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Send more data than only the form (node.js post)

0

Is it possible when someone submits a form it sends additional things instead of only the input fields? In my case it would be the username.

javascript node.js forms post webforms

share|improve this question

asked Mar 7 at 5:57

SW30000SW30000

52

add a comment | 

0

Is it possible when someone submits a form it sends additional things instead of only the input fields? In my case it would be the username.

javascript node.js forms post webforms

share|improve this question

asked Mar 7 at 5:57

SW30000SW30000

52

add a comment | 

Is it possible when someone submits a form it sends additional things instead of only the input fields? In my case it would be the username.

javascript node.js forms post webforms

share|improve this question

asked Mar 7 at 5:57

SW30000SW30000

52

share|improve this question

asked Mar 7 at 5:57

SW30000SW30000

52

share|improve this question

asked Mar 7 at 5:57

SW30000SW30000

52

asked Mar 7 at 5:57

SW30000SW30000

52

asked Mar 7 at 5:57

SW30000SW30000

52

2 Answers
2

active

oldest

votes

0

As @jayarjo as said, you can intercept the form submission and handle it manually, for example you can use Ajax as follows.

Note:; To use Ajax make sure you add its script for example

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script>

<script>


 $('#newform').on('submit', function (event) 
 event.preventDefault();
 var data = 
 username: $('#username').val(),
 ;

 $.ajax(
 url: '/upload',
 data: data,
 method: 'POST'
 ).then(function (response) 

 $('body').append(response);
 ).catch(function (err) 
 console.error(err);
 );
 );
</script>

Hope this helps.

share|improve this answer

answered Mar 7 at 6:12

Ike MawiraIke Mawira

4818

add a comment | 

1

You can always intercept the submission of the form and handle it manually. In fact that's what you should do anyway if you want to validate the form fields beforehand.

The logic is pretty straightforward. You attach an onsubmit event listener to the form. Prevent the auto submission by calling preventDefault() on the event. Then collect the values from all the fields, do the validation (if email is in the proper format, if passwords match, etc.), add arbitrary accompanying properties to the object and post the whole thing to the server via ajax post request.

Since you are asking that kind of question I will recommend to use some JS library to make it easier, jQuery is perfect for such case:

$('#myForm').on('submit', function(e) 
 e.preventDefault();

 const $form = $(this); 
 const method = $form.attr('method');
 const action = $form.attr('action')
 const fields = $form.serializeArray();
 
 fields.push(
 name: 'username',
 value: 'myUsername'
 )
 
 const queryStr = $.param(fields);
 
 $[method](action, queryStr)
 .done(function() 
 // submission was successful - do something, refresh page for exmaple
 )
 .fail(function() 
 // submission failed
 );
);

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>

<form id="myForm" method="post" action="/handle.php">
 My name is: <input type="text" name="fname" /><br />
 <label for="agree">I agree</label> <input type="checkbox" name="agree" value="1" /><br />
 <button type="submit">Submit</button>
</form>

Of course /handle.php is a fake endpoint so submission here obviously won't work.

share|improve this answer

edited Mar 7 at 6:21

answered Mar 7 at 5:59

jayarjojayarjo

6,381136397

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  How does this code look like?
  
  – SW30000
  Mar 7 at 6:00
  

  
  
  
  

add a comment | 

Your Answer

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0

As @jayarjo as said, you can intercept the form submission and handle it manually, for example you can use Ajax as follows.

Note:; To use Ajax make sure you add its script for example

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script>

<script>


 $('#newform').on('submit', function (event) 
 event.preventDefault();
 var data = 
 username: $('#username').val(),
 ;

 $.ajax(
 url: '/upload',
 data: data,
 method: 'POST'
 ).then(function (response) 

 $('body').append(response);
 ).catch(function (err) 
 console.error(err);
 );
 );
</script>

Hope this helps.

share|improve this answer

answered Mar 7 at 6:12

Ike MawiraIke Mawira

4818

add a comment | 

0

As @jayarjo as said, you can intercept the form submission and handle it manually, for example you can use Ajax as follows.

Note:; To use Ajax make sure you add its script for example

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script>

<script>


 $('#newform').on('submit', function (event) 
 event.preventDefault();
 var data = 
 username: $('#username').val(),
 ;

 $.ajax(
 url: '/upload',
 data: data,
 method: 'POST'
 ).then(function (response) 

 $('body').append(response);
 ).catch(function (err) 
 console.error(err);
 );
 );
</script>

Hope this helps.

share|improve this answer

answered Mar 7 at 6:12

Ike MawiraIke Mawira

4818

add a comment | 

As @jayarjo as said, you can intercept the form submission and handle it manually, for example you can use Ajax as follows.

Note:; To use Ajax make sure you add its script for example

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script>

<script>


 $('#newform').on('submit', function (event) 
 event.preventDefault();
 var data = 
 username: $('#username').val(),
 ;

 $.ajax(
 url: '/upload',
 data: data,
 method: 'POST'
 ).then(function (response) 

 $('body').append(response);
 ).catch(function (err) 
 console.error(err);
 );
 );
</script>

Hope this helps.

share|improve this answer

answered Mar 7 at 6:12

Ike MawiraIke Mawira

4818

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script>

<script>


 $('#newform').on('submit', function (event) 
 event.preventDefault();
 var data = 
 username: $('#username').val(),
 ;

 $.ajax(
 url: '/upload',
 data: data,
 method: 'POST'
 ).then(function (response) 

 $('body').append(response);
 ).catch(function (err) 
 console.error(err);
 );
 );
</script>

share|improve this answer

answered Mar 7 at 6:12

Ike MawiraIke Mawira

4818

answered Mar 7 at 6:12

Ike MawiraIke Mawira

4818

answered Mar 7 at 6:12

Ike MawiraIke Mawira

4818

1

You can always intercept the submission of the form and handle it manually. In fact that's what you should do anyway if you want to validate the form fields beforehand.

The logic is pretty straightforward. You attach an onsubmit event listener to the form. Prevent the auto submission by calling preventDefault() on the event. Then collect the values from all the fields, do the validation (if email is in the proper format, if passwords match, etc.), add arbitrary accompanying properties to the object and post the whole thing to the server via ajax post request.

Since you are asking that kind of question I will recommend to use some JS library to make it easier, jQuery is perfect for such case:

$('#myForm').on('submit', function(e) 
 e.preventDefault();

 const $form = $(this); 
 const method = $form.attr('method');
 const action = $form.attr('action')
 const fields = $form.serializeArray();
 
 fields.push(
 name: 'username',
 value: 'myUsername'
 )
 
 const queryStr = $.param(fields);
 
 $[method](action, queryStr)
 .done(function() 
 // submission was successful - do something, refresh page for exmaple
 )
 .fail(function() 
 // submission failed
 );
);

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>

<form id="myForm" method="post" action="/handle.php">
 My name is: <input type="text" name="fname" /><br />
 <label for="agree">I agree</label> <input type="checkbox" name="agree" value="1" /><br />
 <button type="submit">Submit</button>
</form>

Of course /handle.php is a fake endpoint so submission here obviously won't work.

share|improve this answer

edited Mar 7 at 6:21

answered Mar 7 at 5:59

jayarjojayarjo

6,381136397

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  How does this code look like?
  
  – SW30000
  Mar 7 at 6:00
  

  
  
  
  

add a comment | 

1

You can always intercept the submission of the form and handle it manually. In fact that's what you should do anyway if you want to validate the form fields beforehand.

The logic is pretty straightforward. You attach an onsubmit event listener to the form. Prevent the auto submission by calling preventDefault() on the event. Then collect the values from all the fields, do the validation (if email is in the proper format, if passwords match, etc.), add arbitrary accompanying properties to the object and post the whole thing to the server via ajax post request.

Since you are asking that kind of question I will recommend to use some JS library to make it easier, jQuery is perfect for such case:

$('#myForm').on('submit', function(e) 
 e.preventDefault();

 const $form = $(this); 
 const method = $form.attr('method');
 const action = $form.attr('action')
 const fields = $form.serializeArray();
 
 fields.push(
 name: 'username',
 value: 'myUsername'
 )
 
 const queryStr = $.param(fields);
 
 $[method](action, queryStr)
 .done(function() 
 // submission was successful - do something, refresh page for exmaple
 )
 .fail(function() 
 // submission failed
 );
);

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>

<form id="myForm" method="post" action="/handle.php">
 My name is: <input type="text" name="fname" /><br />
 <label for="agree">I agree</label> <input type="checkbox" name="agree" value="1" /><br />
 <button type="submit">Submit</button>
</form>

Of course /handle.php is a fake endpoint so submission here obviously won't work.

share|improve this answer

edited Mar 7 at 6:21

answered Mar 7 at 5:59

jayarjojayarjo

6,381136397

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  How does this code look like?
  
  – SW30000
  Mar 7 at 6:00
  

  
  
  
  

add a comment | 

You can always intercept the submission of the form and handle it manually. In fact that's what you should do anyway if you want to validate the form fields beforehand.

The logic is pretty straightforward. You attach an onsubmit event listener to the form. Prevent the auto submission by calling preventDefault() on the event. Then collect the values from all the fields, do the validation (if email is in the proper format, if passwords match, etc.), add arbitrary accompanying properties to the object and post the whole thing to the server via ajax post request.

Since you are asking that kind of question I will recommend to use some JS library to make it easier, jQuery is perfect for such case:

$('#myForm').on('submit', function(e) 
 e.preventDefault();

 const $form = $(this); 
 const method = $form.attr('method');
 const action = $form.attr('action')
 const fields = $form.serializeArray();
 
 fields.push(
 name: 'username',
 value: 'myUsername'
 )
 
 const queryStr = $.param(fields);
 
 $[method](action, queryStr)
 .done(function() 
 // submission was successful - do something, refresh page for exmaple
 )
 .fail(function() 
 // submission failed
 );
);

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>

<form id="myForm" method="post" action="/handle.php">
 My name is: <input type="text" name="fname" /><br />
 <label for="agree">I agree</label> <input type="checkbox" name="agree" value="1" /><br />
 <button type="submit">Submit</button>
</form>

Of course /handle.php is a fake endpoint so submission here obviously won't work.

share|improve this answer

edited Mar 7 at 6:21

answered Mar 7 at 5:59

jayarjojayarjo

6,381136397

$('#myForm').on('submit', function(e) 
 e.preventDefault();

 const $form = $(this); 
 const method = $form.attr('method');
 const action = $form.attr('action')
 const fields = $form.serializeArray();
 
 fields.push(
 name: 'username',
 value: 'myUsername'
 )
 
 const queryStr = $.param(fields);
 
 $[method](action, queryStr)
 .done(function() 
 // submission was successful - do something, refresh page for exmaple
 )
 .fail(function() 
 // submission failed
 );
);

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>

<form id="myForm" method="post" action="/handle.php">
 My name is: <input type="text" name="fname" /><br />
 <label for="agree">I agree</label> <input type="checkbox" name="agree" value="1" /><br />
 <button type="submit">Submit</button>
</form>

$('#myForm').on('submit', function(e) 
 e.preventDefault();

 const $form = $(this); 
 const method = $form.attr('method');
 const action = $form.attr('action')
 const fields = $form.serializeArray();
 
 fields.push(
 name: 'username',
 value: 'myUsername'
 )
 
 const queryStr = $.param(fields);
 
 $[method](action, queryStr)
 .done(function() 
 // submission was successful - do something, refresh page for exmaple
 )
 .fail(function() 
 // submission failed
 );
);

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>

<form id="myForm" method="post" action="/handle.php">
 My name is: <input type="text" name="fname" /><br />
 <label for="agree">I agree</label> <input type="checkbox" name="agree" value="1" /><br />
 <button type="submit">Submit</button>
</form>

$('#myForm').on('submit', function(e) 
 e.preventDefault();

 const $form = $(this); 
 const method = $form.attr('method');
 const action = $form.attr('action')
 const fields = $form.serializeArray();
 
 fields.push(
 name: 'username',
 value: 'myUsername'
 )
 
 const queryStr = $.param(fields);
 
 $[method](action, queryStr)
 .done(function() 
 // submission was successful - do something, refresh page for exmaple
 )
 .fail(function() 
 // submission failed
 );
);

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>

<form id="myForm" method="post" action="/handle.php">
 My name is: <input type="text" name="fname" /><br />
 <label for="agree">I agree</label> <input type="checkbox" name="agree" value="1" /><br />
 <button type="submit">Submit</button>
</form>

share|improve this answer

edited Mar 7 at 6:21

answered Mar 7 at 5:59

jayarjojayarjo

6,381136397

answered Mar 7 at 5:59

jayarjojayarjo

6,381136397

answered Mar 7 at 5:59

jayarjojayarjo

6,381136397