Javascript 如何使用where子句从firestore中删除文档
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How to delete document from firestore using where clause
提问by Amit Sinha
var jobskill_ref = db.collection('job_skills').where('job_id','==',post.job_id);
jobskill_ref.delete();
Error thrown
抛出错误
jobskill_ref.delete is not a function
jobskill_ref.delete 不是函数
回答by Frank van Puffelen
You can only delete a document once you have a DocumentReferenceto it. To get that you must first execute the query, then loop over the QuerySnapshotand finally delete each DocumentSnapshotbased on its ref.
您只能在拥有文档后才能删除DocumentReference它。要获得它，您必须首先执行查询，然后循环QuerySnapshot并最后DocumentSnapshot根据其ref.
var jobskill_query = db.collection('job_skills').where('job_id','==',post.job_id);
jobskill_query.get().then(function(querySnapshot) {
querySnapshot.forEach(function(doc) {
doc.ref.delete();
});
});
回答by jamstooks
I use batched writesfor this. For example:
我为此使用批量写入。例如：
var jobskill_ref = db.collection('job_skills').where('job_id','==',post.job_id);
let batch = firestore.batch();
jobskill_ref
.get()
.then(snapshot => {
snapshot.docs.forEach(doc => {
batch.delete(doc.ref);
});
return batch.commit();
})
ES6 async/await:
ES6 异步/等待：
const jobskills = await store
.collection('job_skills')
.where('job_id', '==', post.job_id)
.get();
const batch = store.batch();
jobskills.forEach(doc => {
batch.delete(doc.ref);
});
await batch.commit();
回答by Jacob Hixon
the key part of Frank's answer that fixed my issues was the .refin doc.ref.delete()
该固定我的问题，坦率的回答的关键部分是.ref在doc.ref.delete()
I originally only had doc.delete()which gave a "not a function" error.now my code looks like this and works perfectly:
我最初只有doc.delete()which 给出了“不是函数”错误。现在我的代码看起来像这样并且完美运行：
let fs = firebase.firestore();
let collectionRef = fs.collection(<your collection here>);
collectionRef.where("name", "==", name)
.get()
.then(querySnapshot => {
querySnapshot.forEach((doc) => {
doc.ref.delete().then(() => {
console.log("Document successfully deleted!");
}).catch(function(error) {
console.error("Error removing document: ", error);
});
});
})
.catch(function(error) {
console.log("Error getting documents: ", error);
});
回答by HMagdy
And of course, you can use await/async:
当然，您可以使用 await/async：
exports.delete = functions.https.onRequest(async (req, res) => {
try {
var jobskill_ref = db.collection('job_skills').where('job_id','==',post.job_id).get();
jobskill_ref.forEach((doc) => {
doc.ref.delete();
});
} catch (error) {
return res.json({
status: 'error', msg: 'Error while deleting', data: error,
});
}
});
I have no idea why you have to get()them and loopon them, then delete()them, while you can prepare one query with where to delete in one step like any SQL statement, but Google decided to do it like that. so, for now, this is the only option.
我不知道为什么你必须get()它们并循环它们，然后delete()它们，而你可以像任何 SQL 语句一样准备一个查询，在一个步骤中删除哪里，但谷歌决定这样做。所以，就目前而言，这是唯一的选择。
回答by Luis Figueredo
delete(seccion: string, subseccion: string)
{
const deletlist = this.db.collection('seccionesclass', ref => ref.where('seccion', '==', seccion).where('subseccion', '==' , subseccion))
deletlist.get().subscribe(delitems => delitems.forEach( doc=> doc.ref.delete()));
alert('record erased');
}
