Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

Getting results from arbitrary SQL statements with correct binding in SQLite.swift

The SQLite.swift documentation says about executing arbitrary SQL:

let stmt = try db.prepare("SELECT id, email FROM users")
for row in stmt {
    for (index, name) in stmt.columnNames.enumerate() {
        print ("\(name)=\(row[index]!)")
        // id: Optional(1), email: Optional("[email protected]")
    }
}

I wanted to get the values directly like this

let stmt = try db.prepare("SELECT id, email FROM users")
for row in stmt {
    let myInt: Int64 = row[0] // error: Cannot convert value of type 'Binding?' to specified type 'Int64'
    let myString: String = row[1] // error: Cannot convert value of type 'Binding?' to specified type 'String'
}

but the row index is of type Binding? and I can't figure out how to convert that to the type I need. I see there is a Statement.bind method in the source code but I am still not discovering how to apply it.

like image 285
Suragch Avatar asked Aug 15 '16 17:08

Suragch


1 Answers

You can retrieve correctly typed selected columns from a table like this:

// The database.
let db = try Connection(...)

// The table.
let users = Table("users")

// Typed column expressions.
let id = Expression<Int64>("id")
let email = Expression<String>("email")

// The query: "SELECT id, email FROM users"
for user in try db.prepare(users.select(id, email)) {
    let id = user[id]       // Int64
    let mail = user[email]  // String
    print(id, mail)
}

An alternative is to (optionally) cast the Binding values to the correct type:

let stmt = try db.prepare("SELECT id, email FROM users")
for row in stmt {
    if let id = row[0] as? Int64,
        let mail = row[1] as? String {
        print(id, mail)
    }
}
like image 82
Martin R Avatar answered Oct 18 '22 07:10

Martin R