Ionic with SQlite plugin cannot create table

Question

I’m using ionic Sqlite native plugin. When trying to create table I’m getting this error in the console:

“Console MessagesError: {“rows”:{“length”:0},“rowsAffected”:0}”

which I don’t understand cause I think my code is ok.

import { HttpClient } from '@angular/common/http';
import { Injectable } from '@angular/core';
import { SQLite, SQLiteObject } from "@ionic-native/sqlite";


@Injectable()
export class DatabaseProvider {
theConsole: string = "Console Messages";

options: any = {
name: 'db_name.db',
location: 'default'
}

private db: SQLiteObject;
private isOpen: boolean;

constructor(public http: HttpClient,private sqlite: SQLite) {
console.log('Hello DatabaseProvider Provider');
this.connectToDb();
}

private connectToDb():void {

  this.sqlite = new SQLite();
  this.sqlite.create(this.options)
    .then((db: SQLiteObject) => {
      this.db = db;
      this.isOpen = true;
      console.log('Hello DatabaseProvider connected to db');
      this.createTables();
    })
    .catch(e => {
      this.theConsole += JSON.stringify(e);
      console.log(this.getConsoleMessages());
    });

  }

  private createTables():void{
  this.createTableContatti();
   }

 private createTableContatti(): void{
 var sql = 'create table IF NOT EXISTS contatti(id_contatto INTEGER PRIMARY 
 KEY AUTOINCREMENT,nome TEXT, data_nascita TEXT)';
 this.db.executeSql(sql)
  .then(() => {
  this.theConsole += 'Executed SQL' + sql
  console.log(this.getConsoleMessages());
  })
  .catch(e => {
    this.theConsole += "Error: "  + JSON.stringify(e)
    console.log(this.getConsoleMessages());
  });
  }

Can someone help please? Thank you in advance

ashirwadrk · Accepted Answer · 2018-10-25 13:29:49Z

1

The ionic SQLite's executeSql requires 2 parameters query and data. If there is no data pass an empty array object.

var sql = 'create table IF NOT EXISTS contatti(id_contatto INTEGER PRIMARY 
 KEY AUTOINCREMENT,nome TEXT, data_nascita TEXT)';
 this.db.executeSql(sql,{})
  .then(() => {
  this.theConsole += 'Executed SQL' + sql
  console.log(this.getConsoleMessages());
  })
  .catch(e => {
    this.theConsole += "Error: "  + JSON.stringify(e)
    console.log(this.getConsoleMessages());
  });

Your query works fine if you pass the second parameter.

answered Oct 25, 2018 at 13:29

ashirwadrk

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6 Comments

Iva Pesa Over a year ago

If I add the second parameter, I have this compile error: "Argument of type '{}' is not assignable to parameter of type 'any[]'. Property 'length' is missing in type '{}'." Because of than I removed the second parameter. Do you know why doesn't compile?

ashirwadrk Over a year ago

I ran your code with the second parameter and it worked. Can you show me the change you did?

Iva Pesa Over a year ago

private createTableContatti(): void{ var sql = 'create table IF NOT EXISTS contatti(id_contatto INTEGER PRIMARY KEY AUTOINCREMENT,nome TEXT, data_nascita TEXT)'; this.db.executeSql(sql,{}) .then(() => { this.theConsole += 'Executed SQL' + sql; console.log(this.getConsoleMessages()); }) .catch(error => { let errMsg = (error.messageText) ? error.messageText : error.status ; this.theConsole += "Error: " + errMsg + " " + JSON.stringify(error); console.log(this.getConsoleMessages()); }); }

ashirwadrk Over a year ago

Your query works fine. Can you try removing all extra code and then running? just log "success" or "error" in callbacks.

ashirwadrk Over a year ago

Also if this doesn't work try passing empty array [] instead of {}

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