我在我的ionic-3应用程序中使用SQLite原生插件,当我试图查询一些学生时,查询不起作用。这给出的结果是空的。
//studentIds array
let studentIds = [1,2,4,5];
//Query part
this.database.executeSql("Select * from student where id in (?)", [studentIds]).then(data => {
let students = [];
for(let i=0; i < data.rows.length; i++) {
students.push({id: data.rows.item[i].id, name: data.rows.item[i].name});
}
return students;
});发布于 2018-06-22 17:19:21
这对我在Ionic 3中很有效:
//studentIds array
let studentIds = [1,2,4,5];
//Query part
this.database.executeSql(`select * from student where id in (?)`, [studentIds]).then(data => {
let students = [];
for(let i=0; i < data.rows.length; i++)
students.push({id: data.rows.item(i).id, name: data.rows.item(i).name});
return students;
});问题出在data.rows.item[i]的[]上,您必须像这样使用() data.rows.item(i)
发布于 2018-04-03 11:53:30
对于创建数据库
this._platform.ready().then(() => {
this.sqlite = new SQLite();
this.sqlite.create({name: "mymain.db", location: "default"}).then((db: SQLiteObject) => {
db.executeSql('CREATE TABLE IF NOT EXISTS mydata (allvalues TEXT)', {})
.then(() => console.log('Executed SQL'))
.catch(e => console.log(e));
})
.catch(e => console.log(e));
});
}用于将数据添加到sqltie
public saveTablicaToSqlite(DataArray){
this.sqlite.create({
name: 'mymain.db',
location: 'default'
})
.then((db: SQLiteObject) => {
db.executeSql("INSERT INTO mydata (allvalues) VALUES (?)", [DataArray]).then((data) => {
console.log("INSERTED: " + JSON.stringify(data));
}, (error) => {
console.log("ERROR: " + JSON.stringify(error.err));
});
})
.catch(e => console.log(e));
}用于获取数据
fetchdata(datavalue,splashScreen){
console.info('fetch',datavalue);
return new Promise((resolve, reject) => {
this.sqlite.create({
name: 'mymain.db',
location: 'default'
})
.then((db: SQLiteObject) => {
db.executeSql("SELECT * FROM mydata", []).then((data) => {
if(data.rows.length > 0) {
for(let i = 0; i < 1; i++) {
var obj = JSON.parse(data.rows.item(i).allvalues);
console.info(obj);https://stackoverflow.com/questions/49613913
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