Question

我正在Flutter中开发WebApp，但是当我单击<form #formData="ngForm" (ngSubmit)="formData.valid && submit(formData, cardIndex); this.debug.log(formData)" > ... <buttontype="button" (click)="showPartPicker(inputPart)"></button> <input ngModel #inputPart type="text" name="warehouse_part_code" required /> ...或tel:987654321链接进入未找到页面时，要在移动设备中打开默认应用，我使用的是url_launcher依赖关系，但是在后台单击WebView时，也运行相同的链接并转到未找到的页面。

如何处理此任务？

Answer 1

navigationDelegate: (NavigationRequest request) {
  if(request.url.contains("mailto:")) {
    launch(request.url);
    return NavigationDecision.prevent;
  }
  else if (request.url.contains("tel:")) {
    launch(request.url);
    return NavigationDecision.prevent;
  }
},

Answer 2

添加一些细节，需要url_launcher插件

import 'package:webview_flutter/webview_flutter.dart';
import 'package:url_launcher/url_launcher.dart';

  _launchURL(url) async {
    if (await canLaunch(url)) {
      await launch(url);
    } else {
      throw 'Could not launch $url';
    }
  }

in Widget-build ...

WebView(
        initialUrl: 'http://example.com',
        navigationDelegate: (NavigationRequest request) {
          if (request.url.contains("mailto:")) {
            _launchURL(request.url);
            return NavigationDecision.prevent;
          } else if (request.url.contains("tel:")) {
            _launchURL(request.url);
            return NavigationDecision.prevent;
          }
          return NavigationDecision.navigate;
        },
)

如何在WebView Flutter中允许mailto和tel URL方案？

2 个答案: