为什么以下代码中有echo的语句都提示syntax error
while($row=mysql_fetch_array($result)){
$date=$row["出生日期"];
$timeTemp=strtotime($date);
$time=date("Y-n-j",$timeTemp);
//这里提示错误
echo "<tr><td class=STYLE2>$row['学号']</td>";
echo "<td class=STYLE2>$row['姓名']</td>";
echo "<td class=STYLE2>$row['专业名']</td></tr>";
//以上三个都提示错误
if(ord($row['性别']==1)) echo "<td class=STYLE2>男</td>";
else echo "<td class=STYLE2>女</td>";
echo "<td class=STYLE2>$time</td>";
echo "<td class=STYLE2>$row['总学分']</td>";
if($row["照片"]){
?>
<td class=STYLE2><a href="lookPicture.php?id=<?php echo $row['学号'];?>">查看</a></td>
<?php
}
else echo "<td class=STYLE2>暂无</td>";
if($row['备注'])
echo "<td class=STYLE2>$row['备注']</td>";
else
//这里也提示错误
echo "<td class=STYLE2>无</td>";
echo "</tr>";
}
echo "</table>";
echo "<div align=center class=STYLE2>";
if($page>=2){
?>
while($row=mysql_fetch_array($result)){
$date=$row["出生日期"];
$timeTemp=strtotime($date);
$time=date("Y-n-j",$timeTemp);
//这里提示错误
echo "<tr><td class=STYLE2>$row['学号']</td>";
echo "<td class=STYLE2>$row['姓名']</td>";
echo "<td class=STYLE2>$row['专业名']</td></tr>";
//以上三个都提示错误
if(ord($row['性别']==1)) echo "<td class=STYLE2>男</td>";
else echo "<td class=STYLE2>女</td>";
echo "<td class=STYLE2>$time</td>";
echo "<td class=STYLE2>$row['总学分']</td>";
if($row["照片"]){
?>
<td class=STYLE2><a href="lookPicture.php?id=<?php echo $row['学号'];?>">查看</a></td>
<?php
}
else echo "<td class=STYLE2>暂无</td>";
if($row['备注'])
echo "<td class=STYLE2>$row['备注']</td>";
else
//这里也提示错误
echo "<td class=STYLE2>无</td>";
echo "</tr>";
}
echo "</table>";
echo "<div align=center class=STYLE2>";
if($page>=2){
?>
==>
{$row['学号']} 后面的也一样。if($page >=2 ){
while($row=mysql_fetch_array($result)){
$date=$row["出生日期"];
$timeTemp=strtotime($date);
$time=date("Y-n-j",$timeTemp);
echo "<tr><td class=STYLE2>{$row['学号']}</td>";
echo "<td class=STYLE2>{$row['姓名']}</td>";
echo "<td class=STYLE2>{$row['专业名']}</td></tr>";
if(ord($row['性别']==1)) echo "<td class=STYLE2>男</td>";
else echo "<td class=STYLE2>女</td>";
echo "<td class=STYLE2>$time</td>";
echo "<td class=STYLE2>{$row['总学分']}</td>";
if($row["照片"]){
?>
<td class=STYLE2><a href="lookPicture.php?id=<?php echo $row['学号'];?>">查看</a></td>
<?php
}
else echo "<td class=STYLE2>暂无</td>";
if($row['备注'])
echo "<td class=STYLE2>{$row['备注']}</td>";
else
echo "<td class=STYLE2>无</td>";
echo "</tr>";
}
echo "</table>";
echo "<div align=center class=STYLE2>";
if($page>=2){
?>
<a href="StuSearch.php?page=<?php echo "$page-1"; ?>&StuNu=<?if(!$page) echo $StuNumber; else echo $StuNu;?>&StuNa=<? if(!$page) echo $StuName;else echo $StuNa;?>&Pro=<? if(!$page) echo $Project; else echo $Pro;?>"
<? echo $i?></a>
<?php
}
for($i=1;$i<$pages;$i++){
if($page==$i) echo $i;
else {
?><a
href="StuSearch.php?page=<?php echo "$i"; ?>&StuNu=<?if(!$page) echo $StuNumber; else echo $StuNu;?>&StuNa=<? if(!$page) echo $StuName;else echo $StuNa;?>&Pro=<? if(!$page) echo $Project; else echo $Pro;?>"
<? echo $i?></a>
<?php
}
if($page<$pages){
?>
<a
href="StuSearch.php?page=<?php echo "$page+1";?>&StuNu=<?if(!$page) echo $StuNumber; else echo $StuNu;?>&StuNa=<? if(!$page)echo $StuName;else echo $StuNa;?>&Pro=<? if(!$page) echo $Project; else echo $Pro;?>">下一页</a>
<?php
}
echo "共(".$pages.")页</div>"; }
mysql_close();
?>
显示Parse error: syntax error, unexpected $end in E:\Appserve\wamp\www\XsKc\StuQuery.php on line 140
其中line 140为结尾
这是为什么啊