Sqlite3 (node.js) returns [object Object]

Question

I am trying to use SQLITE3 to store data for my discord bot and here i need to save a number that goes up each time a command is used. Here is my code: (creating table and setting the value to 0.)

  db.run("CREATE TABLE gbanCases ( number INT )")
  db.run("INSERT INTO gbanCases (number) VALUES ($number)", {
    $number : 0
});

Trying to see what number returns:

        var caseNumber = db.run("SELECT number FROM gbanCases")
        console.log(caseNumber)

The thing is that caseNumber returns me "[object Object]" and I want it to return the 0 that was setted up earlier. Please help me if you can, i'm still very new to this.

Replace your current log with console.log({ caseNumber }) and see what you get. — code
– code, Commented Aug 5, 2022 at 17:46
Here is what i get: {caseNumber: Database} caseNumber: Database {filename: 'database.sqlite', mode: 65542} [[Prototype]]: Object — CryZz
– CryZz, Commented Aug 5, 2022 at 17:51
Don't use db.run() for a SELECT query, use db.get() or db.all() depending on whether you want 1 row or all rows. They take a callback function that receives the results. — Barmar
– Barmar, Commented Aug 5, 2022 at 17:54

Barmar · Accepted Answer · 2022-08-05 17:57:06Z

2

Use db.get() to get the result of a query, not db.run(). The resulting row is passed as an argument to the callback function.

db.get("SELECT number FROM gbanCases", function(err, row) {
    if (err) {
        throw err;
    }
    console.log(row.number);
})

answered Aug 5, 2022 at 17:57

Barmar

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15 Comments

CryZz Over a year ago

console.log(row.number); returns : "{}1"

Barmar Over a year ago

I suspect {} is coming from something else, not this. row.number shouldn't contain that.

CryZz Over a year ago

How can I fix it? I'm supposed to get 0.

Barmar Over a year ago

I can't think of any reason why it doesn't work. The {} is not coming from my code, it must be coming from something else you're doing.

CryZz Over a year ago

Maybe i didn't do this correctly : ` db.run("CREATE TABLE gbanCases ( number INT )") db.run("INSERT INTO gbanCases (number) VALUES ($number)", { $number : 0 });`

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