How can I check if key in array exist without error code?

Question

I made this variable "generateError" to not work on purpose but I encounter this same issue in another context when checking for keys in an array. So my question, how do I get the variable status to just show "array not working" instead of generating error code?

var numbers = [1, 2, 3, 4];
var generateError = numbers['badKey'][2];

if (typeof(generateError)==undefined){
 var status=("array not working");
}
else
{
    var status=("array is working");
}

document.getElementById("status").innerHTML=status;

<div id="status"></div>

Oops, I linked to a wrong page, should have been developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/… — Teemu
– Teemu, Commented Mar 5, 2021 at 12:09

Sandro · Accepted Answer · 2021-03-05 18:27:25Z

1

Side Info - typeof method will return a string so when using it try it like this

if(typeof generateError == 'undefined')

You could make use of the try/catch methods. it will try to run everything in the try block and if theres an error the catch block will catch the error and do whatever you want with it.

var numbers = [1, 2, 3, 4];
var status;

try{
    var tryToGetIndex = numbers['badKey'][2];
    status = tryToGetIndex;
}
catch(error){
    status="array is not working";
}

document.getElementById("status").innerHTML=status;

edited Mar 5, 2021 at 18:27

answered Mar 5, 2021 at 12:01

Sandro

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Comments

Ihor Nebeschenko · Accepted Answer · 2021-03-05 12:00:20Z

1

var numbers = [1, 2, 3, 4];
var generateError = numbers['badKey'][2];
var status;
if (typeof(generateError)=='undefined'){
  status=("array not working");
} else {
  status=("array is working");
}

document.getElementById("status").innerHTML=status;

Try this and read this page its help you in future.

edited Mar 5, 2021 at 12:00

answered Mar 5, 2021 at 11:53

Ihor Nebeschenko

112 bronze badges

4 Comments

sw_ Over a year ago

Still get error code. I believe that the actual errorcode is generated in this line: var generateError = numbers['badKey'][2];

Teemu Over a year ago

@Ihor The problem lies in numbers['badKey'][2] where retrieving index 2 from ['badKey'] fires an error. OP doesn't want an error message, they want to continue the execution.

sw_ Over a year ago

@Teemu Exactly... and how do I get the variable "status" to tell me that without generating error message?

Teemu Over a year ago

You're the one who is answering here = ).

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