Solution:
You have to create a function
to call again within jQuery.post response. try the below code.
// build data
var dataURL = dataURLs[0],
data = {
email: email,
name: name,
linkforsharedURL: linkforsharedURL
};
SendDataToPhp( data );
function SendDataToPhp(){
// send data
jQuery.post("<?php echo admin_url('admin-ajax.php'); ?>", data, function(response) {
if (response.share_id !== undefined) {
var pattern = new RegExp('(share_id=).*?(&|$)'), shareUrl = window.location.href;
if (shareUrl.search(pattern) >= 0) {
shareUrl = shareUrl.replace(pattern, '$1' + response.share_id + '$2');
linkforsharedURL = shareUrl;
data.linkforsharedURL = 'updated value to send';
SendDataToPhp( data );
}
}
});
}