创建一个临时表并从中选择

当我运行此命令时，出现错误“未声明的变量：temp”...

<?php 

$maketemp = "CREATE TEMPORARY TABLE temp(`itineraryId` int NOT NULL, `live` varchar(1), `shipCode` varchar(10), `description` text, `duration` varchar(10), PRIMARY KEY(itineraryId))"; 

mysql_query( $maketemp, $connection ) or die ( "Sql error : " . mysql_error ( ) );

$inserttemp = "SELECT live, id AS itineraryId, ship AS shipCode, description AS description, duration AS length FROM cruises WHERE live ='Y' INTO temp";

mysql_query( $inserttemp, $connection ) or die ( "Sql error : " . mysql_error ( ) );

$select = "SELECT intineraryId, shipCode, description, duration FROM temp";

$export = mysql_query ( $select, $connection ) or die ( "Sql error : " . mysql_error( ) );

有任何想法吗 ？

请不要使用mysql_*新代码中的函数 https://stackoverflow.com/q/12859942/。它们不再被维护并已被正式弃用 https://wiki.php.net/rfc/mysql_deprecation。请参阅red box https://www.php.net/manual/en/function.mysql-connect.php？学习关于准备好的陈述 https://en.wikipedia.org/wiki/Prepared_statement相反，并使用PDO http://php.net/pdo or MySQLi http://php.net/mysqli - 本文 https://php.net/manual/en/mysqlinfo.api.choosing.php将帮助您决定哪个。如果您选择PDO，这是一个很好的教程 http://wiki.hashphp.org/PDO_Tutorial_for_MySQL_Developers.

这段代码应该可以工作：

<?php

  $maketemp = "
    CREATE TEMPORARY TABLE temp_table_1 (
      `itineraryId` int NOT NULL,
      `live` varchar(1),
      `shipCode` varchar(10),
      `description` text,
      `duration` varchar(10),
      PRIMARY KEY(itineraryId)
    )
  "; 

  mysql_query($maketemp, $connection) or die ("Sql error : ".mysql_error());

  $inserttemp = "
    INSERT INTO temp_table_1
      (`itineraryId`, `live`, `shipCode`, `description`, `duration`)
    SELECT `id`, `live`, `ship`, `description`, `duration`
    FROM `cruises`
    WHERE `live` = 'Y'
  ";

  mysql_query($inserttemp, $connection) or die ("Sql error : ".mysql_error());

  $select = "
    SELECT `itineraryId`, `shipCode`, `description`, `duration`
    FROM temp_table_1
  ";
  $export = mysql_query($select, $connection) or die ("Sql error : ".mysql_error());

我猜你会用临时表做更多的事情，或者只是玩弄它，但如果不知道整个代码可以总结为：

<?php

  $query = "
    SELECT `id` AS 'itineraryId', `ship`, `description`, `duration`
    FROM `cruises`
    WHERE `live` = 'Y'
  ";
  $export = mysql_query($query, $connection) or die ("Sql error : ".mysql_error());