在appcelerator钛sql左连接不起作用

Pra*_*Raj 1 sql database titanium appcelerator

我正在研究APPCELERATOR TITANIUM.

我正在使用一个数据库,其中奇怪的是一些SQL查询不能正常工作,例如:

db.execute("SELECT Location_Diners.Dining_Time,Location_Diners.First_Name,Location_Diners.Last_Name,Location_Diners.PartySize,Location_Diners.Diner_ID,Location_Diners.DinerStatusColor_ID,Location_Diners.Notes,Location_Diners.Diner_ID,Location_SeatedDiners.Table_ID FROM Location_Diners LEFT JOIN Location_SeatedDiners ON Location_Diners.Diner_ID=Location_SeatedDiners.Diner_ID");
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这是我得到的错误,请帮助我在这里.....

2011-12-21 11:47:29.416 abc[5254:ac03] [ERROR] A SQLite database error occurred on database '/Users/../iPhone Simulator/4.3.2/Applications/C377123C-7A2A-4FDF-9314-428713C885FD/Library/Application Support/database.sql': Error Domain=com.plausiblelabs.pldatabase Code=3 "An error occured parsing the provided SQL statement." UserInfo=0x6ad4330 {com.plausiblelabs.pldatabase.error.vendor.code=1, NSLocalizedDescription=An error occured parsing the provided SQL statement., com.plausiblelabs.pldatabase.error.query.string=SELECT Location_Diners.Dining_Time,Location_Diners.First_Name,Location_Diners.Last_Name,Location_Diners.PartySize,Location_Diners.Diner_ID,Location_Diners.DinerStatusColor_ID,Location_Diners.Notes,Location_Diners.Diner_ID,Location_SeatedDiners.Table_ID FROM Location_Diners LEFT JOIN Location_SeatedDiners ON Location_Diners.Diner_ID=Location_SeatedDiners.Diner_ID, com.plausiblelabs.pldatabase.error.vendor.string=no such table: Location_Diners} (SQLite #1: no such table: Location_Diners) (query: 'SELECT Location_Diners.Dining_Time,Location_Diners.First_Name,Location_Diners.Last_Name,Location_Diners.PartySize,Location_Diners.Diner_ID,Location_Diners.DinerStatusColor_ID,Location_Diners.Notes,Location_Diners.Diner_ID,Location_SeatedDiners.Table_ID FROM Location_Diners LEFT JOIN Location_SeatedDiners ON Location_Diners.Diner_ID=Location_SeatedDiners.Diner_ID')
[ERROR] invalid SQL statement. Error Domain=com.plausiblelabs.pldatabase Code=3 "An error occured parsing the provided SQL statement." UserInfo=0x6ad4330 {com.plausiblelabs.pldatabase.error.vendor.code=1, NSLocalizedDescription=An error occured parsing the provided SQL statement., com.plausiblelabs.pldatabase.error.query.string=SELECT Location_Diners.Dining_Time,Location_Diners.First_Name,Location_Diners.Last_Name,Location_Diners.PartySize,Location_Diners.Diner_ID,Location_Diners.DinerStatusColor_ID,Location_Diners.Notes,Location_Diners.Diner_ID,Location_SeatedDiners.Table_ID FROM Location_Diners LEFT JOIN Location_SeatedDiners ON Location_Diners.Diner_ID=Location_SeatedDiners.Diner_ID, com.plausiblelabs.pldatabase.error.vendor.string=no such table: Location_Diners} in -[TiDatabaseProxy execute:] (TiDatabaseProxy.m:136)
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任何帮助都非常感谢................

The*_*ero 5

嘿Prateek我认为你应该使用LEFT OUTER JOIN.检查一下,

SQLite支持哪些加入?

根据这个LEFT JOIN = LEFT OUTER JOIN,但SQLite3支持关键字LEFT OUTER JOIN.