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我有一个使用 instascan 扫描 QR 码的应用程序,但是每次您在 android 手机上扫描 QR 码时,都会弹出带有解码结果的警报。例如,如果我生成了一个带有“我喜欢香蕉”的二维码,那么警报会显示“https://mywebsite.com 说我喜欢香蕉”。

有什么方法可以阻止此警报显示?它不会发生在我的笔记本电脑或平板电脑上,只有我的手机。如果您想查看,这是我的代码:

    <video id="preview" visible="true"></video>
    <script type="text/javascript">
        let scanner = new Instascan.Scanner({ video: document.getElementById('preview') });
        //When a QR code is detected
        scanner.addListener('scan', function (content) {
            //If it is a url
            if (content.match(/^http?:\/\//i)) {
                //Redirect to new page
                window.open(content);
            } else {                    
                var options = {};
                options.url = "CameraList.aspx/GetContent";
                options.type = "POST";
                options.data = JSON.stringify({ content });
                options.dataType = "json";
                options.contentType = "application/json";
                options.success = function (result) { };
                options.error = function (err) { };

                $.ajax(options);
            }
        });
    </script>

任何帮助表示赞赏。

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1 回答 1

0

我根据这篇文章弄清楚了:Stop alert javascript popup in webbrowser c# control

我想从一个名为 GetContent 的方法返回一个字符串,并仅在字符串不为空时向用户发出警报:

        //write a new function for alert that expects a boolean value
        var fnAlert = alert;
        alert = function (message, doshow) {
        //only if you pass true show the alert
            if (doshow === true) {
                fnAlert(message);
            }
        }
        let scanner = new Instascan.Scanner({ video: document.getElementById('preview') });
        //When a QR code is detected
        scanner.addListener('scan', function (content) {
            //If it is a url
            if (content.match(/^http?:\/\//i)) {
                //Redirect to new page
                window.open(content);
            } else {                    
                var options = {};
                options.url = "CameraList.aspx/GetContent";
                options.type = "POST";
                options.data = JSON.stringify({ content });
                options.dataType = "json";
                options.contentType = "application/json";
                options.success = function (result) {
                    //If my c# method GetContent returns a value
                    if (result != "") {
                        //then alert the user by passing true
                        alert(result, true);
                    }
                };
                options.error = function (err) { };

                $.ajax(options);
            }
        });
于 2021-03-11T23:01:13.333 回答