跪求解答，while嵌套if语句问题

<?php
  $radio_buttons = 0;
//读取一个存放地址的表
  $addresses_query = tep_db_query("select address_book_id, entry_firstname as firstname, entry_lastname as lastname, entry_company as company, entry_street_address as street_address, entry_suburb as suburb, entry_city as city, entry_postcode as postcode, entry_state as state, entry_zone_id as zone_id, entry_country_id as country_id from " . TABLE_ADDRESS_BOOK . " where customers_id = '" . (int)$customer_id . "'");
  while ($addresses = tep_db_fetch_array($addresses_query)) { //tep_db_fetch_array() 调用了mysql_fetch_array()函数
  //判断 如果表的ID等于5 就打印“显示”否则“隐藏”
  if($addresses['address_book_id'] == 5){
  echo '显示'
?>
<table style="display:none;" id="electAddress" name="electAddress" border="0" width="100%" cellspacing="0" cellpadding="2">
<?php
  }
  else {
 
  echo '隐藏'
   
 
<?php 
  }
   
  }
?>
</table>



为什么结果是 隐藏 隐藏 隐藏 隐藏 显示

应该是什么结果？

我想要的是 只显示 “显示”隐藏不显示出来

看不明白，你还不如这样写：

PHP code

$addresses_query = tep_db_query("select * from " . TABLE_ADDRESS_BOOK . " where customers_id ='{$customer_id}' and ");
while ($addresses = tep_db_fetch_array($addresses_query)) 
{ 
    if($addresses['address_book_id'] == 5)
   {
       $display="block"
   }
   else
   {
      $display="none"
  }

echo "<table style=\"display:{$display};\"......";//

}

写的有些错误。。。自己改

把else那块去掉就行了.

我代码简化了 还有个按钮 可以把隐藏的显示的来的

点击按钮控制显示隐藏，请用js

呵呵 这也是个办法

还有其他的办法没

还有就是 5 之后的全部隐藏了 5 之前的就没隐藏掉 这是怎么回事呢