Why is debugging not working in Visual Studio Code?

Question

JavaScript

Error when launching the debugger in Visual Studio Code:
Debugger listening on ws://127.0.0.1:25010/79b1de70-5458-4f9a-af24-90aab2d4060f
For help, see: https://nodejs.org/en/docs/inspector .
Until recently, everything worked well, but recently I installed an antivirus, and debugging stopped working after it. Tried turning off antivirus and firewall. What can be done?

Answer 1

Look at what has lately been in the "Quarantine" of the antivirus.
There can be a ton of problems. As an example, the Node JS executable files may have been corrupted.

Answer 2

So this, there is no error in the debugger output, this is a regular output. Here I brought Hello world

C:\Program Files\nodejs\node.exe --inspect-brk=44476 index.js 
Debugger listening on ws://127.0.0.1:44476/fe1dcce1-77f5-4ca5-88bd-f2ef15262148
For help, see: https://nodejs.org/en/docs/inspector
Hello world

those. These logs show that my code was executed.
The problem is in your code which is an eternal loop. You are writing a loop that runs while both of these variables are equal to one, and you always have them equal to one, so there is an eternal loop.
If you take into account what you have in the comments. So there are no promt and alert functions in the node.

Answer 3

if (mysqli_num_rows($res)>0)
   echo 'ДА';
else 
   echo 'НЕТ';

Answer 4

Good evening.
From mysqli_query documentation
It turns out that
is incorrect.
From the same documentation

if (mysqli_query($link, "CREATE TEMPORARY TABLE myCity LIKE City") === TRUE) {
    printf("Таблица myCity успешно создана.\n");
}