如何修复'{}类型的参数不能分配给'any []'类型的参数。 [ng]类型“ {}”中缺少属性“长度”。

时间:2019-03-27 20:06:15

标签: angularjs typescript sqlite ionic-framework error-handling

我将借助本教程https://www.techiediaries.com/ionic-cordova-sqlite-barcode-scanner-product-inventory-manager/

使用条形码扫描仪和SQLite构建Ionic库存管理应用程序

当我添加此代码时:

async createTables(){
    try {
        await this.database.executeSql(this.familyTable, {});
        await this.database.executeSql(this.locationTable,{});
        await this.database.executeSql(this.productTable,{});
        await this.database.executeSql(this.transactionTable,{});
    }catch(e){
        console.log("Error !");
    }
}

...到data-service.service.ts我收到此错误:

  src / app / data-service.service.ts(54,64)中的

ERROR:错误TS2345:   类型“ {}”的参数不能分配给类型“ any []”的参数。   [ng]类型“ {}”中缺少属性“长度”。 [ng]   src / app / data-service.service.ts(55,65):错误TS2345:类型的参数   不能将“ {}”分配给类型“ any []”的参数。 [ng]属性   类型“ {}”中缺少“长度”。 [ng]   src / app / data-service.service.ts(56,64):错误TS2345:类型的参数   不能将“ {}”分配给类型“ any []”的参数。 [ng]属性   类型“ {}”中缺少“长度”。 [ng]   src / app / data-service.service.ts(57,68):错误TS2345:类型的参数   不能将“ {}”分配给类型“ any []”的参数。 [ng]属性   类型“ {}”中缺少“长度”。

这是整个data-service.service.ts代码:

import { Injectable } from '@angular/core';

import 'rxjs/add/operator/map';
import { SQLite, SQLiteObject } from '@ionic-native/sqlite/ngx';



@Injectable({
  providedIn: 'root'
})
export class DataServiceService {
  public database: SQLiteObject;

  productTable : string = `CREATE TABLE IF NOT EXISTS  products (
    id INTEGER PRIMARY KEY,
    sku TEXT,
    barcode TEXT,
    title TEXT NOT NULL,
    description TEXT,
    quantity REAL,
    unit VARCHAR,
    unitPrice REAL,
    minQuantity INTEGER,
    familly_id INTEGER,
    location_id INTEGER,
    FOREIGN KEY(familly_id) REFERENCES famillies(id),
    FOREIGN KEY(location_id) REFERENCES locations(id)
    );`;

familyTable : string = `CREATE TABLE IF NOT EXISTS famillies (
    id INTEGER PRIMARY KEY,
    reference VARCHAR(32) NOT NULL,
    name TEXT NOT NULL,
    unit VARCHAR);`;

locationTable : string = `CREATE TABLE IF NOT EXISTS locations (
        id INTEGER PRIMARY KEY,
        name TEXT NOT NULL);`;
//Date , Quantity , Unit Cost , Reason (New Stock - Usable Return - Unusable Return ) ,UPC (Universal Product Code ) Comment    
transactionTable : string = `CREATE TABLE IF NOT EXISTS transactions (
        id INTEGER PRIMARY KEY,
        date TEXT,
        quantity REAL,
        unitCost REAL,
        reason VARCHAR,
        upc TEXT,
        comment TEXT,
        product_id INTEGER,
        FOREIGN KEY(product_id) REFERENCES products(id));`;

        async createTables(){
          try {
              await this.database.executeSql(this.familyTable, {});
              await this.database.executeSql(this.locationTable,{});
              await this.database.executeSql(this.productTable,{});
              await this.database.executeSql(this.transactionTable,{});
          }catch(e){
              console.log("Error !");
          }
      }

        constructor(public sqlite :SQLite) {
          console.log('Hello DataServiceProvider Provider')

              this.sqlite.create({name: "data.db", location: "default"}).then((db : SQLiteObject) => {
                      this.database = db;
                  }, (error) => {
                      console.log("ERROR: ", error);
              }); 
    }

}

有人知道如何解决吗?

1 个答案:

答案 0 :(得分:2)

executeSql的签名是:

executeSql(statement: string, params?: any[]): Promise<any>;

如果您以{}作为第二个参数调用executeSql,我知道您不想传递任何参数。因此,您应该像这样调用方法:

await this.database.executeSql(this.familyTable);
await this.database.executeSql(this.locationTable);
await this.database.executeSql(this.productTable);
await this.database.executeSql(this.transactionTable);