An “Unexpected identifier” error occurs when you have put an identifier where it is not supposed to be in your JavaScript code, it results in a syntax error that makes the interpreter fail to read the code, mostly, it happens because of incorrect identification or location in relation to expressions, variables and other structures found within the code.

Understanding an error

Misplacement or misapplication of an identifier is the cause of such an error, to keep to JavaScript syntax rules, identifiers must be suitably implemented and located in lines of code, when it is placed where it should not be like directly after another identifier without appropriate operator or separator, this violates the grammatical rules of the language and causes SyntaxError, such mistake can be avoided through correct usage and placement of identifiers.

Case 1: Error Cause: Variable Declaration

• let declares a variable named name.
• “GeeksForGeeks” appears to be intended as a string or a value, but it’s not properly formatted as either.
• JavaScript interprets GeeksForGeeks as an identifier because it’s not enclosed in quotes (‘ or “).

Example: In the example below an identifier is placed directly after another identifier without any operator or separator, causing the error.

let name = GeeksForGeeks;
console.log(name);

Output:

SyntaxError: Unexpected identifier

Resolution of error

To resolve this error, you need to clarify whether GeeksForGeeks should be treated as a string literal or as a variable.

let name = 'GeeksForGeeks';
console.log(name); 

GeeksForGeeks

Note: Here, ‘GeeksForGeeks’ is a string literal assigned to the variable name. It will print GeeksForGeeks to the console.

Case 2: Error Cause: assigning the wrong value

The error in the provided JavaScript code occurs because resultValue is being used within the expression to initialize result without being properly defined or assigned beforehand. resultValue is used in the expression (5 + resultValue), but it has not been previously defined or initialized. JavaScript does not know what resultValue refers to at this point.

Example: In the example below an identifier is incorrectly placed within an expression without proper separation, causing the error.

let result = 5 + resultValue;
console.log(result);

Output:

SyntaxError: Unexpected identifier

Resolution of error

If resultValue is meant to hold a numeric value, you should define it before using it in the expression. Here, resultValue is initialized with the value 10, which is then used in the expression 5 + resultValue. The value of result becomes 15, which is then logged to the console.

let resultValue = 10; 
let result = 5 + resultValue;
console.log(result); 

15

Conclusion

To avoid “Unexpected identifier” errors in JavaScript, make sure identifiers are properly defined and used as intended, either as quoted strings or initialized variables according to JavaScript syntax rules. Know their purpose and placement within expressions to maintain code integrity and functionality.

FAQs

Why am I getting an “Unexpected identifier” error in my JavaScript code?

Ensure strings are enclosed in quotes (`’` or `”`), and variables are properly declared and initialized before use.

What causes a “SyntaxError: Unexpected identifier” when declaring variables?

Use quotes (`’` or` "`) around strings and stick to JavaScript’s variable naming rules (no spaces, start with a letter).

How can I fix “Unexpected identifier” when using variables in expressions?

Declare and initialize variables before using them in expressions to ensure they are recognized by JavaScript.

How do I prevent “Unexpected identifier” errors when working with objects or functions?

Use correct syntax for object keys (key: value) and function names (ensure proper parentheses for invocation).