PHP: Undefined index by retrieving $_POST data

Question

It's a very simple code, I was trying to do an ajax submit to work. Until here, ajax is working correct, but Why cannot print out $_POST data?

console.log

 Notice: Undefined index: fieldText in C:\xampp\htdocs\rajax.php on line 4 

sendajax.php

<form method="POST">
    <input type="text" name="fieldText" value="">
      <button type="submit" id="save">Send</button>     
</form>

<script type="text/javascript">
    $(document).ready(function(){
        //alert("Jquery's Working");
        $("#save").click(function(e){
            e.preventDefault();
            //alert("Click Event is working");
            $.ajax({
                type:"POST",
                url:'rajax.php',
                data: {field: $("input[name=fieldText]").val()},
                success: function(result){
                    console.log(result);
                    //alert($("input[name=fieldText]").val()); #Print Value is working
                },
                error: function(result){
                    console.log(result);
                }
            });
        });

    });
</script>

recajax.php

<?php 

    if($_SERVER["REQUEST_METHOD"]=="POST"){
        $test = $_POST['fieldText'];
        echo $test;
    }

?>

Possible duplicate of PHP: "Notice: Undefined variable", "Notice: Undefined index", and "Notice: Undefined offset" — Scuzzy
– Scuzzy, Commented Jul 1, 2018 at 23:25

Nicolas · Accepted Answer · 2018-07-01 23:10:36Z

2

That's because your variable name isn't fieldText, it's field. try changing it in your PHP.

<?php 

    if($_SERVER["REQUEST_METHOD"]=="POST"){
       $test = $_POST['field'];
       echo $test;
    }

?>

answered Jul 1, 2018 at 23:10

Nicolas

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5 Comments

Shinomoto Asakura Over a year ago

hmm... I'm retired from php. It worked. Can you explain where I overlook? Because, the name that I got in input is fieldText

Nicolas Over a year ago

Yea but you passed it as field in your data attribute. the $_POST variable is an associative array (in your case), and it reflect the value of the data attribute of your AJAX call.

Nicolas Over a year ago

which is data: {field: $("input[name=fieldText]").val()},. not data: {fieldText: $("input[name=fieldText]").val()}, Notice the key of your value has changed.

Shinomoto Asakura Over a year ago

Oh man, you're right! I forgot completely about this. Thank you man

Nicolas Over a year ago

No problem, have a nice day !

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