从tb_xmpcps 可以找到用户名$myrow2['xmpcps_user'],然后再通过用户名去tb_user中找到姓名,两个表嵌套查询还真是第一次做,劳大家指点指点!为什么中说$query2有错误?
<?php
//查找评分表中相应批次评审状态
$query2=mssql_query("select * from tb_xmpcps where xmpcps_pcid='$xmpc_id'");
$x=1;
while($myrow2=mssql_fetch_array($query2)){
?>
<tr>
<td height="22" align="center" bgcolor="#FFFFFF"><span class="STYLE1"><?php echo $x;?>.</span></td>
<?php //找到与用户名匹配的评委姓名
$query3=mssql_query("select * from tb_user where online_user={$myrow2['xmpcps_user']} limit 1");
$myrow3=mssql_fetch_array($query3);
?>
<?php
//查找评分表中相应批次评审状态
$query2=mssql_query("select * from tb_xmpcps where xmpcps_pcid='$xmpc_id'");
$x=1;
while($myrow2=mssql_fetch_array($query2)){
?>
<tr>
<td height="22" align="center" bgcolor="#FFFFFF"><span class="STYLE1"><?php echo $x;?>.</span></td>
<?php //找到与用户名匹配的评委姓名
$query3=mssql_query("select * from tb_user where online_user={$myrow2['xmpcps_user']} limit 1");
$myrow3=mssql_fetch_array($query3);
?>
SELECT B.* FROM tb_xmpcps AS A LEFT OUVER JOIN tb_user AS B ON A.xmpcps_user=B.online_user WHERE A.xmpcps_pcid='$xmpc_id'
$query2=mssql_query("SELECT B.* FROM tb_xmpcps AS A LEFT OUTER JOIN tb_user AS B ON A.xmpcps_user=B.online_user WHERE A.xmpcps_pcid='{$xmpc_id}'");
<?php
//查找评分表中相应批次评审状态
$query2=mssql_query("SELECT B.* FROM tb_xmpcps AS A LEFT OUTER JOIN tb_user AS B ON A.xmpcps");
$x=1;
while($myrow2=mssql_fetch_array($query2)){
?>
<tr>
<td height="22" align="center" bgcolor="#FFFFFF"><span class="STYLE1"><?php echo $x;?>.</span></td>
<td height="22" align="center" bgcolor="#FFFFFF" class="STYLE1"><?php echo $myrow2[online_name];?></td>
<td align="center" bgcolor="#FFFFFF" class="STYLE1"><?php echo $myrow2[xmpcps_user];?></td>
<td align="center" bgcolor="#FFFFFF"><span class="STYLE1"><?php echo $myrow2[xmpcps_state];?></span></td>
</tr>
<?php
$x++;
}
?>
看看都有什么
//查找评分表中相应批次评审状态
$query2=mssql_query("select * from tb_xmpcps where xmpcps_pcid='$xmpc_id'");
$x=1;
while($myrow2=mssql_fetch_assoc($query2)){
?>
<tr class="STYLE1">
<td height="22" align="center" bgcolor="#FFFFFF"><span class="STYLE1"><?php echo $x;?>.</span></td>
<?php //找到与用户名匹配的评委姓名
$query3=mssql_query("select * from tb_user where online_user='$myrow2[xmpcps_user]'");
$myrow3=mssql_fetch_assoc($query3);
?>