Type Errors - learn how to fix these

TypeError related to console.log()

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Error message:

• “TypeError: Property ‘log’ of object # (Chrome)
• “TypeError: console.log is not a function” (Firefox)
• “TypeError: ‘your string’ is not a function (evaluating ‘console.log(“your string”)’)” (Safari)
• “TypeError: Function expected” (IE)

What has gone wrong?

You have changed console.log() from a function into a string or a number.

Explanation:

console.log() is a function, it will log/print on the console/screen the message or value you put in the ( ).

console.log() is not a variable - it is not expected that you will want to make console.log() EQUAL to something. The interpreter won’t stop you from doing so but you should understand the consequences.

At some point recently, you ran your code with a statement resembling this:

console.log = "You are allowed to play.";
console.log = 18;

It may not have been one of those exact values but somewhere you told the interpreter to make console.log equal to something. You assigned a new value to console.log(). The result of doing so was that the usual code that is invoked when you use the console.log() function was overwritten by whatever it was you put on the right hand side of the = sign.

Normally console.log() is a function but now you have turned it into a string or a number.

If you are curious you can check to see if that is true. You can run these commands in the console to see how you have changed console.log() from a function to a string:

Normal:

   typeof(console.log)
=> 'function'

Changed to a string:

   console.log = "Hello?"
=> 'Hello?'
   typeof(console.log)
=> 'string'

Changed to a number:

   console.log = 18;
=> 18
   typeof(console.log)
=> 'number'   

How to fix this:

1. Correct your syntax by making sure you don’t have console.log = anywhere in your code.
2. Then reload/refresh the webpage. That is the only way to restore console.log() to its normal state.

TypeError related to prompt()

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Error message:

• “TypeError: Property ‘prompt’ of object [object Object] is not a function” (Chrome)
• “TypeError: prompt is not a function” (Firefox)
• “TypeError: ‘a string, this could vary’ is not a function (evaluating ‘prompt(“your question”)’)” (Safari)
• “TypeError: Function expected” (IE)

What has gone wrong?

You have changed prompt() from a function into a string or a number.

Explanation:

prompt() is a function, it is waiting for you to put something inside those ( ). It will then put a popup on the screen with the message you put in the ( ).

prompt() is not a variable - it is not expected that you will want to make prompt() EQUAL to something. The interpreter won’t stop you from doing so but you should understand the consequences.

At some point recently, you ran your code with a statement resembling this:

prompt = "What is your age?";
prompt = 18;

It may not have been those exact values but somewhere you told the interpreter to make prompt equal to something. You assigned a new value to prompt(). The result of doing so was that the usual code that is invoked when you use the prompt() function was overwritten by whatever it was you put on the right hand side of the = sign.

Normally prompt() is a function but now you have turned it into a string or a number.

If you are curious you can check to see if that is true by running this code that checks what prompt() is currently set at:

console.log("Before I messed up prompt() it looked like this: \n " + prompt + "\n");
prompt = "What is your age?";
console.log("Now prompt() looks like this: " + prompt);    

How to fix this:

1. Correct your syntax by making sure you don’t have prompt = anywhere in your code.
2. Then reload/refresh the webpage. That is the only way to restore prompt() to its normal state.

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Popup Blockers

You may also get the error if you are running a strict popup blocker. Check to see if your popup blocker is blocking codecademy.com

TypeError related to confirm()

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Error message:

• “TypeError: Property ‘confirm’ of object [object Object] is not a function” (Chrome)
• “TypeError: confirm is not a function” (Firefox)
• “TypeError: ‘a string, this could vary’ is not a function (evaluating ‘confirm(“your question”)’)” (Safari)
• “TypeError: Function expected” (IE)

What has gone wrong?

You have changed prompt() from a function into a string or a number.

Explanation:

confirm() is a function, it is waiting for you to put something inside those ( ). It will then put a popup on the screen with the message you put in the ( ).

confirm() is not a variable - it is not expected that you will want to make confirm() EQUAL to something. The interpreter won’t stop you from doing so but you should understand the consequences.

At some point recently, you ran your code with a statement resembling this:

confirm = "Ready to play?";

It may not have been that exact string but somewhere you told the interpreter to make confirm equal to something. The effect of doing so was that the usual code that is invoked when you use the confirm() function was overwritten by whatever it was you put on the right hand side of the = sign. If you are curious you can check to see if that is true by running this code that checks what confirm() is currently set at:

console.log("Before I messed up confirm() it looked like this: " + confirm);
confirm = "Ready to play?";
console.log("Now confirm() looks like this: " + confirm);

How to fix this:

1. Correct your syntax by making sure you don’t have confirm = anywhere in your code.
2. Reload the webpage. That is the only way to restore confirm() to its normal state.

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Popup Blockers

You may also get the error if you are running a strict popup blocker. Check to see if your popup blocker is blocking codecademy.com

Thank you so much for posting this, being a person completely new to coding all of this helped a ton!

TypeError: ‘undefined’ is not an object

TypeError: Cannot call method ‘substring’ of undefined

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I guess I’ll go here for answers. I’m having trouble with substrings.

Here is my code for exercise 23:

console.log("january").substring(0,3);

TypeError: ‘undefined’ is not an object (evaluating ‘console.log(“january”).substring’)

I have no idea what I’m doing wrong, and I don’t see this in any of your examples above.

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I figured it out, here is the working code if anyone needs it.

console.log("January".substring(0,3));

TypeError: Cannot read property ‘firstName’ of undefined

I am doing the “Building an Address Book” Exercise 6/6. Here is the code:

var add = function(firstName,lastName,email,phoneNumber) {
    contacts[contacts.length] = {
        firstName: firstName,
        lastName: lastName,
        email: email,
        phoneNumber: phoneNumber
    };
};

add("John","Doe","[email protected]","999-999-9999");
list();

If somewhere along the way you assigned prompt or confirm to a var, an easy solution to this is to type:

delete prompt;
delete confirm;

somewhere in the CodeAcademy console. You only have to do this once, and that should clear up the problem.

“yourName”.length will not work for me. Help please!

I had a similar problem. The error is in the ‘printPerson’ function. Check that you have the following code: function printPerson(person) { console.log(person.firstName + “ “ + person.lastName); }

I had a syntac problem