how to submit a form after validating by javascript

Question

I have done a registration page in php.I performed validation using javascript I have done the validation but i dont know the syntax for submitting the details to next page.This is my code i am a very new beginner to php so need your help. Thank you in advance..sorry for my poor english

register.php
    <!DOCTYPE html>
    <html>
    <head>
     <script>
    function validateForm() {

        var a = document.getElementById('n1').value;
        if(a==""){
        alert("please enter your name");
        return false;
        }
        var b = document.getElementById('u1').value;
        if(b==""){
        alert("please enter the username");
        return false;
        }
        var p3 = document.getElementById('p1').value;
        if(p3==""){
        alert("please enter the password");
        return false;
        }

        var p4 = document.getElementById('p2').value;
        if(p4==""){
        alert("please enter the confirmation password");
        return false;
        }

         //if(p3.value!=p4.value){
         //alert("hh");
         //alert("Passwords do no match");
         //return false;
        //}else{
        //return true;
        //}

          var c = document.getElementById('a1').value;
            if(c==""){
        alert("please enter your address");
        return false;
        }
          var d = document.getElementById('ph').value;
        if(d==""){
        alert("please enter your phone no:");
        return false;
        }
    (48)    document.registrationform.submit;


    }
    </script>
    </head>
    <body>
    <form action ="#" name="registrationform" id=="formid" class="formclass" action="registerbackend.php" method ="post" >
    name    :<input type="text" id="n1" name="name"><br><br>
    username:<input type="text" id="u1" name="username"><br><br>
    password:<input type="password" id="p1" name="password"><br><br>
    confirm password:<input type="password" id="p2" name="password2"><br><br>
    Address   :<textarea name="address" id= "a1"rows="4" cols="40"></textarea><br><br>
    phone   :<input id="ph" type="numbers" name="phone"><br><br>
 (62)   <input  type="button" name="register" onclick="validateForm()" value="Register">
    <a href="login.php"><input id="button"  type="button"  value="cancel"> </a>
    </form>
    </body>
    </html>

Possible duplicate of submit the form using ajax

Zibellino
– Zibellino

2015-11-12 12:02:30 +00:00
Commented Nov 12, 2015 at 12:02 — Zibellino
– Zibellino, Commented Nov 12, 2015 at 12:02

Rajdeep Paul · Accepted Answer · 2015-11-12 12:09:13Z

0

Your form submission syntax is wrong.

Change this

document.registrationform.submit;

to this:

document.getElementsByTagName("form")[0].submit();

answered Nov 12, 2015 at 12:09

Rajdeep Paul

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1 Comment

vimal kumar Over a year ago

yes i did it.I didn't reached 15 reputations so please upvote my question it will be displayed with in 2 days.. :-)

AeJey · Accepted Answer · 2015-11-12 11:47:27Z

0

Try the below code for submitting the form

document.getElementById("formid").submit();

answered Nov 12, 2015 at 11:47

AeJey

1,45720 silver badges41 bronze badges

4 Comments

vimal kumar Over a year ago

still getting below error while trying this code.Uncaught TypeError: Cannot read property 'submit' of null.alidateForm @ register.php:48onclick @ register.php:62

AeJey Over a year ago

could you please update the question with the error you get and the codes on line 48 and 63 so that we can check and provide you the solution

vimal kumar Over a year ago

i have found the solution.its because of the wrong syntax used in the submit.Rajdeep Paul provided me the write syntax see it above.

vimal kumar Over a year ago

If you found my question is worth..please upvote my question

Alina Anjum · Accepted Answer · 2015-11-12 11:58:16Z

**change this :-**
 <form action ="#" name="registrationform" id=="formid" class="formclass" action="registerbackend.php" method ="post" >
**to this:->**
 <form action ="#" name="registrationform" id="formid" class="formclass" action="registerbackend.php" method ="post" >
**change your javascript code to:**
 function validateForm() {
   var status="true";
        var a = document.getElementById('n1').value;
        var b = document.getElementById('u1').value;
        var p4 = document.getElementById('p2').value;
        var c = document.getElementById('a1').value;
        var d = document.getElementById('ph').value;
        var p3 = document.getElementById('p1').value;
        if(a==""){
        alert("please enter your name");
        status="false";
        }        
        else if(b==""){
        alert("please enter the username");
                status="false";
        }       
        else if(p3==""){
        alert("please enter the password");
        status="false";
        }        
        else if(p4==""){
        alert("please enter the confirmation password");
        status="false";
        }
        else if(c==""){
        alert("please enter your address");
        status="false";
        }          
        else if(d==""){
        alert("please enter your phone no:");
        status="false";
        }
       if(status=="true"){
     document.getElementById("fromid").submit();
}
}

Madara's Ghost · Accepted Answer · 2015-11-28 16:50:18Z

0

Use this :-

document.getElementById("formid").submit();

change this :-

document.registrationform.submit;

to this :-

document.getElementById("formid").submit();

edited Nov 28, 2015 at 16:50

Madara's Ghost

176k52 gold badges274 silver badges314 bronze badges

answered Nov 12, 2015 at 11:42

Abhishek Sharma

6,6241 gold badge17 silver badges20 bronze badges

3 Comments

vimal kumar Over a year ago

while using the above code am getting error message from chrome browser. Uncaught TypeError: Cannot read property 'submit' of nullvalidateForm @ register.php:48onclick @ register.php:62

Abhishek Sharma Over a year ago

document.getElementById("formid").submit();

vimal kumar Over a year ago

I've tried it am getting error in two places.that is mentioned as 48 and 62 in the above code

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