Thursday 26 October 2017

Asynchronous Javascript Variable Overwrite

itemprop="text">

The code has an issue that the
variable gets over written when the asynchronous function is called. How can it be
fixed?




Code:



for
(x in files) {

asynchronousFunction(var1, var2,
function(){
console.log(x.someVaraible);
});


}



Now
the issue is that when the callback function in the asynchronousFunction is called the
x.files variable has been updated to the next varaible in the json array files. I want
that the variable should contain the previous
value.



The number of variables in the callback
function can not be changes, so variable name can not be sent in the call back
function.



Answer




The problem wtih using 'local' variables in
javascript is that your variables actually have function scope, rather than block scope
- like java and c# etc has



One way to get around
this is using let which has block scope, but only firefox
supports this currently.



So this code would work
in firefox only :




for
(var x in files) {
// This variable has BLOCK scope
let file =
files[x];
asynchronousFunction(var1, var2, function(){

console.log(file.someVaraible);
});

}



For
other browsers, the alternative is to use
closures



for (var x in files)
{
var file = files[x];
asynchronousFunction(var1, var2,
(function(file){
return function() {

console.log(file.someVaraible);
};
})(file);

}



Another
way you could do this is using map/forEach, assuming that the datatype of files is an
array.



files.forEach(function(file)
{
asynchronousFunction(var1, var2, function(){

console.log(file.someVaraible);

});
});



If
it's not an array, then you can always use this technique



 [].forEach.call(files,
function(file) {
asynchronousFunction(var1, var2, function(){

console.log(file.someVaraible);

});
});



The
fuller way of writing this is of
course



Array.prototype.forEach.call(files,
function(file) {
// As
before


But I feel
[].forEach is nicer on the eyes



No comments:

Post a Comment

php - file_get_contents shows unexpected output while reading a file

I want to output an inline jpg image as a base64 encoded string, however when I do this : $contents = file_get_contents($filename); print &q...