CODEWITHSUNDEEP
javascriptphpjqueryajaxforms
SFY
SFY

Reputation: 61

how to get ajax form submit return value

I use ajax to submit a form, however when i finished the whole process, it stopped at the api.php and did get back to the ajax success. Here the code:

<form method="POST" id="form_1" action="../api.php" enctype="multipart/form-data" novalidate="novalidate">
  <input type="text" name="name" id="name" value="Amy" readonly="readonly">
  <input id="fileBox" class="fileUpload" type="file" name="img" accept="image/*" required>
  <input type="hidden" name="isSubmit" value="1"/>
</form>

<a id="btnSubmit" href="javascript:void(0)">Submit</a>

$("#form_1").submit(function(){
  var formData = new FormData(($this)[0]);
  $ajax({
    type: "POST",
    url: "../../api.php",
    data: { action:"formSubmit", formData:formData},
    processData: false,
    contentType: false,
    success: function(){
      console.log('success return');
    }
  })
});

$("#btnSubmit").click(function(){
  if($("#form_1").valid()){ 
    // inside click event handler you trigger form submission
    $("#form_1").trigger('submit');
  }
});

in my api.php

if($action=='formSubmit'){
  $name = $_POST['name'];
  if(createUser($name)){
    return true;
  }else{
    return false;
  }
}

I checked the database , record is successfully created. But I can't get the console in success. The url show on the browser there is the path of the api.php. Why I stopped there can't get the return?

Thanks for your help!!

I finallly did it !I am missing "return false" after the ajax; But actually I dont really know the function of return false;

$("#form_1").submit(function(){
var formData = new FormData((this)[0]);
$.ajax({
type: "POST",
url: "../../api.php",
data: { action: "formSubmit",  formData: formData},
processData:false,
contentType:false,
success: function(data){
$("#thankyou").show();
}
})
return false;  
});

Upvotes: 0

Views: 2453

Answers (2)

Praveen Kumar Purushothaman
Praveen Kumar Purushothaman

Reputation: 167182

Redirecting to other page will have the header() to use:

if ($action == 'formSubmit') {
  $name = $_POST['name'];
  if (createUser($name)) {
    // Redirect to right.htm
    header("Location: right.htm");
  } else {
    // Redirect to wrong.htm
    header("Location: wrong.htm");
  }
  die();
}

Upvotes: 3

RNK
RNK

Reputation: 5792

javascript doesn't recognize return value from php. Use echo instead.

if($action=='formSubmit'){
    $name = $_POST['name'];
    if(createUser($name)){
       echo 'true';
    }else{
       echo 'false';
    }
}

If you want to see actual output coming from api.php then use below code in your ajax.

success: function(res){
   console.log(res);
}

Upvotes: 1

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