显示从MySQL查询返回的记录

问题描述

| 我无法显示发送给用户的消息。 假设用户是我的表“ messages”中的id列，那么我将如何显示发送给该用户的消息？

session_start();
require \"bidcon.PHP\";
$userfinal=$_SESSION[\'username\'];
$id = MysqL_query(\"SELECT id FROM searchengine\") or die(MysqL_error());

// get the messages from the table.
$get_messages = MysqL_query(
   \"SELECT messages_id FROM messages
   WHERE to_user=\'$id\' ORDER BY messages_id DESC\")
  or die(MysqL_error());

$get_messages2 = MysqL_query(        
    \"SELECT * FROM messages        
    WHERE to_user=\'$id\' ORDER BY messages_id DESC\")      
  or die(MysqL_error());

$num_messages = MysqL_num_rows($get_messages);
// display each message title,with a link to their content
echo \'<ul>\';

for($count = 1; $count <= $num_messages; $count++)
{   
    $row = MysqL_fetch_array($get_messages2);

    

解决方法

        这个：

$id = mysql_query(\"SELECT id FROM searchengine\");

将使“ 2”成为MySQL资源，而不是您要查找的整数/字符串。 而是这样做：

$res = mysql_query(\"SELECT id FROM searchengine\");
$id = mysql_result($res,0);

    ,        我不清楚您为什么需要两个不同的查询。但是您的主要问题是如何显示记录。您可以在while循环内进行操作，如下所示：

$rs = mysql_query(\"SELECT id FROM searchengine\") or die(mysql_error());
$rec = mysql_fetch_assoc($rs);
$id = $rec[\'id\'];

// get the messages from the table.
$get_messages = mysql_query(
    \"SELECT * FROM messages
    WHERE to_user=\'$id\' ORDER BY messages_id DESC\")
  or die(mysql_error());
$num_messages = mysql_num_rows($get_messages);

// display each message title,with a link to their content
echo \'<ul>\';
while($row=mysql_fetch_assoc($get_messages)){
    // do stuff with each row
    echo \"<li>\".$row[\'somefield\'].\"</li>\";
}
echo \'</ul>\';