我有 2 个具有相同结构的集合,我想将文档从一个复制到另一个。当我第一次运行它时,它会复制其中一个项目,如果我再运行它多次,它就不会再复制任何东西了。exports.copyDocs = functions.https.onRequest((request, response) => { let terminalIdToCopy = `123`; // terminal to be copied let terminalId = `456`; // terminal for the copies to be added let docsToCopy = db.collection(`/terminal/${terminalIdToCopy}/folderA/`); let docsNewDir = db.collection(`/terminal/${terminalId}/folderB/`); docsToCopy.get().then(snapshot => { let result = []; snapshot.forEach(doc => { result.push(doc.data()); let promise = docsNewDir.doc(doc.id).set(doc.data()) }); response.status(200).send(`Copied items: ${JSON.stringify(result)}. From terminal ${terminalIdToCopy} to terminal ${terminalId}`); }).catch(error => { response.status(500).send(`Error: ${error}`); });});我也试过添加一个 .then (不打印日志消息):docsNewDir.doc(doc.id).set(doc.data()) .then( i => { console.log(`i (${i})`)}, i2 => { console.log(`i2: ${i2}`);}) .catch(error => console.log(`Could NOT copy doc with id ${doc.id}: ${error}`) );而且我还尝试将所有承诺结合起来:docsToCopy.get().then(snapshot => { let result = []; let promises = []; snapshot.forEach(doc => { result.push(doc.data()); promises.push(docsNewDir.doc(doc.id).set(doc.data())); }); Promise.all(promises).then(values => {console.log("all done")}); // this won't print response.status(200).send(`Copied items: ${JSON.stringify(result)}. From terminal ${terminalIdToCopy} to terminal ${terminalId}`); }).catch(error => { response.status(500).send(`Error: ${error}`); });在result包含了应被复制的文件。
1 回答
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墨色风雨
TA贡献1853条经验 获得超6个赞
您使用 Promise.all() 走在正确的轨道上,但您的最终解决方案仍然需要在所有承诺解决后才发送响应。您现在所拥有的仍然在任何事情完全解决之前发送响应,这会在它们完成之前终止该功能。你应该做更多这样的事情:
Promise.all(promises).then(values => {
console.log("all done")
response.status(200).send(`Copied items: ${JSON.stringify(result)}. From terminal ${terminalIdToCopy} to terminal ${terminalId}`);
});
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