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sqllite-我有问题，在Android的Ionic3与选择
this.platform.ready().then(() => {
this.sqlite.create({
name: 'temp.db',
location: 'default'
}).then((db: SQLiteObject) => {
console.log('Querying for temp user '+user.userName+'Password '+user.password);
console.log('User queried'+user.userName);
db.executeSql("SELECT * FROM USER where USER_NAME = ? and USER_PWD=?", [user.userName,password]).then(
response => {
let records='';
for (let i = 0; i < response.rows.length; i++) {
records = records+ JSON.stringify(response.rows.item(i))+'\n'; //Prints row correctly
}
this._util.presentAlert('Records selected like from- USR-',records);
})
.catch(
e => this._util.presentAlert('Fail- Select like from- USER-Temp DBUSER',e));
db.executeSql("SELECT * FROM USER where USER_NAME = ? and USER_PWD=? ", [user.userName,password ]).then(
response => {
if (response && response.rows && response.rows.length > 0) {
for (let i = 0; i < response.rows.length; i++) {
let access = {
firstName :response.rows.item[i].FIRST_NAME, //This is undefined.
lastName :response.rows.item[i].LAST_NAME,
userName:response.rows.item[i].USER_NAME,
userId:response.rows.item[i].USER_ID
}
observer.next(access);
}
observer.complete();
} else {
let access = {status:'Fail',msg:'Bad credentials for Temp DB login'};
console.log('No record for the user from- USER'+user.userName);
observer.next(access);
observer.complete();
}
})
.catch(
e => {
console.log('Fail- Select query gone wrong * from- USER FOR Temp DB LOGIN' + e);
let access = {status:'Fail',msg:'Bad credentials for Temp DB login'};
observer.next(access);
observer.complete();
});
问题用科尔多瓦从SQL精简版数据库查询记录，这是一个正确打印记录
JSON.stringify(response.rows.item(i))
O/P
{'USER_ID':1,'FIRST_NAME':'Temp','LAST_NAME':'User','USER_NAME':'TEMPUSER','USER_PWD':'TEMPPWD'}
下面是投掷undefined error
firstName :response.rows.item[i].FIRST_NAME
Fail- Select query gone wrong * from- USER FOR Temp DB LOGIN TypeError: Cannot read property 'FIRST_NAME' of undefined
有人有这个问题吗？我不知道为什么我无法像JSON那样获取它。 请告知
你可以尝试用'user.password'尽管在您的查询密码？ – ericminio
实际上没有密码被加密。所以我正在使用一个新的变量。但是如果你看到这里，当我提醒下面的声明它工作正常，并给出了JSON。记录+ JSON.stringify（response.rows.item（i））+'\ n'; 。问题只有当我使用：response.rows.item [i] .FIRST_NAME – jslearner07
console.log（'用户在DB中找到 - >响应项目类型'+ typeof（response.rows.item [i]））;给出的类型为undefined。 – jslearner07