Question

我正在尝试使用以下代码。我的目标是显示一个id ='$ id'的页面。但是，如果$ id的值为空，我想在我的数据库中显示最新的条目。

<?php
mysql_connect("localhost", "root", "password");
mysql_select_db("database");
if(!(isset($_GET['id']))){ 
     $query = mysql_query("SELECT * FROM mytable WHERE name='somename' ORDER BY id DESC LIMIT 1");
      while($var1 = mysql_fetch_assoc($query)){
      $y = $var1['id'];
      }
}
mysql_data_seek($query, 0);

$id = (isset($_GET['id'])) ? (int)$_GET['id'] : '$y';
$query1 = mysql_query("SELECT * FROM mytable WHERE id='$id'");
while($query_row = mysql_fetch_assoc($query1)){
//some code 
}
?>

问题是当$ id没有值时，首先执行循环。但第二个while循环不起作用。有人请告诉我一个解决方案

Answer 1

尝试删除第一个if语句

周围的括号

if(!isset($_GET['id'])){ 
     $query = mysql_query("SELECT * FROM mytable WHERE name='somename' ORDER BY id DESC LIMIT 1");
      while($var1 = mysql_fetch_assoc($query)){
      $y = $var1['id'];
      }
}

Answer 2

试试此代码

<?php
    mysql_connect("localhost", "root", "password");
    mysql_select_db("database");

    $query = mysql_query("SELECT * FROM mytable WHERE name='somename' ORDER BY id DESC LIMIT 1");
    $var1 = mysql_fetch_assoc($query); 
    $y = $var1['id'];

    $id = (isset($_GET['id'])) ? $_GET['id'] : $y ;
    $query1 = mysql_query("SELECT * FROM mytable WHERE id='$id'");
    while($query_row = mysql_fetch_assoc($query1)){
     //some code 
    }
?>

Answer 3

尝试

$query1 = mysql_query("SELECT * FROM mytable WHERE id='".$id."'");

而不是

$query1 = mysql_query("SELECT * FROM mytable WHERE id='$id'");

此外，清理您的数据以避免SQL注入攻击并使用PDO。

如何在单个代码中使用两个mysql_query和两个mysql_fetch_assoc

3 个答案: