Javascript Can';无法在JS中进行表单验证
所以我一直在学习基本JS,我无法让表单验证用于数字。我将在下面发布一段代码片段。请告诉我,我是一个noob,所以我可能有一个更容易理解的答案,从一个刚刚学习语言的人那里。前三个if语句工作正常,但第四个我有问题Javascript Can';无法在JS中进行表单验证,javascript,forms,Javascript,Forms,所以我一直在学习基本JS,我无法让表单验证用于数字。我将在下面发布一段代码片段。请告诉我,我是一个noob,所以我可能有一个更容易理解的答案,从一个刚刚学习语言的人那里。前三个if语句工作正常,但第四个我有问题 var checkbox = function() { var error = ""; var firstname = document.getElementById("fn").value; var lastnam
var checkbox = function() {
var error = "";
var firstname = document.getElementById("fn").value;
var lastname = document.getElementById("ln").value;
var email = document.getElementById("Email").value;
var age = parseInt(document.getElementById("age").value);
var address = document.getElementById("A").value;
var phone = parseInt(document.getElementById("pn").value);
if(firstname.length < 1){
error = "Enter a valid first name!";
}
if(lastname.length < 1){
error += "\nEnter a valid last name!";
}
if(email.length <1){
error += "\nEnter a valid email!";
}
if(age.length < 1 ){
error += "\nEnter a valid age!";
}
if(error.length){
alert(error)
return false;
}
return true;
}
var checkbox=function(){
var误差=”;
var firstname=document.getElementById(“fn”).value;
var lastname=document.getElementById(“ln”).value;
var email=document.getElementById(“email”).value;
var age=parseInt(document.getElementById(“age”).value);
var address=document.getElementById(“A”).value;
var phone=parseInt(document.getElementById(“pn”).value);
如果(firstname.length<1){
error=“输入有效的名字!”;
}
如果(lastname.length<1){
错误+=“\n请输入有效的姓氏!”;
}
如果(email.length你可以试试这个
if(firstname.trim() == ""){
error = "Enter a valid first name!";
}
if(lastname.trim()==""){
error += "\nEnter a valid last name!";
}
if(email.trim() ==""){
error += "\nEnter a valid email!";
}
if(age.trim() == "" ){
error += "\nEnter a valid age!";
}
if(error.length>0){
alert(error)
return false;
}
您在检查上的年龄字段是什么?您在检查它是否存在?如果是,您可以执行以下操作:
if(!age) {
error += "\nThis field is required";
}
if(typeof age !== "number") {
error += "\nPlease enter a valid number!";
}
if(age !== 30) {
error += "\nYou are under 30!";
}
如果要检查“年龄”字段中的值是否为数字,可以执行以下操作:
if(!age) {
error += "\nThis field is required";
}
if(typeof age !== "number") {
error += "\nPlease enter a valid number!";
}
if(age !== 30) {
error += "\nYou are under 30!";
}
如果要检查“年龄”字段是否在特定年龄范围内,可以执行以下操作:
if(!age) {
error += "\nThis field is required";
}
if(typeof age !== "number") {
error += "\nPlease enter a valid number!";
}
if(age !== 30) {
error += "\nYou are under 30!";
}
在调用parseInt
之前,您应该检查document.getElementById(“age”).value的长度。否则变量age
可以是NaN
。例如:
var ageVal = document.getElementById("age").value;
var age;
if (ageVal){ //same as ageVal.length > 0
age = parseInt(age);
if (!age){// it can be NaN
error += "\nEnter a valid age!";
}
}
else
{
error += "\nEnter an age!"; //here we have age not set
}
请试试这个
if(isNaN(age)){
error += "\nEnter a valid age!";
}
isNan()是一个内置javascript函数。有关详细信息,请访问