Answer the question
In order to leave comments, you need to log in
Laptop freezing during games with and without sound, what to do?
Hello everyone, I would like to ask you
. I started freezing from the moment I just turned off the laptop through the start, everything is as it should, I turn it on the next day.
The first thing I noticed was a freeze (rare) of programs like Opera, Zoom, and so purely if without work costs. I didn't do anything turned it off! Then a Windows update came out and installed it and everything seemed to be ok as soon as I started playing games such as
Samp (GTA San Andreas multiplayer)
My Summer Car
I played MSC calmly, everything was ok as soon as the sharp sounds like Trrr-Trr began to jam, I turned off the game (The temperature of the processor was 75-85% with games 15-25% without games)
I went into another game to play and then bam Hanged !!
Nothing works but note The laptop was warm and it was 50-100fps smoothly
As a result (5-10 minutes of the game freezing, it happens every other time) please help. Info below!!!
Notebook character
Brand: Asus Vivobook Max
Processor: Intel Celeron 128MB (may be wrong) 1.10GHz 1.10GHz
(2-Core)
64-bit System
Windows 10 Pro
RAM (RAM): 4.00GB of which available
3.86GB Model: X541NA
Answer the question
In order to leave comments, you need to log in
because AJAX is an asynchronous call - this time
getData() will always return a void value. - these two
need either a synchronous call, or pass the search value to another function and hang a handler there that processes the data.
You can test like this
function getData() {
console.log('1. вхождение в функцию');
$.ajax({
url: '../elem/search.php',
type: 'POST',
cache:false,
success : function(search) {
console.log('3. результат вызова');
console.log(search);
return search;
}
})
console.log('2. выход из функции');
}
var data = getData();
console.log(data);
function getData() {
console.log('1. вхождение в функцию');
var result;
$.ajax({
url: '../elem/search.php',
type: 'POST',
cache:false,
async: false,
success : function(search) {
result = search;
console.log('3. результат вызова');
console.log(data);
}
})
console.log('2. выход из функции');
return result;
}
var data = getData();
console.log(data);
Didn't find what you were looking for?
Ask your questionAsk a Question
731 491 924 answers to any question