JQuery Why my array.every is not a function?

Question

I want to know why my array.every is not a function, im searching for hours but i dont understand why it is not. Can someone help me ?

function controlUserInput(inputText, appLang) {
 const regex = /\$[^$]*\$/gm;
const str = $('#formulaire-preview-textarea').val();
let m;
 var array =  populateVariable(appLang);
 
while ((m = regex.exec(str)) !== null) {
    // This is necessary to avoid infinite loops with zero-width matches
    if (m.index === regex.lastIndex) {
        regex.lastIndex++;
    }
    // The result can be accessed through the `m`-variable.
    m.forEach((match, groupIndex) => {
        console.log(`Found match, group ${groupIndex}: ${match}`);
  var isEqual = match.length==array.length;
  for (i=0; i<=array.length-1;i++){
   if(displayCpt == 4 && isEqual && match.toArray().every(function(paramInMatch){
    return $.inArray(paramInMatch, array) != -1; 
   })){
    osapi.jive.core.container.sendNotification({
    "message": "Toutes les valeurs rentrées sont correctes",
    "severity": "success"
    });
   }else{
    osapi.jive.core.container.sendNotification({
    "message": "Vous n'avez pas utilisé toutes les valeurs",
    "severity": "error"
    });
   } 
  }
 })
    };
}

Asons · Accepted Answer · 2018-07-01T13:37:46.760

3

The .toArray() is a jQuery object method, and as your match is not a jQuery object, it won't work.

You could use the plain javascript Array.from(match).

edited Jul 01 '18 at 13:37

answered Jul 01 '18 at 13:27

Asons

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like that ? Array.from(match) – Willy Nguyen Jul 01 '18 at 13:31
@WillyNguyen Correct...and check my update, as you might not need it either – Asons Jul 01 '18 at 13:32
apparently, when I do match.every it also returns me the error "match.every is not a function" – Willy Nguyen Jul 01 '18 at 13:36
@WillyNguyen Then simply use `Array.from()` ... and rolled back my answer – Asons Jul 01 '18 at 13:37

JQuery Why my array.every is not a function?

1 Answers1