了保存mysql表。但它仍然不會輸出我想要的結果。 – Wax