(求助)Mysql库表存在,且内有记录,用PHP为什么显示不出 if ($result) { $result在哪儿定义的 解决方案 » 免费领取超大流量手机卡,每月29元包185G流量+100分钟通话, 中国电信官方发货 哦,不好意思啊。是没有对Sresult赋值但是它还是没有查到记录,因为“无上载记录”是在前面就有显示的。再帮我看看吧,谢谢。 修改一下//执行查询SQL语句 $$result =@mysql_query($selectquery, $link); $count1=@mysql_num_rows($result); if($count1<1) { echo( '无上载记录'); //关闭数据库 } elseif ($result) { echo "Found these entries in the database:<br><p></p>"; echo "<table width=90% align=center border=1><tr> <td align=center bgcolor=#00FFFF>User Name</td> <td align=center bgcolor=#00FFFF>Last Name</td> </tr>"; while ($r = mysql_fetch_array($result)) { $idx = $r["name"]; $user = $r["messages"]; echo "<tr> <td>$idx</td> <td>$user</td> </tr>"; } echo "</table>"; } else { echo "No data."; } 参考如下:$conn=mysql_connect("localhost","root","");mysql_select_db("kmbest");$query=mysql_query("select * from messages",$conn);echo "<table>";while($value=mysql_fetch_array($query){ echo "<tr>"; echo "<td>".$value["column1"]."</td><td>".$value ["column2"]."</td>"; echo "</tr>";}echo "</table>"; 多谢各位了,可是还是:无上载记录。改成下面这面还是不行。<?php$conn=mysql_connect("localhost","root","");mysql_select_db("kmbest");$query=mysql_query("select * from messages",$conn);$result =@mysql_query($query, $conn); $count1=@mysql_num_rows($result); if($count1<1) { echo( '无上载记录'); //关闭数据库 } elseif ($result) { echo "Found these entries in the database:<br><p></p>"; echo "<table width=90% align=center border=1><tr> <td align=center bgcolor=#00FFFF>User Name</td> <td align=center bgcolor=#00FFFF>Last Name</td> </tr>"; while ($r = mysql_fetch_array($result)) { $idx = $r["name"]; $user = $r["messages"]; echo "<tr> <td>$idx</td> <td>$user</td> </tr>"; } echo "</table>"; } else { echo "No data."; }?> $query=mysql_query("select * from messages",$conn);$result =@mysql_query($query, $conn); 用一句就行!同一功能!用这一句替换上两句:$result=mysql_query("select * from messages",$conn); 更标准的写法$sql="select * from messages";$query=mysql_query($sql)or die(mysql_error); web根目录里有两个index.html 请高手指点 PHP分页问题 怎样在Zend Studio发布项目? sqlsrv_query() expects parameter 1 to be resource, boolean given in 急啊!急啊~~!ChartDirector能不能画拓扑图! 在一个新站上配置别人写好的代码,该先看哪部分 新手求页面传值的代码,在线等! 帮人做的第一个php站,看看行吗? 如何php+mysql实现身份验证(散分。。。。) 去者给分! 我准备在单机上装个开发系统,要装什么东西? 我对网站一翘不通,如何跟客户谈?
但是它还是没有查到记录,因为“无上载记录”是在前面就有显示的。
再帮我看看吧,谢谢。
//执行查询SQL语句
$$result =@mysql_query($selectquery, $link);
$count1=@mysql_num_rows($result);
if($count1<1) { echo( '无上载记录');
//关闭数据库
}
elseif ($result) {
echo "Found these entries in the database:<br><p></p>";
echo "<table width=90% align=center border=1><tr>
<td align=center bgcolor=#00FFFF>User Name</td>
<td align=center bgcolor=#00FFFF>Last Name</td>
</tr>"; while ($r = mysql_fetch_array($result))
{
$idx = $r["name"];
$user = $r["messages"]; echo "<tr>
<td>$idx</td>
<td>$user</td>
</tr>";
}
echo "</table>";
}
else
{
echo "No data.";
}
$conn=mysql_connect("localhost","root","");
mysql_select_db("kmbest");
$query=mysql_query("select * from messages",$conn);
echo "<table>";
while($value=mysql_fetch_array($query)
{
echo "<tr>";
echo "<td>".$value["column1"]."</td><td>".$value ["column2"]."</td>";
echo "</tr>";
}
echo "</table>";
改成下面这面还是不行。<?php
$conn=mysql_connect("localhost","root","");
mysql_select_db("kmbest");
$query=mysql_query("select * from messages",$conn);
$result =@mysql_query($query, $conn);
$count1=@mysql_num_rows($result);
if($count1<1) { echo( '无上载记录');
//关闭数据库
}
elseif ($result) {
echo "Found these entries in the database:<br><p></p>";
echo "<table width=90% align=center border=1><tr>
<td align=center bgcolor=#00FFFF>User Name</td>
<td align=center bgcolor=#00FFFF>Last Name</td>
</tr>"; while ($r = mysql_fetch_array($result))
{
$idx = $r["name"];
$user = $r["messages"]; echo "<tr>
<td>$idx</td>
<td>$user</td>
</tr>";
}
echo "</table>";
}
else
{
echo "No data.";
}?>
$result =@mysql_query($query, $conn); 用一句就行!同一功能!用这一句替换上两句:
$result=mysql_query("select * from messages",$conn);
$sql="select * from messages";
$query=mysql_query($sql)or die(mysql_error);