Android端app拉起微信小程序跳转提示“跳转失败” 不知道为何？ - WXOPEN Club

IWXAPI api = WXAPIFactory.createWXAPI(context, appId, false);
            WXLaunchMiniProgram.Req req = new WXLaunchMiniProgram.Req();
            req.userName = userId;
            if (TextUtils.isEmpty(path)) {
                req.path = “”;
            } else {
                req.path = path;
            }
            if (TextUtils.isEmpty(mini)) {
                req.miniprogramType = WXLaunchMiniProgram.Req.MINIPTOGRAM_TYPE_RELEASE;
            } else {
                req.miniprogramType = WXLaunchMiniProgram.Req.MINIPROGRAM_TYPE_PREVIEW;
            }
            api.sendReq(req);


微信版本7.0.6

appId: wx516f09ef989c1298
userId，req.userName: gh_b52e04a0100c
path=“”;
mini=“”;
req.miniprogramType = WXLaunchMiniProgram.Req.MINIPTOGRAM_TYPE_RELEASE;

compile ‘com.tencent.mm.opensdk:wechat-sdk-android-with-mta:+’

3 回复

wei321 楼6 年前

我也遇这个问题，但确认appid没用错，应用签名，大小写都试过，帮我看原因

package com.example.myapplication3;

import android.content.BroadcastReceiver;

import android.content.Context;

import android.content.Intent;

import android.content.IntentFilter;

import android.os.Bundle;

import android.util.Log;

import android.view.View;

import android.widget.Toast;

import androidx.appcompat.app.AppCompatActivity;

import com.tencent.mm.opensdk.constants.ConstantsAPI;

import com.tencent.mm.opensdk.modelbiz.WXLaunchMiniProgram;

import com.tencent.mm.opensdk.openapi.IWXAPI;

import com.tencent.mm.opensdk.openapi.WXAPIFactory;

public class MainActivity extends AppCompatActivity {

public static final String TAG = MainActivity.class.getSimpleName();

// APP_ID 替换为你的应用从官方网站申请到的合法appID

private static final String APP_ID = "wx82d0****fe25eb40";

// IWXAPI 是第三方app和微信通信的openApi接口

private IWXAPI api;

private BroadcastReceiver broadcastReceiver;

private void regToWx() {

// 通过WXAPIFactory工厂，获取IWXAPI的实例

api = WXAPIFactory.createWXAPI(this, APP_ID, true);

// 将应用的appId注册到微信

api.registerApp(APP_ID);

broadcastReceiver = new BroadcastReceiver() {

@Override

public void onReceive(Context context, Intent intent) {

// 将该app注册到微信

api.registerApp(APP_ID);

}

};

//建议动态监听微信启动广播进行注册到微信

registerReceiver(broadcastReceiver, new IntentFilter(ConstantsAPI.ACTION_REFRESH_WXAPP));

}

// 调启我的小程序

private void goMiniProgram(){

WXLaunchMiniProgram.Req req = new WXLaunchMiniProgram.Req();

req.userName = "wxbd6534d10790****"; // 填小程序原始id

req.path = "pages/index/index"; //拉起小程序页面的可带参路径，不填默认拉起小程序首页

req.miniprogramType = WXLaunchMiniProgram.Req.MINIPTOGRAM_TYPE_RELEASE; //.MINIPTOGRAM_TYPE_RELEASE; // 可选打开开发版，体验版和正式版

api.sendReq(req);

}

@Override

protected void onCreate(Bundle savedInstanceState) {

super.onCreate(savedInstanceState);

setContentView(R.layout.activity_main);

// 注册到微信

this.regToWx();

}

public void clickToWX(View view){

Log.i(TAG,TAG+"clickToWX");

Toast.makeText(this, "点击跳转到小程序", Toast.LENGTH_LONG).show();

this.goMiniProgram();

}

@Override

protected void onDestroy() {

super.onDestroy();

Log.i(TAG,TAG+"onDestroy***********************************************");

if(broadcastReceiver!= null){

unregisterReceiver(broadcastReceiver);

}

lidong2 楼6 年前

你好，可以先对照文档排除一下问题：https://open.weixin.qq.com/cgi-bin/showdocument?action=dir_list&t=resource/res_list&verify=1&id=21526646437Y6nEC&token=&lang=zh_CN

chengguiying3 楼5 年前

我这边解决了 appid用错了

Android端应该用应用appid而不是小程序appid来WXAPIFactory.createWXAPI(context, appId, false);

但是ios端只需要小程序原始id 这两个appid（应用appid和小程序appid）都不需要用到

但是什么错误提示都没有！！！就提示一个“跳转失败”！！！

这点设计的真的太次了！！！