Sqlite 可观察的fork-join问题
嗨,我有三张桌子,每一张都是另一张的孩子。我编写了一个从sqllite数据库获取数据的方法,如下所示Sqlite 可观察的fork-join问题,sqlite,angular,reactjs,observable,ionic3,Sqlite,Angular,Reactjs,Observable,Ionic3,嗨,我有三张桌子,每一张都是另一张的孩子。我编写了一个从sqllite数据库获取数据的方法,如下所示 public downloadFromOfflineDB(db,testSO){ var observableBatch = []; observableBatch.push(db.executeSql("select * from TMP_AUD WHERE CRE_BY=? AND AUD_NUMBER=? ",
public downloadFromOfflineDB(db,testSO){
var observableBatch = [];
observableBatch.push(db.executeSql("select * from TMP_AUD WHERE CRE_BY=? AND AUD_NUMBER=? ",
[localStorage.getItem("user_name"), testSO.auditNumber]).then(
response => {
this._util.logData('In downloadPendingInstancesForSyncFromOfflineDB- folder'+response.rows.length+'ID= '+response.rows.item(0).FLD_NUMBER);
if (response && response.rows && response.rows.length > 0) {
if (response && response.rows && response.rows.length > 0) {
var FLD_NUMBER = response.rows.item(0).FLD_NUMBER;
var folderArray = []
observableBatch.push(db.executeSql("select * from TMP_FOLDER WHERE CRE_BY=? AND FLD_NUMBER=? ",
[localStorage.getItem("user_name"), FLD_NUMBER]).then(
a => {
this._util.logData('In downloadPendingInstancesForSyncFromOfflineDB-TMP_FOLDER'+a.rows.length);
if (a && a.rows && a.rows.length > 0) {
for (let i = 0; i < a.rows.length; i++) {
var folderObj = {
folderName: a.rows.item(i).FLD_NAME,
files:[]
}
var FLD_NAME = a.rows.item(i).FLD_NAME
this._util.logData('In downloadPendingInstancesForSyncFromOfflineDB-TMP_FOLDER '+FLD_NAME);
observableBatch.push( db.executeSql("select * from TMP_FILES WHERE CRE_BY=? AND FLD_NAME=? ",
[localStorage.getItem("user_name"), FLD_NAME]).then(
b => {
this._util.logData('In downloadPendingInstancesForSyncFromOfflineDB-TMP_FILES'+b.rows.length);
var fileArray = [];
if (b && b.rows && b.rows.length > 0) {
for (let j = 0; j < b.rows.length; j++) {
var fileSO = {
compliance: b.rows.item(j).COMPLIANCE,
remarks: b.rows.item(j).REMARKS,
fileName: b.rows.item(j).FILE_NAME,
title: b.rows.item(j).TITLE
}
);
fileArray.push(fileSO);
}}
folderObj.files=fileArray;
}).catch(
e => {
this._util.logData('For sync error'+JSON.stringify(e));
return Observable.throw("An error occurred during sync");
})
);
folderArray.push(folderObj);
}}
}).catch(
e => {
this._util.logData('For sync error'+JSON.stringify(e));
return Observable.throw("An error occurred during sync");
})
);
}
}
testSO.folderArray = folderArray;
this._util.logData('Candidate for selected for sync' + JSON.stringify(testSO));
})
);
return Observable.forkJoin(observableBatch);
}
第一个方法正在执行,而第二个方法在第一个执行完成之前返回,我不会让我的对象testSO填充。有人能指导我并告诉我我做错了什么。我使用了observable fork Join。看起来你调用的是
observable.forkJoin(observableBatch)
,只有一项-结果是db.executeSql
。稍后添加更多项时,它不会影响forkJoin
observeBatch是我将所有db.executeSql函数推送到的数组。因为db.executeSql返回承诺。所以我用下面的方式完成了observableBatch.push(Observable.fromPromise(db.executeSql(…))并尝试了。但是没有luck@jslearn07,再次说明,当forkJoin已经被调用时,您正在向observatebatch
添加承诺,这样它就不会等待它们了。那么您的意思是说我应该在最后一条语句的末尾返回observatable.forkJoin(observatebatch)?我遵循这个例子。如果可能的话,请您发布代码。@jslearner07,实际上我是说,当您的下一个任务取决于上一个任务的结果时,您不能使用forkJoin
。如果你在计算机上重现这个问题,或者类似的问题,那么我将为你解决。也许你所说的是对的。然而,我将我的实现改为这样,我为3个表保留了一个公共标识符。我使用这个公共标识符对3个表进行顺序查询,每个表都返回promise,最后我将通过比较父id在内部合并所有结果并创建json。这样我就得到了上面的impl工作。谢谢
public getFiles(testSO) {
return Observable.create(observer => {
this.platform.ready().then(() => {
this.sqlite.create({
name: 'offline.db',
location: 'default'
}).then((db: SQLiteObject) => {
this.downloadFromOfflineDB(db, testSO).subscribe(c => {
observer.next(c[0]);//This is undefined
observer.complete();
},
error => {
observer.error("An error occurred sync files.");
});
});
});
});
}