Making If Statements Easier

Question

Making If Statements Easier

Currently, I'm tackling an assignment that involves utilizing if statements and switch statements. Here is a snippet of code that I am working on:


if (validateField(document.forms[0].name) == false) {
    isValid = false;
}

if (validateField(document.forms[0].email) == false) {
    isValid = false;
}

if (validateField(document.forms[0].phone) == false) {
    isValid = false;
}

if (validateField(document.forms[0].message) == false) {
    isValid = false;
}

if (checkPattern(document.forms[0].username, /^(?!.*admin$).*$/) == false) {
    isValid = false;
}

if (checkPattern(document.forms[0].password, /^.{8,}$/) == false) {
    isValid = false;
}

if (checkPattern(document.forms[0].zipcode, /^\d{5}$/) == false) {
    isValid = false;
}

if (checkPattern(document.forms[0].ssn, /^\d{3}-\d{2}-\d{4}$|\d{9}$/) == false) {
    isValid == false;
}

I am looking for ways to streamline or simplify the string of if statements as shown above since they are all interconnected. Switching between different conditions doesn't seem feasible in this scenario. Is there a more elegant solution available?

javascript if-statement switch-statement

Answer 1

Answer №1

To validate form fields, one option is to create an array and use boolean functions as callbacks.

isValid = [
    testLength(document.forms[0].lname),
    testLength(document.forms[0].fname),
    testLength(document.forms[0].address),
    testLength(document.forms[0].summary),
    testPattern(document.forms[0].account, /^ACT\d{6}$/),
    testPattern(document.forms[0].department, /^DEPT\d{3}$/),
    testPattern(document.forms[0].project, /^PROJ\d{5}$/),
    testPattern(document.forms[0].ssn, /^\d{3}-\d{2}-\d{4}$|\d{9}$/)
].every(Boolean);

Another approach involves using arrays for property names with corresponding patterns for validation.

var keys = ['name', 'fname', 'address', 'summary'],
    patterns = [
        ['account', /^ACT\d{6}$/],
        ['department', /^DEPT\d{3}$/],
        ['project', /^PROJ\d{5}$/],
        ['ssn', /^\d{3}-\d{2}-\d{4}$|\d{9}$/]
    ],
    isValid = keys.every(k => testLength(document.forms[0][k]))
        && patterns.every((k, p) => testPattern(document.forms[0][k], p));

Answer 2

To validate form fields, one option is to create an array and use boolean functions as callbacks.

isValid = [
    testLength(document.forms[0].lname),
    testLength(document.forms[0].fname),
    testLength(document.forms[0].address),
    testLength(document.forms[0].summary),
    testPattern(document.forms[0].account, /^ACT\d{6}$/),
    testPattern(document.forms[0].department, /^DEPT\d{3}$/),
    testPattern(document.forms[0].project, /^PROJ\d{5}$/),
    testPattern(document.forms[0].ssn, /^\d{3}-\d{2}-\d{4}$|\d{9}$/)
].every(Boolean);

Another approach involves using arrays for property names with corresponding patterns for validation.

var keys = ['name', 'fname', 'address', 'summary'],
    patterns = [
        ['account', /^ACT\d{6}$/],
        ['department', /^DEPT\d{3}$/],
        ['project', /^PROJ\d{5}$/],
        ['ssn', /^\d{3}-\d{2}-\d{4}$|\d{9}$/]
    ],
    isValid = keys.every(k => testLength(document.forms[0][k]))
        && patterns.every((k, p) => testPattern(document.forms[0][k], p));

Answer 3

Answer №2

If you start with the assumption that isValid = true, you can achieve it in this way:

var form = document.forms[0];
var props = ["lname", "fname", "address", "summary"];
var isValid = props.every(prop => testLength(form[prop]))
           && testPattern(form.account, /^ACT\d{6}$/)
           && testPattern(form.department, /^DEPT\d{3}$/)
           && testPattern(form.project, /^PROJ\d{5}$/) 
           && testPattern(form.ssn, /^\d{3}-\d{2}-\d{4}$|\d{9}$/);

You can also implement an array approach for the patterns, like so:

var form = document.forms[0];
var props = ["lname", "fname", "address", "summary"];
var patts = [["account", /^ACT\d{6}$/], ["department", /^DEPT\d{3}$/],
             ["project", /^PROJ\d{5}$/], ["ssn", /^\d{3}-\d{2}-\d{4}$|\d{9}$/]];
var isValid = props.every(prop => testLength(form[prop]))
           && patts.every(([prop, regex]) => testPattern(form[prop], regex));

The concept here is to initialize the variables form, props, and patts only once instead of doing it every time validation is needed, although it wouldn't be a problem if done otherwise.

Answer 4

If you start with the assumption that isValid = true, you can achieve it in this way:

var form = document.forms[0];
var props = ["lname", "fname", "address", "summary"];
var isValid = props.every(prop => testLength(form[prop]))
           && testPattern(form.account, /^ACT\d{6}$/)
           && testPattern(form.department, /^DEPT\d{3}$/)
           && testPattern(form.project, /^PROJ\d{5}$/) 
           && testPattern(form.ssn, /^\d{3}-\d{2}-\d{4}$|\d{9}$/);

You can also implement an array approach for the patterns, like so:

var form = document.forms[0];
var props = ["lname", "fname", "address", "summary"];
var patts = [["account", /^ACT\d{6}$/], ["department", /^DEPT\d{3}$/],
             ["project", /^PROJ\d{5}$/], ["ssn", /^\d{3}-\d{2}-\d{4}$|\d{9}$/]];
var isValid = props.every(prop => testLength(form[prop]))
           && patts.every(([prop, regex]) => testPattern(form[prop], regex));

The concept here is to initialize the variables form, props, and patts only once instead of doing it every time validation is needed, although it wouldn't be a problem if done otherwise.

Answer 5

Answer №3

When dealing with boolean values, you can utilize the logical operator &&.

var isValid = testLength(document.forms[0].lname) && 
              testLength(document.forms[0].fname) &&
              .....
              testPattern(document.forms[0].ssn, /^\d{3}-\d{2}-\d{4}$|\d{9}$/);

Answer 6

When dealing with boolean values, you can utilize the logical operator &&.

var isValid = testLength(document.forms[0].lname) && 
              testLength(document.forms[0].fname) &&
              .....
              testPattern(document.forms[0].ssn, /^\d{3}-\d{2}-\d{4}$|\d{9}$/);

Making If Statements Easier

Answer №1

Answer №2

Answer №3

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