PHP问题，很简单的。。

<?php
$sql='select table_name from information_schema.tables where table_name like "%efolio%" ';
$alltbl = mysql_query($sql);
echo '<form method="post" action="search_admin_database.php">' . "\n\n";
echo "<h3>List of Tables in Database</h3>"
. '<input type="text" name="searname" />' . "\n"
. '<input type="submit" name="submit" value="Search" />' . "\n";
echo '<table>'
.'<tr bgcolor="#CCFF99">'
.'<td width="300"><strong>TableName</strong></td>'
.'<td><strong>Export</strong></td>'
.'</tr>';
while ($result=mysql_fetch_array($alltbl)){
$tname = $result[0];
echo '<

在<input type="submit" name="submit" value="Search" />中加上

<input type="submit" name="submit" value="Search" onclick="check">
然后再代码开始加个java脚本
<script language="javascript">
function check(){
if(document.form.searname.value==""){////////////////这里要求你的表单名称叫form////////////
alert('名称不能为空');
return fasle;
}
}
</script>

在search_admin_database.php页面里写
<?php
if(!empty($_POST['searname'])){
print "<script>alert('searname不能为空');history.back();</script>";
}
?>
这样可以了么？

<?php
if($_POST[searname]){
print"<script>alert('用户名不能为空');</script>";
}
else{
$sql='select table_name from information_schema.tables where table_name like "%efolio%" ';