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site with youtube api player cannot watch video from phone

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On my website there is a youtube api player and users can watch the video by entering the link in the input field. everything works fine from the computer where the program itself is located. But I can’t play the video from other devices.

I thought it might be a cookie problem because every time I tried to upload a video from my phone, awarning message popped up.

Is the problem with playing videos really related to cookies? If yes, how can this be fixed in asp.net core 7?

I'm a newbie, so don't judge me harshly if I said something wrong

function loadVideoPlayer (){    player = new YT.Player('player', {        height: '360',        width: '640',        events: {'onReady': onPlayerReady,'onStateChange': onPlayerStateChange,        },        playerVars: {'origin': 'https://192.168.1.39:7269'        },    });}// 4. The API will call this function when the video player is ready.function onPlayerReady(event) {    event.target.playVideo();}// 5. The API calls this function when the player's state changes.//    The function indicates that when playing a video (state=1),//    the player should play.var done = false;function onPlayerStateChange(event) {    if (event.data == YT.PlayerState.PLAYING && !done) {        done = true;    }}function stopVideo() {    player.stopVideo();}document.getElementById("videoSrc").addEventListener("click", function () {    var userVideoUrl = document.getElementById("userUrl").value;    var userVideoId = getVideoId(userVideoUrl);    if (userVideoId == "") {        alert('Enter correct url!');    } else {        connection.invoke("SendVideoData", userVideoId).catch(function (err) {            return console.error(err.toString());        });    }});connection.on("ReceiveVideoData", function (userVideoId) {    player.loadVideoById(userVideoId);});function getVideoId(url) {    var videoId = "";    if (url.indexOf("youtube.com") !== -1) {        videoId = url.split("v=")[1];    } else if (url.indexOf("youtu.be") !== -1) {        videoId = url.split(".be/")[1];    }    return videoId;}

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