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Arkadaşlar yine aynı sorun ama bir kez hata verdi direk yazmıyorum yani 1.5-2 saat uğraştım ama olmuyor

DEFAULT CHARSET=latin1 kısımlarının hepsini sildim yine olmadı...

Eski host
MySQL - 4.1.11-Debian_4sarge4-log
phpMyAdmin - 2.8.2
MySQL client version: 4.1.11

Yeni host
MySQL - 4.0.27-standard
phpMyAdmin - 2.8.0.2
MySQL client version: 4.1.10

Sürekli bu veya buna benzer sorunlar alıyorum


Hata

SQL sorgusu:



CREATE TABLE IF NOT EXISTS `phpbb_auth_access` (
`group_id` mediumint( 8 ) NOT NULL default '0',
`forum_id` smallint( 5 ) unsigned NOT NULL default '0',
`auth_view` tinyint( 1 ) NOT NULL default '0',
`auth_read` tinyint( 1 ) NOT NULL default '0',
`auth_post` tinyint( 1 ) NOT NULL default '0',
`auth_reply` tinyint( 1 ) NOT NULL default '0',
`auth_edit` tinyint( 1 ) NOT NULL default '0',
`auth_delete` tinyint( 1 ) NOT NULL default '0',
`auth_sticky` tinyint( 1 ) NOT NULL default '0',
`auth_announce` tinyint( 1 ) NOT NULL default '0',
`auth_vote` tinyint( 1 ) NOT NULL default '0',
`auth_pollcreate` tinyint( 1 ) NOT NULL default '0',
`auth_attachments` tinyint( 1 ) NOT NULL default '0',
`auth_mod` tinyint( 1 ) NOT NULL default '0',
KEY `group_id` ( `group_id` ) ,
KEY `forum_id` ( `forum_id` )
) ENGINE = MYISAM DEFAULT CHARSET = latin1;

MySQL çıktısı: Yardım
#1064 - You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near 'DEFAULT CHARSET=latin1' at line 18

CHARSET = latin1

sil

"DEFAULT CHARSET=latin1" leri sildim bu hatayı verdi Bende "collate latin1_bin NOT NULL default" vardı 1 tane onu sildim.Doğru mu yaptım acaba?[deniyorum şimdi]

Hata

SQL sorgusu:

-- --------------------------------------------------------
--
-- Table structure for table `phpbb_search_wordlist`
--
CREATE TABLE `phpbb_search_wordlist` (
`word_text` varchar( 50 ) CHARACTER SET latin1 COLLATE latin1_bin NOT NULL default '',
`word_id` mediumint( 8 ) unsigned NOT NULL AUTO_INCREMENT ,
`word_common` tinyint( 1 ) unsigned NOT NULL default '0',
PRIMARY KEY ( `word_text` ) ,
KEY `word_id` ( `word_id` )
) ENGINE = MYISAM AUTO_INCREMENT =51865;

MySQL çıktısı: Yardım
#1064 - You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near 'collate latin1_bin NOT NULL default '',
`word_id` mediumint(

CHARSET = latin1 leri silince böyle diyor

Hata

SQL sorgusu:

CREATE TABLE `phpbb_auth_access` (
`group_id` mediumint( 8 ) NOT NULL default '0',
`forum_id` smallint( 5 ) unsigned NOT NULL default '0',
`auth_view` tinyint( 1 ) NOT NULL default '0',
`auth_read` tinyint( 1 ) NOT NULL default '0',
`auth_post` tinyint( 1 ) NOT NULL default '0',
`auth_reply` tinyint( 1 ) NOT NULL default '0',
`auth_edit` tinyint( 1 ) NOT NULL default '0',
`auth_delete` tinyint( 1 ) NOT NULL default '0',
`auth_sticky` tinyint( 1 ) NOT NULL default '0',
`auth_announce` tinyint( 1 ) NOT NULL default '0',
`auth_vote` tinyint( 1 ) NOT NULL default '0',
`auth_pollcreate` tinyint( 1 ) NOT NULL default '0',
`auth_attachments` tinyint( 1 ) NOT NULL default '0',
`auth_mod` tinyint( 1 ) NOT NULL default '0',
KEY `group_id` ( `group_id` ) ,
KEY `forum_id` ( `forum_id` )
) ENGINE = MYISAM DEFAULT;

MySQL çıktısı: Yardım
#1064 - You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near 'DEFAULT' at line 18

üstat ben forumu yeni tasıdım

DEFAULT CHARSET = latin1

Bunu sileceksin


CREATE TABLE `phpbb_search_wordlist` (
`word_text` varchar( 50 ) CHARACTER SET latin1 COLLATE latin1_bin NOT NULL default '',
`word_id` mediumint( 8 ) unsigned NOT NULL AUTO_INCREMENT ,
`word_common` tinyint( 1 ) unsigned NOT NULL default '0',
PRIMARY KEY ( `word_text` ) ,
KEY `word_id` ( `word_id` )
) ENGINE = MYISAM AUTO_INCREMENT =51865;

Yukardakinde veren hatayı düzeltmek içinde

CREATE TABLE phpbb_search_wordlist (
word_text varchar(50) binary NOT NULL default '',
word_id mediumint(8) unsigned NOT NULL auto_increment,
word_common tinyint(1) unsigned NOT NULL default '0',
PRIMARY KEY (word_text),
KEY word_id (word_id)
) TYPE=MyISAM;

Bu sorguyu olustur Sorunsuz Çalışacaktır İyi Günler