По значениям from и took таблицы request получаю ники из таблицы users
Код: Выделить всё
$query = "SELECT `treq`.`uid`, `tusr1`.`nick` AS `from`, `tusr2`.`nick` AS `took`
FROM `requests` AS `treq`
LEFT JOIN `users` AS `tusr1` ON `tusr1`.`uid` = `treq`.`from`
LEFT JOIN `users` AS `tusr2` ON `tusr2`.`uid` = `treq`.`took`
WHERE `treq`.`uid` = '".$_GET['rid']."'";
$result = mysql_query($query) or die ("ERROR: ".mysql_error());
if (mysql_num_rows($result) > 0) {
while ($row = mysql_fetch_array($result)) {
// Берём результаты из каждой строки
echo $row['from'].'~'.$row['took'];
}
}
Понадобилось вывести из таблицы request значение поля from как оно есть (число) и также по нему вывести ник из таблицы users
Код: Выделить всё
$query = "SELECT `treq`.`uid`,`treq1`.`from` AS `id`, `tusr1`.`nick` AS `from`, `tusr2`.`nick` AS `took`
FROM `requests` AS `treq`
LEFT JOIN `requests` AS `treq1` ON `treq1`.`uid` = `treq`.`from`
LEFT JOIN `users` AS `tusr1` ON `tusr1`.`uid` = `treq`.`from`
LEFT JOIN `users` AS `tusr2` ON `tusr2`.`uid` = `treq`.`took`
WHERE `treq`.`uid` = '".$_GET['rid']."'";
$result = mysql_query($query) or die ("ERROR: ".mysql_error());
if (mysql_num_rows($result) > 0) {
while ($row = mysql_fetch_array($result)) {
// Берём результаты из каждой строки
echo $row['id'].'~'.$row['from'].'~'.$row['took'];
}
}
From пытаюсь вывести как есть вот так
[php]`treq1`.`from` AS `id`
LEFT JOIN `requests` AS `treq1` ON `treq1`.`uid` = `treq`.`from`[/php]
Решено
Код: Выделить всё
$query = "SELECT `treq`.`uid`, `treq`.`from`, `tusr1`.`nick` AS `nick_from`, `tusr2`.`nick` AS `took`
FROM `requests` AS `treq`
LEFT JOIN `users` AS `tusr1` ON `tusr1`.`uid` = `treq`.`from`
LEFT JOIN `users` AS `tusr2` ON `tusr2`.`uid` = `treq`.`took`
WHERE `treq`.`uid` = '".$_GET['rid']."'";
$result = mysql_query($query) or die ("ERROR: ".mysql_error());
if (mysql_num_rows($result) > 0) {
while ($row = mysql_fetch_array($result)) {
// Берём результаты из каждой строки
echo $row['from'].'~'.$row['nick_from'].'~'.$row['took'];
}
}