xmlhttprequest Обработка ошибок
const xhttp = new XMLHttpRequest();
xhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
console.log( xhttp.responseText );
}
};
// Error Handling:
xhttp.onerror = function(error){
console.error( error );
}
xhttp.open("GET", "filename", true);
xhttp.send();
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