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Sergio [31]
2 years ago
6

How to trigger watchers on initialization in vue.js ?

Computers and Technology
1 answer:
ale4655 [162]2 years ago
3 0

Answer:

variable when the component is mounted. This triggers the watchers which then triggers an emit i have in the watcher. I do not want the emit to happen when the variable is only being initialized. The variable is from the data section of the vue component.

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Hidden costs of computers into our schools
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There are certainly many hidden costs, and you need to find them out. I am listing some. The GPU can cost a lot if you are using them for complex computing like in the case of Bitcoin. You need to pay heavy electricity bills as well. And if you want to install the webserver then as well, you need to keep your computer open all the time, and pay a good sum as an electricity bill. Many more hidden costs can be found. And one out of above is used in Schools, the webserver. Some more hidden costs can be Network cost, as the school is big, and you need to connect all through LAN, and at times we also need WAN set up. And these are another hidden cost. Various education licenses come for free, and smart classes cost as well. The video conferencing, VOIP, etc costs as well. Smart classes training by various computer training institute for teachers like one from adhesive.

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Please check the answer section.

6 0
4 years ago
What is a digital pen called<br><br>A digital pen is called a ____
steposvetlana [31]
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4 0
3 years ago
Read 2 more answers
Consider the recursive method whose definition appear below. Why? public static String mysteryString (String s){ if(s.length()==
Marizza181 [45]

Answer:

retupmoc

Explanation:

1.) Anwser will be retupmoc

because

public static String mysteryString(String s){

if(s.length() == 1){

return s;

}

else{

return s.substring(s.length() -1) + mysteryString(s.substring(0, s.length()-1));

}

}

In this program input is "computer" . So the function mysteryString(String s) it does

return s.substring(s.length() -1) + mysteryString(s.substring(0, s.length()-1));

so when it enters the first time ??s.substring(s.length() -1) and it will be give you 'r' then it calls the function recursively by reducing the string length by one . So next time it calls the mysteryString function with string "compute" and next time it calls return s.substring(s.length()-1)? + mysteryString(s.substring(0,s.length-1)) so this time it gives "e" and calls the function again recursively . It keeps on doing till it matched the base case.

so it returns "retupmoc".

6 0
4 years ago
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