Use JavaScript to assign a value to a variable based on the value of a drop-down menu

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Release: 2024-01-25 08:27:20
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Use JavaScript to assign a value to a variable based on the value of a drop-down menu

javascript determines the value of the drop-down menu and assigns it to the variable

Personal testing is available.

You need to put the definition of variable i in the initialization function, rather than defining it as a global variable.

What is the difference between assigning anonymous functions to variables in JavaScript and directly naming functions

Originally I didn’t like to answer such a general question. But I can't bear to see other wrong answers misleading people, so I'll just answer it briefly.

The biggest difference between the two writing methods is:

var init = function () { }; is a function expression. The function is only executed when the code execution reaches the current line, and init is assigned a value.

function init() {} is the declaration of a function. Like var, it will be defined at the front of the code.

So, the following two examples:

foo(); // Calling foo after the function declaration can be called normally. Because foo is defined to the front.

function foo() {

return true;

}

foo(); // Call the function before the function expression and report an error. Because there is no foo variable at this time.

var foo = function() {

return foo;

};ES5 stipulates that functions can only be declared in the top-level scope and function scope, otherwise it is illegal. For example:

if( true ) {

// It is wrong to define functions outside the top-level and function scope. Although the browser may not necessarily throw an error, it is not standardized.

function foo() {

return true;

}

}

ES6 introduced the concept of block-level scope, and this definition method is allowed. A function declared in a block-level scope has a scope similar to a variable declared using let and cannot be called outside the block-level scope.

For example:

{

function foo() {

return true;

}

}

foo(); // Calling a function outside the code block where the function is declared will throw an error.

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source:docexcel.net
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