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sqllite-我有問題，在Android的Ionic3與選擇
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();
});
問題用科爾多瓦從SQL精簡版數據庫查詢記錄，這是一個正確打印記錄
JSON.stringify(response.rows.item(i))
O/P
{'USER_ID':1,'FIRST_NAME':'Temp','LAST_NAME':'User','USER_NAME':'TEMPUSER','USER_PWD':'TEMPPWD'}
下面是投擲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
有人有這個問題嗎？我不知道爲什麼我無法像JSON那樣獲取它。 請告知
你可以嘗試用'user.password'儘管在您的查詢密碼？ – ericminio
實際上沒有密碼被加密。所以我正在使用一個新的變量。但是如果你看到這裏，當我提醒下面的聲明它工作正常，並給出了JSON。記錄+ JSON.stringify（response.rows.item（i））+'\ n'; 。問題只有當我使用：response.rows.item [i] .FIRST_NAME – jslearner07
console.log（'用戶在DB中找到 - >響應項目類型'+ typeof（response.rows.item [i]））;給出的類型爲undefined。 – jslearner07