需要此錯誤的幫助。 看不到有什麼問題。mysql_query()期望參數
錯誤:
Warning: mysql_query() expects parameter 1 to be string, resource given in C:\wamp\www\pm website\php\bulletin_board\bulletin_board.php on line 48
代碼:
<?php
$con = mysql_connect("localhost","root",""); //Databse connection
if(!$con)
{
die ('Could not connect to DB' . mysql_error()); //Error promt
}
mysql_select_db("profound_master", $con); //Selecting the DB
$view = mysql_query("SELECT * FROM bulletin ORDER BY pro_no DESC"); //Selecting the table from the DB
if (!mysql_query ($view, $con))
{
die ('Error Sir' . mysql_error()); //Error promt
}
while ($row = mysql_fetch_array($view))
echo "<table width=\"1000\">";
echo "<tr id=\"boardLetter\">";
echo "<th width=\"46\">".$row['pro_no']."</th>";
echo "<th width=\"56\">".$row['date']."</th>";
echo "<th width=\"138\">".$row['project']."</th>";
echo "<th width=\"138\">".$row['task']."</th>";
echo "<th>".$row['originated']."</th>";
echo "<th>".$row['incharge']."</th>";
echo "<th>".$row['deadline']."</th>";
echo "<th width=\"139\">".$row['status']."</th>";
echo "<th width=\"151\">".$row['comment']."</th>";
echo "<th>".$row['din']."</th>";
echo "</tr>";
echo "</table>";
?>
非常感謝你先生,錯誤消失了!跟進先生問題先生。我看不到任何東西。該表必須與來自數據庫的數據一起出來。 :( –
非常感謝您的先生。現在我解決了這個問題。 –