【问题标题】:get user id from petition and inserted in other table从请愿书中获取用户 ID 并插入到其他表中
【发布时间】:2013-02-13 03:19:41
【问题描述】:

帮助我如何从该表中获取用户 ID,例如请愿书 ID = 52,然后如果获取我想要插入到 SYSTEMID 中的第二个表(自动增量)的所有用户 ID

这是我的命令:

$result = $db->Execute("选择请愿 ID,MT4ACCOUNTID, phpfox_petition_text.petition_goal,user_id " . "来自 phpfox_petition_text,phpfox_tradeusers " . "其中 phpfox_petition_text.petition_goal = $mt4acctid 和 user_id=$systemid "); $symbol1 = $result->fields[0]; $symbol14 = $result->字段[3]; $result212 = $db->Execute("SELECT request_id, user_id " . "from phpfox_petition_sign " . "wherepetition_id=".$symbol1." and MT4ACCOUNTID=$mt4acctid ORDER BY time_stamp DESC"); $symbol122 = $result212->字段[1]; $result111 = $db->Execute("SELECT full_name " . "from phpfox_user " . "where user_id=$systemid "); $symbol1user = $result111->fields[0]; $记录 = 数组(); $记录[“符号”] = $符号; $record["CMD"] = $cmd; $record["VOLUME"] = $volume; $记录[“价格”] = $价格; $record["SLIPPAGE"] = $slippage; $record["STOPLOSS"] = $stoploss; $record["TAKEPROFIT"] = $takeprofit; $record["COMMENT"] = 'MyfaceFx>复制自 ::'.$symbol1user; $记录[“魔法”] = $魔法; $record["EXPIRATION"] = $expiration; $record["MT4SIGNALID"] = $signalid; $record["BROKERNAME"] = $brokername; $record["MT4ACCTID"] = $mt4acctid; $record["TIMESTAMP"] = time(); $record["SYSTEMID"] =$symbol122; $record["ORDERUNITS"] = $basecurrencyunits; $record["ACCTBALANCE"] = $acctbalance; $record["ACCTCURRENCY"] = $acctcurrency; $insertSQL = $db->AutoExecute('phpfox_tradesignals', $record, 'INSERT'); $insertsignalid = $db->Insert_ID();

它的工作,但只得到 1 个这样的结果:

过期:systemid ..................................................... ...... : 27

但我想得到这样的结果:

过期:systemid ..................................................... ...... : 27 : 29 : 26 : 34 等等....

表请愿

请愿ID:用户ID ..................................................... ...... 50 : 35 52 : 27 40 : 23 52 : 29 50 : 35 52 : 26 40 : 23 52 : 34 等等....

表系统ID

过期:系统ID ..................................................... ......

【问题讨论】:

  • 你的代码太乱了!!
  • 这个问题的状态让我头晕目眩
  • 你能安排一下代码吗?在每行末尾添加 2 个空格,然后按 enter 以显示 2 个分隔行。请阅读faq

标签: mysql database


【解决方案1】:

因此,如果您想将一个表中的值插入到另一个表中:

INSERT INTO table2 (systemid)
SELECT user_id
FROM   table1
WHERE  petition_id = 52

【讨论】:

    猜你喜欢
    • 1970-01-01
    • 2019-11-19
    • 1970-01-01
    • 1970-01-01
    • 2019-11-04
    • 2014-11-11
    • 1970-01-01
    • 2021-08-25
    • 1970-01-01
    相关资源
    最近更新 更多