我正在使用MaNGOS数据库。目前我已经拥有自己的身份，但我想使用官方一次。问题是我已经在我想要更新的数据库中获得了一些值。
我要更新的表格如下所示。
mysql> SELECT * FROM `character` LIMIT 10;
+----+---------------------------------+--------------+--------------+-------+
| id | char_name_de | char_name_en | id_new | id_old|
+----+---------------------------------+--------------+--------------+-------+
| 1 | | NULL | 2 | NULL |
| 2 | Abgesandter des Schattenhammers | NULL | 18201 | NULL |
| 3 | Abgesplittertes Skelett | NULL | 10478 | NULL |
| 4 | Adept aus Scholomance | NULL | 10469 | NULL |
| 5 | Aduscha | NULL | 18204 | NULL |
| 6 | Aggressiver Bluthund | NULL | 8922 | NULL |
| 7 | Akolyt aus Scholomance | NULL | 10471 | NULL |
| 8 | Akolyt der Schmetterschilde | NULL | 9045 | NULL |
| 9 | Althena | NULL | 18208 | NULL |
| 10 | Amadelis | NULL | 18209 | NULL |
+----+---------------------------------+--------------+--------------+-------+
10 rows in set (0.00 sec)
我想做的事情： 替换＆＃34; id＆＃34; by＆＃34; new_id＆＃34;
问题： ＆＃34; id_new＆＃34;不是某些多语言字符串的唯一原因。 我希望将它们匹配到一行，保存＆＃34; id＆＃34;进入＆＃34; id_old＆＃34;，替换＆＃34; id＆＃34; by＆＃34; id_new＆＃34;并删除＆＃34; id_new＆＃34;。
这看起来像这样：
mysql> SELECT *
-> FROM `character`
-> WHERE id_new IN(
-> SELECT id_new
-> FROM `character`
-> GROUP BY `character`.id_new
-> HAVING COUNT(`character`.id_new) > 1
-> )
-> ORDER BY id_new ASC;
+-----+--------------------------+--------------+--------------+-------+
| id | char_name_de | char_name_en | id_new | id_old|
+-----+--------------------------+--------------+--------------+-------+
| 258 | Natter | NULL | 2914 | NULL |
| 342 | Snake | NULL | 2914 | NULL |
| 290 | Rat | NULL | 4075 | NULL |
| 292 | Ratte | NULL | 4075 | NULL |
| 276 | Orakel der Hakkari | NULL | 11346 | NULL |
| 152 | Hakkari Oracle | NULL | 11346 | NULL |
| 148 | Gurubashi Warrior | NULL | 11355 | NULL |
| 201 | Krieger der Gurubashi | NULL | 11355 | NULL |
| 344 | Sohn von Hakkar | NULL | 11357 | NULL |
| 347 | Son of Hakkar | NULL | 11357 | NULL |
| 47 | Bloodseeker Bat | NULL | 11368 | NULL |
| 51 | Blutsucherfledermaus | NULL | 11368 | NULL |
| 560 | Molten Giant | NULL | 11658 | NULL |
| 123 | Geschmolzener Riese | NULL | 11658 | NULL |
| 545 | Flamewaker | NULL | 11661 | NULL |
| 93 | Feuerschuppe | NULL | 11661 | NULL |
| 546 | Flamewaker Priest | NULL | 11662 | NULL |
| 94 | Feuerschuppenpriester | NULL | 11662 | NULL |
| 553 | Lava Annihilator | NULL | 11665 | NULL |
| 214 | Lavavernichter | NULL | 11665 | NULL |
| 540 | Firelord | NULL | 11668 | NULL |
| 92 | Feuerlord | NULL | 11668 | NULL |
| 543 | Flame Imp | NULL | 11669 | NULL |
| 104 | Flammenwichtel | NULL | 11669 | NULL |
| 536 | Core Hound | NULL | 11671 | NULL |
| 194 | Kernhund | NULL | 11671 | NULL |
| 534 | Ancient Core Hound | NULL | 11673 | NULL |
| 384 | Uralter Kernhund | NULL | 11673 | NULL |
| 549 | Golemagg the Incinerator | NULL | 11988 | NULL |
| 129 | Golemagg der Verbrenner | NULL | 11988 | NULL |
| 558 | Majordomo Executus | NULL | 12018 | NULL |
| 240 | Majordomus Exekutus | NULL | 12018 | NULL |
| 554 | Lava Elemental | NULL | 12076 | NULL |
| 212 | Lavaelementar | NULL | 12076 | NULL |
| 564 | Sulfuron Harbinger | NULL | 12098 | NULL |
| 359 | Sulfuronherold | NULL | 12098 | NULL |
| 541 | Firesworn | NULL | 12099 | NULL |
| 90 | Feueranbeter | NULL | 12099 | NULL |
| 557 | Lava Surger | NULL | 12101 | NULL |
| 215 | Lavawoger | NULL | 12101 | NULL |
| 556 | Lava Spawn | NULL | 12265 | NULL |
| 211 | Lavabrut | NULL | 12265 | NULL |
| 297 | Razzashi Raptor | NULL | 14821 | NULL |
| 301 | Razzashiraptor | NULL | 14821 | NULL |
+-----+--------------------------+--------------+--------------+-------+
44 rows in set (0.23 sec)
如何解决： 这个Query可以做我想要的，但是有一个问题，MySQL不能更新你用于选择子查询的表。
UPDATE `character`
SET `character`.id_old = (
SELECT `character`.id
FROM `character`
GROUP BY `character`.id_new
HAVING COUNT(`character`.id_new) > 1
)
WHERE `character`.id_new IN (
SELECT `character`.id_new
FROM `character`
GROUP BY `character`.id_new
HAVING COUNT(`character`.id_new) > 1
);
所以我必须创建一个包含我需要的所有值的temp_table：
CREATE TABLE `tmp_character` (
`id` INT(11) NULL DEFAULT NULL,
`id_new` INT(11) NULL DEFAULT NULL
)
COLLATE='latin1_swedish_ci'
ENGINE=InnoDB
;
并填写值：
INSERT INTO tmp_character (id, id_new)
SELECT id, id_new
FROM `character`
WHERE id_new IN(
SELECT id_new
FROM `character`
GROUP BY `character`.id_new
HAVING COUNT(`character`.id_new) > 1
)
ORDER BY id_new ASC;
现在我收到了这个查询：
UPDATE `character`
SET `character`.id_old = (
SELECT tmp_character.id
FROM tmp_character
GROUP BY tmp_character.id_new
HAVING COUNT(tmp_character.id_new) > 1
)
WHERE `character`.id_new IN (
SELECT tmp_character.id_new
FROM tmp_character
GROUP BY tmp_character.id_new
HAVING COUNT(tmp_character.id_new) > 1
);
现在出现问题：ERROR 1242（21000）：子查询返回的行数超过1行 这是真的，我有22个不同的身份要更新，我有22个团体，不知道为什么它不会工作。
在此查询之后，我的表应如下所示：
+-----+--------------------------+--------------+--------------+-------+
| id | char_name_de | char_name_en | id_new | id_old|
+-----+--------------------------+--------------+--------------+-------+
| 258 | Natter | NULL | 2914 | 258 |
| 342 | Snake | NULL | 2914 | 342 |
| 290 | Rat | NULL | 4075 | 290 |
| 292 | Ratte | NULL | 4075 | 292 |
| 276 | Orakel der Hakkari | NULL | 11346 | 276 |
| 152 | Hakkari Oracle | NULL | 11346 | 152 |
| 148 | Gurubashi Warrior | NULL | 11355 | 148 |
| 201 | Krieger der Gurubashi | NULL | 11355 | 201 |
| 344 | Sohn von Hakkar | NULL | 11357 | 344 |
| 347 | Son of Hakkar | NULL | 11357 | 347 |
+-----+--------------------------+--------------+--------------+-------+
.
.
.
现在＆＃34; id_new＆＃34;应该是这样的独特：
+-----+--------------------------+--------------+--------------+-------+
| id | char_name_de | char_name_en | id_new | id_old|
+-----+--------------------------+--------------+--------------+-------+
| 258 | Natter | NULL | 2914 | 258 |
| 290 | Rat | NULL | 4075 | 290 |
| 276 | Orakel der Hakkari | NULL | 11346 | 276 |
| 148 | Gurubashi Warrior | NULL | 11355 | 148 |
| 344 | Sohn von Hakkar | NULL | 11357 | 344 |
+-----+--------------------------+--------------+--------------+-------+
.
.
.
现在我要删除列＆＃34; id_new＆＃34;得到我的决赛桌。
+-------+--------------------------+--------------+-------+
| id | char_name_de | char_name_en | id_old|
+-------+--------------------------+--------------+-------+
| 2914 | Natter | NULL | 258 |
| 4075 | Rat | NULL | 290 |
| 11346 | Orakel der Hakkari | NULL | 276 |
| 11355 | Gurubashi Warrior | NULL | 148 |
| 11357 | Sohn von Hakkar | NULL | 344 |
+-------+--------------------------+--------------+-------+
.
.
.
有人可以帮我修改我的查询以使其运行吗？我需要在突出显示之后修复查询＆＃34;现在我已经得到了这个查询：＆＃34;，休息很简单，我自己得到它，我只是发布它背景我想要的东西结束。
答案 0 :(得分：1)
这是我的脚本将值从id移动到id_old，其中我有多行：
<?php
function executeQuery($dbConnection, $Query) {
$result = $dbConnection -> query($Query);
return $result;
}
function openDB() {
$dbConnection= new mysqli($mServer, $mUser, $mPwd, $mDatabase);
return $dbConnection;
}
// open Database
$DBconnection=openDB();
// 1st query
$query="SELECT *
FROM `character`
WHERE id_new IN(
SELECT id_new
FROM `character`
GROUP BY `character`.id_new
HAVING COUNT(`character`.id_new) > 1
)
ORDER BY id_new ASC;";
// send query to database
$result=executeQuery($DBconnection, $query);
$numRows=$result->num_rows;
// as long as you've got rows
for(;$numRows>0;$numRows--) {
// go to next row
$currentRow=$result->fetch_object();
for($i=0;$i<2;$i++) {
if($i==0) {
// skip 1st line from each row which is doubled
$currentRow=$result->fetch_object();
}
// delete 2nd line from each row which is doubled
$id_new=$currentRow->id;
$query="delete from `character`where id=" .$id_new .";";
$result2=executeQuery($DBconnection, $query);
}
}
?>
这是第二个修改版本，如果一行加倍，将删除第二行。如果一行被摧毁或更多，它将无效！
do{
while ( scanner.hasNext() ) {
String command = scanner.nextLine();
System.out.println("line:"+command);
}
} while ( conditionFlag)
但正如我所说，这是一个肮脏的剧本！
它是如何工作的？ 我在第二个查询中使用一个查询的一些值。我承诺它可以在没有PHP的情况下工作，但有时快速而肮脏的解决方案比一个需要几天的好的解决方案更好！
//关闭