Load Js Script After Another One Is Loaded
Solution 1:
If you use jQuery, there is a very easy way to do this through getScript function. Just add the part of script that you need to be executed after the load and pass it as a parameter to the function.
$.getScript( "app.js" )
.done(function( script, textStatus ) {
console.log( textStatus );
//your remaining code
})
.fail(function( jqxhr, settings, exception ) {
//script fail warning if you want it
});
Solution 2:
Defer instructs the contents of the script tag to not execute until the page has loaded. So I would actually expect alert popup first and then app.js loaded.
Without defer the sripts should be synchronously loaded in the order you put them.
Solution 3:
You can achieve that by async-await
create a JavaScript function wrapping them both.
consttask1 = ()=>{
// app.js content
}
consttask2 = ()=>{
// alert(1);
}
asyncfunctionasyncCall(){
awaittask1(); // then task2 want start until the task1 finishestask2();
};
asyncCall();
You can use async functions to execute one function after another one asynchronously
additional resources: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Statements/async_function
you can also solve the same problem using JavaScript Promises but I prefer async-await functions. they are less painful.
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