因此，我一直在努力实现从速卖通获得的蓝牙指纹读取器。我已经花了两天的时间进行操作，目前处于可以连接和发送内容的阶段。但是我不知道如果没有适当的UUID和服务怎么办。这是外围设备信息
{
"characteristics":[
{
"properties":{
"Read":"Read"
},
"characteristic":"2a00",
"service":"1800"
},
{
"properties":{
"Read":"Read"
},
"characteristic":"2a01",
"service":"1800"
},
{
"properties":{
"Read":"Read"
},
"characteristic":"2a04",
"service":"1800"
},
{
"properties":{
"Read":"Read"
},
"characteristic":"2a29",
"service":"180a"
},
{
"properties":{
"Read":"Read"
},
"characteristic":"2a24",
"service":"180a"
},
{
"properties":{
"Read":"Read"
},
"characteristic":"2a25",
"service":"180a"
},
{
"properties":{
"Read":"Read"
},
"characteristic":"2a27",
"service":"180a"
},
{
"properties":{
"Read":"Read"
},
"characteristic":"2a26",
"service":"180a"
},
{
"properties":{
"Read":"Read"
},
"characteristic":"2a28",
"service":"180a"
},
{
"properties":{
"Read":"Read"
},
"characteristic":"2a23",
"service":"180a"
},
{
"properties":{
"Read":"Read"
},
"characteristic":"2a2a",
"service":"180a"
},
{
"properties":{
"Write":"Write",
"Read":"Read"
},
"characteristic":"49535343-6daa-4d02-abf6-19569aca69fe",
"service":"49535343-fe7d-4ae5-8fa9-9fafd205e455"
},
{
"descriptors":[
{
"value":null,
"uuid":"2902"
}
],
"properties":{
"Notify":"Notify",
"Write":"Write"
},
"characteristic":"49535343-aca3-481c-91ec-d85e28a60318",
"service":"49535343-fe7d-4ae5-8fa9-9fafd205e455"
},
{
"descriptors":[
{
"value":null,
"uuid":"2902"
}
],
"properties":{
"Indicate":"Indicate",
"Notify":"Notify"
},
"characteristic":"fff1",
"service":"fff0"
},
{
"properties":{
"Write":"Write",
"WriteWithoutResponse":"WriteWithoutResponse"
},
"characteristic":"fff2",
"service":"fff0"
}
],
"services":[
{
"uuid":"1800"
},
{
"uuid":"180a"
},
{
"uuid":"49535343-fe7d-4ae5-8fa9-9fafd205e455"
},
{
"uuid":"fff0"
}
],
"advertising":{
"txPowerLevel":2,
"serviceData":{
},
"serviceUUIDs":[
"fff0"
],
"localName":"FGT19100003",
"isConnectable":true,
"manufacturerData":{
"bytes":[
2,
1,
2,
12,
9,
70,
71,
84,
49,
57,
49,
48,
48,
48,
48,
51,
3,
3,
240,
255,
2,
10,
2,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0
],
"data":"AgECDAlGR1QxOTEwMDAwMwMD8P8CCgIAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA=",
"CDVType":"ArrayBuffer"
}
},
"rssi":-48,
"id":"88:1B:99:25:AD:54",
"name":"FGT19100003"
但是我看到只有4个字符的特征。据我所知，它们的格式为XXXXXXXX-XXXX-XXXX-XXXX-XXXXXXXXXXXX。我已经在JAVA中获得了一些源代码，如果有人想看一下，我可以上载（我安装了它，它可以扫描指纹并将其BitMatrix返回给我。）（src java项目：{{ 3}}）
这是我在RN中使用的功能的代码。
test(peripheral: { connected: any; id: string; }) {
if (peripheral) {
if (peripheral.connected) {
BleManager.disconnect(peripheral.id);
} else {
BleManager.connect(peripheral.id).then(() => {
let peripherals = this.state.peripherals;
let p = peripherals.get(peripheral.id);
if (p) {
p.connected = true;
peripherals.set(peripheral.id, p);
this.setState({ peripherals });
}
console.log('Connected to ' + peripheral.id);
setTimeout(() => {
BleManager.retrieveServices(peripheral.id).then((peripheralInfo) => {
console.log(JSON.stringify(peripheralInfo))
var service = '49535343-6daa-4d02-abf6-19569aca69fe';
var crustCharacteristic = '49535343-fe7d-4ae5-8fa9-9fafd205e455';
setTimeout(() => {
BleManager.startNotification(peripheral.id, service, crustCharacteristic).then(() => {
console.log('Started notification on ' + peripheral.id);
setTimeout(() => {
BleManager.write(peripheral.id, service, crustCharacteristic, [0]).then(() => {
console.log('write succes');
});
}, 500);
}).catch((error) => {
console.log('Notification error', error);
});
}, 200);
});
}, 900);
}).catch((error) => {
console.log('Connection error', error);
});
}
}
}
有人有经验吗？我还得到一张表，该表指出我认为其中包含所有可能的命令（https://filebin.net/7jeo1o0lkgd8x2oy）。
答案 0 :(得分：1)
BLE服务和特征是Bluetooth SIG设置的标准服务。可以分别在here和here中找到它们。该列表中的大多数特性都是标准特性，我敢打赌唯一具有128位UUID（0x49535343-fe7d-4ae5-8fa9-9fafd205e455）的特性是指纹数据的自定义服务，即0xFFF0服务。它们每个都有一些带有指示/通知的特征，所以我想这些将具有数据。 0xFFF0不是标准的UUID，但通常在示例代码和实现示例中使用，但通常不应该在运输产品中使用，但它具有与其他产品相同的正确性。我会尝试制作一个超大的数据缓冲区，并设置特征以在两种服务中进行通知/指示，然后进行指纹测试，然后看看您得到了什么。
或者，您可以使用Nordic（NRFconnect）或Cypress（CySmart）的手机BLE调试应用程序之一进行相同的测试，并通过使用手机执行相同的测试来了解正确的特性。< / p>