Trying to use the javascript variabe as php variable and store it in mysql

Question

I have two JS variables I want to use them as php variables and store them in the database. Here is sample

var apiKey = 'apikey';
var pageToken = '';
var numOfResult = 0;
var maxResults = 200;
function searchVideo(){
   var separator = ",";
   $.getJSON('https://www.googleapis.com/youtube/v3/playlistItems?part=snippet&maxResults=50&pageToken=' + pageToken + '&playlistId=PLrEnWoR732-BHrPp_Pm8_VleD68f9s14-&key=apikey&callback=?',function(data){
        var l = data.items.length;
        pageToken = data.nextPageToken;
        numOfResult += l;
        var itemUrl = '';
        for(var i = 0; i < l; i++) {
                if( i == 0) {
                        separator = ',';
                }
                else {
                        separator = ',';
                }
                var videoid = data.items[i].snippet.resourceId.videoId;
                var title = data.items[i].snippet.title;
                console.log(videoid);

        }

        if( numOfResult < maxResults) {
                searchVideo();
        }

    });
}

How can I store variables title and videoid into the database using php mysql

You can use [`ajax`](https://stackoverflow.com/questions/5004233/jquery-ajax-post-example-with-php) for that — Carl Binalla, Dec 18 '17 at 06:48
I dont have much idea about that and data is in loop is it possible to use that — sravya naidu, Dec 18 '17 at 06:50
[You can pass array to ajax](https://stackoverflow.com/questions/8890524/pass-array-to-ajax-request-in-ajax). Store what you need to store in an array and pass that array — Carl Binalla, Dec 18 '17 at 06:51

score 0 · Accepted Answer · answered Dec 18 '17 at 07:07

0

You can use ajax for this, here is how that would work.

Create an endpoint in PHP where you will accept this data via POST and insert it into DB.
Send an ajax request from JS to that endpoint to store data into the DB.

So, if the php endpoint for example is add.php, then Ajax code will look something like this(this is just an example, look into AJAX for better understanding):

$.ajax({
  method: "POST",
  url:    "add.php",
  data: { title: title, videoid: videoid }
})
.done(function(data) {
  alert( "Data Saved");
});

answered Dec 18 '17 at 07:07

mega6382

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by Using [`POST`](http://php.net/manual/en/reserved.variables.post.php) – mega6382 Dec 18 '17 at 07:14
yes am using post but how can i print the title and videoid to save them using mysql php – sravya naidu Dec 18 '17 at 07:23
The same way you would if the data was coming from a form instead. – mega6382 Dec 18 '17 at 07:27

score 0 · Answer 2 · answered Dec 18 '17 at 07:29

You can add value in the input hidden field and his name will be separate by name key and value.Hidden field be append in the form using any tag, id or class.I have used form tag here. You can submit that form and get all data in php POST variable. See below code snippet:

var apiKey = 'apikey';
var pageToken = '';
var numOfResult = 0;
var maxResults = 200;
function searchVideo(){
         var separator = ",";
         $.getJSON('https://www.googleapis.com/youtube/v3/playlistItems?part=snippet&maxResults=50&pageToken=' + pageToken + '&playlistId=PLrEnWoR732-BHrPp_Pm8_VleD68f9s14-&key=apikey&callback=?',function(data){

                var l = data.items.length;
                pageToken = data.nextPageToken;
                numOfResult += l;
                var itemUrl = '';
                for(var i = 0; i < l; i++) {
                        if( i == 0) {
                                separator = ',';
                        }
                        else {
                                separator = ',';
                        }
                        var videoid = data.items[i].snippet.resourceId.videoId;
                        var title = data.items[i].snippet.title;
                        console.log(videoid);

                        var input_data = "<input type='hidden' value='" + videoid + "' name='videoid[" + i + "]'><input type='hidden' value='" + title + "' name='title[" + i + "]'>";
                        $("form").append(input_data);

                }

                if( numOfResult < maxResults) {
                        searchVideo();
                }
        });
}

This code contains a cross-site-scripting (XSS) vulnerability — Erik van Velzen, Dec 18 '17 at 07:34
Yes it is just a workaround to the solution needed. One needs to apply security steps to prevent any injection/attack at browser/server level while processing form. — Patrick R, Dec 19 '17 at 04:59

Trying to use the javascript variabe as php variable and store it in mysql

2 Answers2