How to run ajax and post to PHP

Question

I am trying to run an ajax script from javascript inside a PHP file :

    $(document).on('click', '.goto_date', function(){
          var  datechosen= $('#pickdate').val();
          alert(datechosen);
           $.ajax({
                     url:"vdater.php",
                     method:"POST",
                     dataType:"json",
                     data:{'datechosen':datechosen},
                     success:function(data){
                      alert('success');
                     }
               });
            });

 <html>
      <head>

           <meta charset="UTF-8">
           <meta name="viewport" content="width=device-width, initial-scale=1.0">

           <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.2.0/jquery.min.js"></script>
           </head>
<body>
                       <div align="center">
                          
                          <input type="button" name="gotodate" value="Go to date" id="gotodate" class="btn btn-danger btn-xs goto_date" />
                          <input type="date" name='pickdate' id='pickdate'  />

                     </div>

</body>
 </html>

But when I run the code nothing happens apart from the alert box referring that the ajax script is reachable in the code.. The PHP file may contain the following code:

 echo $_POST['datechosen'];

You can start by [debugging the ajax request](https://stackoverflow.com/a/21617685/2453432) to find out where it fails. — M. Eriksson, Jun 11 '22 at 11:36

score 1 · Answer 1 · answered Jun 11 '22 at 11:58

1

Surprisingly, It worked by removing the line:

 dataType:"json",

answered Jun 11 '22 at 11:58

lonecoder

11
4

Your answer could be improved with additional supporting information. Please [edit] to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers [in the help center](/help/how-to-answer). – Ethan Jun 11 '22 at 15:00

score 0 · Answer 2 · answered Jun 11 '22 at 12:29

Try this one. It must be worked!

<!DOCTYPE html>
<html lang="en">
<head>
    <meta charset="UTF-8">
    <meta http-equiv="X-UA-Compatible" content="IE=edge">
    <meta name="viewport" content="width=device-width, initial-scale=1.0">
    <title>Get Ajax Response</title>
</head>
<body>
    <div align="center">           
        <input type="button" name="gotodate" value="Go to date" id="gotodate" class="btn btn-danger btn-xs goto_date" />
        <input type="date" name='pickdate' id='pickdate'  />
   </div>

   <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.2.0/jquery.min.js"></script>

<script>

let dateBtn = document.querySelector('#gotodate');
let getDate    = document.querySelector('#pickdate');
dateBtn.addEventListener('click',(event)=>{
    event.preventDefault();
    if(getDate.value.trim() !== ""){
      console.log(getDate.value.trim());
      dateBtn.style.borderColor = "transparent"
      jQuery.ajax({
        type: "POST",
        url: "vdater.php",
        data:{
          datechosen: getDate.value.trim()
        },
        success: ((response,status, object) => {
          response = JSON.parse(response);
            if(response.status){
              alert(`Date Chosen: ${response.PickedDate}`);
            }
        })
      });
    }else{
      dateBtn.style.border = "1px solid #fb5757";
    }
  });
</script>
</body>
</html>


<?php 

if(isset($_POST['datechosen'])){
    $date = $_POST['datechosen'];
    echo json_encode(["PickedDate"=> $date,"status"=>true]);
}

How to run ajax and post to PHP

2 Answers2