2

我正在尝试执行左连接

SELECT TProgram.ProgramName, TProgram.ProgramPath From TProgram
LEFT JOIN TIcons ON TIcons.ProgramName = TProgram.ProgramName
WHERE TIcons.ProgramName IS NULL
GROUP BY TProgram.ProgramName

迅速。所以我做了以下。

let program = db["TProgram"]
let icons = db["TIcons"]
let programNameDB = Expression<String>("ProgramName")
let programNameDBI = Expression<String>("ProgramName")
let programPathDB = Expression<String>("ProgramPath")
let all = program.select(programNameDB, programPathDB).join(Query.JoinType.LeftOuter, icons, on: programNameDB == icons[programNameDBI]).filter(programNameDBI == "").group(programNameDB)

我无法检查它是否正常工作,因为以下结果会导致fatal error: ambiguous column name:

let item = all.first!
let arr = Array(all)

如果我尝试println(all),我可以在调试控制台中看到“TProgram”。

我如何访问数据?

更新 我尝试了以下。错误消失了,但那里似乎没有值。

func readTProgramNames() -> (Query){
    let program = db["TProgram"]
    let icons = db["TIcons"]
    let programNameDB = Expression<String>("ProgramName")
    let programNameDBI = Expression<String>("ProgramName")
    let programPathDB = Expression<String>("ProgramPath")
    let all = program
        .select(program[programNameDB], program[programPathDB])
        .join(.LeftOuter, icons, on: program[programNameDB] == icons[programNameDBI])
        .filter(icons[programNameDBI] == "")
        .group(program[programNameDB])

    let item = all.first!

    println(item[program[programNameDB]])

    return all
}
4

1 回答 1

3

知道了。最String?重要的和最重要的.leftouter

func readTProgramNames(){
    let program = db["TProgram"]
    let icons = db["TIcons"]
    let programNameDB = Expression<String?>("ProgramName")
    let programNameDBI = Expression<String?>("ProgramName")
    let programPathDB = Expression<String?>("ProgramPath")
    //let all = table.select(programNameDB, programPathDB).group(programNameDB)
    //let all = program.join("LEFT OUTER", icons, on: programNameDB == icons[programNameDBI])
    let all = program
        .select(program[programNameDB], program[programPathDB])
        .join(.LeftOuter, icons, on: program[programNameDB] == icons[programNameDBI])
        .filter(icons[programNameDBI] == nil)
        .group(program[programNameDB])

    for item in all{
        println("programName: \(item[program[programNameDB]]!)")
    }
}
于 2015-07-24T14:49:36.543 回答