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having-clause

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mysql - 拉维尔 5.3 : Syntax error or access violation: 1463 Non-grouping field 'distance' is used in HAVING clause

这个错误是在将整个源代码移动到5.3版本后出现的,我现在挠头了两个多小时。所以我有这种Eloquent查询:POI::select('*',DB::raw("SQRT(POW((x-{$this->x}),2)+POW((y-{$this->y}),2))ASdistance"))->where('status',Config::get('app.poi_state.enabled'))->whereNotIn('id',$excludePOIList)->having('distance','orderBy('distance')->get();它在升级之前找到了,现在它抛出:Syn

mysql - 拉维尔 5.3 : Syntax error or access violation: 1463 Non-grouping field 'distance' is used in HAVING clause

这个错误是在将整个源代码移动到5.3版本后出现的,我现在挠头了两个多小时。所以我有这种Eloquent查询:POI::select('*',DB::raw("SQRT(POW((x-{$this->x}),2)+POW((y-{$this->y}),2))ASdistance"))->where('status',Config::get('app.poi_state.enabled'))->whereNotIn('id',$excludePOIList)->having('distance','orderBy('distance')->get();它在升级之前找到了,现在它抛出:Syn

mysql - 选择具有最大 id 计数的 id

有一个包含item_id和color_id的产品表。我正在尝试获取具有最多非空实例的color_id。这失败了:SELECTcolor_idFROMproductsWHEREitem_id=1234GROUPBYitem_idHAVINGMAX(COUNT(color_id))与Invaliduseofgroupfunction这个SELECTcolor_id,COUNT(color_id)FROMproductsWHEREitem_id=1234GROUPBYitem_id返回color_idcount1,3232,1223,554我正在寻找具有最多实例的color_id3。是否有一

mysql - 选择具有最大 id 计数的 id

有一个包含item_id和color_id的产品表。我正在尝试获取具有最多非空实例的color_id。这失败了:SELECTcolor_idFROMproductsWHEREitem_id=1234GROUPBYitem_idHAVINGMAX(COUNT(color_id))与Invaliduseofgroupfunction这个SELECTcolor_id,COUNT(color_id)FROMproductsWHEREitem_id=1234GROUPBYitem_id返回color_idcount1,3232,1223,554我正在寻找具有最多实例的color_id3。是否有一

php - 违反完整性约束 : 1052 Column 'id' in where clause is ambiguous

我正在做一个简单的查询,但它不起作用,我也不知道为什么。我最近开始熟悉PDO与数据库的连接。代码如下:连接是:define("HOST","localhost");define("USER","root");define("PASS","password");define("BASE","portugalforcedb");try{$conexao='mysql:host='.HOST.';dbname='.BASE;$connect=newPDO($conexao,USER,PASS);$connect->setAttribute(PDO::ATTR_ERRMODE,PDO::ERR

php - 违反完整性约束 : 1052 Column 'id' in where clause is ambiguous

我正在做一个简单的查询,但它不起作用,我也不知道为什么。我最近开始熟悉PDO与数据库的连接。代码如下:连接是:define("HOST","localhost");define("USER","root");define("PASS","password");define("BASE","portugalforcedb");try{$conexao='mysql:host='.HOST.';dbname='.BASE;$connect=newPDO($conexao,USER,PASS);$connect->setAttribute(PDO::ATTR_ERRMODE,PDO::ERR

MySQL "Unknown Column in On Clause"

这个问题在这里已经有了答案:MySQLunknowncolumninONclause(4个回答)关闭6年前。我有以下MySQL查询:SELECTposts.id,posts.name,LEFT(posts.content,400),posts.author,posts.date,users.display_name,GROUP_CONCAT(tags.tagORDERBYtag_linking.pidASCSEPARATOR",")update_tagsFROMposts,tag_linking,tagsINNERJOIN`users`ONposts.author=users.id;W

MySQL "Unknown Column in On Clause"

这个问题在这里已经有了答案:MySQLunknowncolumninONclause(4个回答)关闭6年前。我有以下MySQL查询:SELECTposts.id,posts.name,LEFT(posts.content,400),posts.author,posts.date,users.display_name,GROUP_CONCAT(tags.tagORDERBYtag_linking.pidASCSEPARATOR",")update_tagsFROMposts,tag_linking,tagsINNERJOIN`users`ONposts.author=users.id;W

sql - 错误 : The used SELECT statements have a different number of columns

为什么我得到ERROR1222(21000):TheusedSELECTstatementshaveadifferentnumberofcolumnsfromthefollowing?SELECT*FROMfriendsLEFTJOINusersASu1ONusers.uid=friends.fid1LEFTJOINusersASu2ONusers.uid=friends.fid2WHERE(friends.fid1=1)AND(friends.fid2>1)UNIONSELECTfid2FROMfriendsWHERE(friends.fid2=1)AND(friends.fid1用

sql - 错误 : The used SELECT statements have a different number of columns

为什么我得到ERROR1222(21000):TheusedSELECTstatementshaveadifferentnumberofcolumnsfromthefollowing?SELECT*FROMfriendsLEFTJOINusersASu1ONusers.uid=friends.fid1LEFTJOINusersASu2ONusers.uid=friends.fid2WHERE(friends.fid1=1)AND(friends.fid2>1)UNIONSELECTfid2FROMfriendsWHERE(friends.fid2=1)AND(friends.fid1用