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JavaScript Node.js Basics (2014) Building a Command Line Application Using try and catch

Posted by Erik Sayce
Erik Sayce
4,704 Points

Missing a variable declaration in my try block?

I'm struggling with this one, when I run the code I'm getting an error that I'm missing my variable declaration in my try block. I'm new to this so I'm sure it's a simple oversight but I've put a half and hour in trying to figure it out to no avail. Help!!

app.js 
var jsonString = 'This is not a JSON String';
try {
    var jasonObject = JSON.parse(jsonString);
} catch (error){
    console.log(error.message);
};

1 Answer

Robert Richey
Robert Richey
Courses Plus Student 16,352 Points

Hi Erik,

This was tricky. Your variable inside the try block is spelled jasonObject and needs to be jsonObject

Erik Sayce
Erik Sayce
4,704 Points

ugh I've done this like 12 times but didn't catch it this time around.

thanks!

Andrew Chalkley
Andrew Chalkley
Treehouse Guest Teacher
Andrew Chalkley
Andrew Chalkley
Treehouse Guest Teacher

Erik Sayce - A second pair of eyes always help.

Jason Seifer and I have a long running joke of saying "JSON not Jason" or was it "Jason not JSON" :)