可以在频道上没有主播同伴吗?

时间:2018-06-01 06:14:56

标签: hyperledger-fabric

在官方结构样本中的balance-transfer sample (typescript version)中,客户端没有在通道上设置任何锚定对等体。我检查了客户端中exports.getQRCode = functions.https.onRequest((req, res) => { admin.database().ref('lockers/LocationA/empty').limitToFirst(1).once("value", snap=> { // Get the name of the first available door and use a transaction to ensure it is not occupied console.log('QR Code for door:',snap.val()); var door = Object.keys(snap.val())[0]; console.log('door:',door); // var door = snap.key(); var occupiedRef = admin.database().ref('lockers/LocationA/occupied/'+door); occupiedRef.transaction(currentData=> { if (currentData === null) { console.log("Door does not already exist under /occupied, so we can use this one."); return snap.child(door).val(); // Save the chosen door to /occupied } else { console.log('The door already exists under /occupied.'); return nil; // Abort the transaction by returning nothing } }, (error, committed, snapshot) => { console.log('snap.val():',snap.val()); if (error) { console.log('Transaction failed abnormally!', error); res.send("Unknown error."); // This handles any abormal error } else if (!committed) { console.log('We aborted the transaction (because the door is already occupied).'); res.redirect(req.originalUrl); // Refresh the page so that the request is retried } else { // The door is not occupied, so can be given to this user admin.database().ref('lockers/LocationA/empty/'+door).remove(); // Delete the door from /empty console.log('QR Code for door:',snapshot.val()); var qrCodesForDoor = snapshot.val(); res.send(qrCodesForDoor); // Send the chosen door as the response } }); }); }); 对象的_anchor_peers属性是否为空。

那么,通道上没有主播同伴可以吗?或者,当我们没有明确指定它们时,是否会自动设置默认锚点对等体?

1 个答案:

答案 0 :(得分:3)

锚定对等体为对等体启用跨组织通信。没有默认的锚点对等体,您必须明确声明一个。在频道上没有一个仍然允许同一个组织的同伴能够进行通信。