我试图设置一个处理程序,所以当我得到0个结果时,它显示一条消息,说没有找到结果。我有这种工作,但我目前面临2个问题。PHP空处理
,当我得到0的结果确实显示发现了没有结果,但它也显示了以下错误:
Warning: mysqli_query() expects parameter 2 to be string, object given in /customers/b/6/d/herculesdjs.co.uk/FETCH.PHP on line 74
线74是这样的:
$fetch_row = mysqli_query($con,$result);
这是完整的脚本减连接信息。 $ City由下拉框定义。这已经过测试,工作正常:
$result = mysqli_query($con,
sprintf("SELECT * FROM `SouthYorkshire` WHERE `city_id` = '%s'",
preg_replace("[^0-9]","", $city)));
// Handle Null Results.
$fetch_row = mysqli_query($con,$result);
$numrows = $fetch_row[0];
if($numrows==0)
{
echo "<div id=\"NoResults\">";
echo "There are no clubs in this area, Why not come back another day. Know of one in this area? Why not submit it via Contact Us.";
echo "</div>";
}
while($row = mysqli_fetch_array($result))
{
echo "<div id=\"Results\">";
echo "<div id=\"theclub\">";
echo "<div class=\"ClubName\">";
echo $row['EstName'];
echo "</div><br>";
echo "<div class=\"Location\">";
echo $row['EstAddress2'];
echo "</div>";
echo "<br>";
echo "<div id=\"website\"><a href=\"#\"><img src=\"photos/more-info.png\" width=\"75\" height=\"25\"/></a> <a href=\"" . $row['EstWebsite'] ."\" target=\"_blank\"><img src=\"photos/visit-website-button.png\" width=\"75\" height=\"25\" /></div></a></div>";
echo "<br>";
}
echo "</div>";`
我遇到的另一个问题是,如果有结果不显示警告数据库。它显示找到的结果,但它也显示有没有俱乐部消息。
有什么建议吗?
可能是因为你已经宣布在''$结果= mysqli_query($ CON,...'$ con'和你使用它再次在'$ fetch_row = mysqli_query($ con,$ result);'删除一个。 –
看看你在做什么!你正在将'mysqli_query'的结果提供给'mysqli_query'。这就错了。 – deceze
正确的用户mysqli_query()':'mysqli_query($ con,$ query);'你正在向它传递结果集。 –