Why is this function not visible in this other function? [closed] - javascript

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I'm trying to use a function inside another function. Right now, setPanodatas() appears as not defined:
function setPoints(obj, prop) {
// Some code
if (unzippedNewPanodatas)
unzippedNewPanodatas = unzippedNewPanodatas.slice(scope.panodatas.length)
_.each(unzippedNewPanodatas, function(point) {
var Panodata = AV.Object.extend('PanoramaData')
var panodata = new Panodata()
obj[prop].push(setPanodatas(panodata.toJSON(), point))
})
}
function setPanodata(panodata, point) {
panodata.set('index', point.index)
panodata.set('x', offsetX(point.x))
panodata.set('y', offsetY(point.y))
panodata.set('roomModelId', scope.pano.id)
panodata.set('panoDataRotate', 0)
panodata.set('differentLayout', false)
panodata.set('panoCount', 6)
panodata.set('type', 'VRoom')
console.log('PANODATA', panodata)
return panodata
}
Why isn't setPanodata() visible inside setPoints()?

It is because setPanodatas is undefined.
Rather it is setPanodata, you added s.
function setPanodata(panodata, point)

Related

Req.params undefined [closed]

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So i have this code
router.get('/invite/:invitecode', async(res,req) => {
return console.log(req.params)
if(!req.params.invitecode) {
return res.render('404')
}
if(!req.user) {
res.redirect(`/login?state=invite/${req.params.invitecode}`)
} else {
res.render('join')
}
})
And when I go to
http://localhost:3000/invite/fs
It is supposed to return fs as a param, but instead, it console.logs undefined
Why is this happening?
It does not work because you are accessing the Response object, you should flip the arguments
from:
async(res,req)
to
async(req,res)

Get value from an array using find in JavaScript [closed]

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Closed 11 months ago.
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I've got an array:
test = [{'id':'73','name':'bar', 'contact': [{'name':'barFoo','tel':'3333'}]},{'id':'45','name':'foo', 'contact':[]}]
I try to get contact.tel in first object with 'id':'73'
I'm using find method :
let contactTel = test.find(x => x.id === '73').contact.tel;
but it dosen't work. What I'm doing wrong ?
it should be like this
const test = [{'id':'73','name':'bar', 'contact': [{'name':'barFoo','tel':'3333'}]},{'id':'45','name':'foo', 'contact':[]}]
let contactTel = test.find(x => x.id === '73').contact[0].tel;
console.log(contactTel)

javascript for loop does not work [closed]

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Closed 6 years ago.
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I'm creating a site and i needed to do some js. I'm not that good with it but tought i would figer it out. Not. I created a for loop but it does not run.
function order(user,product){
var index;
for(var i = 0; i<users.lenght; i++){
if(user == users[i]){
index = i;
break;
}
}
var budget = budgets[index];
alert(budget);
}
the creation of the users and budgets arrays are done with php and after checking with alert() it was how it should be.
Can anyone help me please?
lenght is spelt length. The misspelt property does not exist, so it undefined, which is equivalent to 0.

Return the first "n" count of characters of a string ( Javascript ) [closed]

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Closed 7 years ago.
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I've used this :
String.Prototype.left = function left(count){
return this.substr(0,count);
}
And apply it this way :
var string = "Hello Stack!";
console.log(string.left(5));
And the console tells me :
TypeError: example.left is not a function
console.log(example.left(5));
How can I fix it? And where is the problem?
String.prototype.left = function(index){
return this.substring(0,index);
}
var str = "Hello World";
console.log(str.left(5));

$(...).offest is not a function [closed]

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This question was caused by a typo or a problem that can no longer be reproduced. While similar questions may be on-topic here, this one was resolved in a way less likely to help future readers.
Closed 7 years ago.
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I am baffled to why I am experiencing this error, maybe its a bug with Jquery or maybe I'm blind, but Jquery has loaded and I can access the offset function after selecting an element?
Uncaught TypeError: $(...).offest is not a function
According to w3, I've used the function correctly.
Current code
var off = $("#canvas").offest();
var xPos = off.left + ( jqo.outerWidth(true)/2 );
var yPos = off.top + ( jqo.outerHeight(true)/2 );
console.log(yPos, xPos);
You have typo. Use offset instead of offest:
var off = $("#canvas").offset();

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