Question

需要有关此错误的帮助。看不出有什么不对。

错误：

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

代码：

<?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>";
?>

Answer 1

你写了这个：

$view = mysql_query("SELECT * FROM bulletin ORDER BY pro_no DESC"); //Selecting the table from the DB
if (!mysql_query ($view, $con))

注意如何执行查询，将结果资源分配给$view，然后尝试使用$view作为SQL运行另一个查询？但是$view不是SQL，它是结果资源，因此是错误。

请写下：

$view = mysql_query("SELECT * FROM bulletin ORDER BY pro_no DESC");
if (!$view)

Answer 2

您无法查询查询结果。试试这个

$view = mysql_query("SELECT * FROM bulletin ORDER BY pro_no DESC"); //Selecting the table from the DB

if (!$result)

它应该做魔术

Answer 3

您实际上有两个查询。第二个查询在if语句中执行：

if (!mysql_query ($view, $con))

请注意，此处的$ view是resource类型，而不是string类型。如果要检查查询是否正确执行，只需写：

if(!$view)

mysql_query（）需要参数

3 个答案: