PHP not getting POST data from $.ajax

Question

I have a JavaScript that runs a POST method once my datepicker has been out of focus (I also tried this on a regular submit button) and runs the script rent-fetch-pick-up-point.php. The PHP runs, however it doesn't get past the if-statement because my it's not getting the POST data. The datepicker is tied to a input field time-period-from

datepickerTo.blur(function(){
  if (selectedDateFrom.length > 0) {

    datepickerFrom.delay(500).queue(function(){

      $.ajax({
        type: "POST",
        url: "include/rent-fetch-pick-up-point.php",
        data: {action: selectedDateFrom},
        success: function(data) {
          $("#pick-up-point-container").html(data);
        }
      });
    });
  }
});

Here is the PHP code:

if (isset($_POST['time-period-from'])) {
  require '../include/connection.php';

  $dateFrom = $_POST['time-period-from'];
  $sql = "SELECT * FROM order WHERE $dateFrom BETWEEN date_from AND date_to";
  $result = mysqli_query($connection, $sql);
  $numRows = mysqli_num_rows($result);

  echo $sql; // For testing purposes
}

And here's the HTML:

  <input type="text" name="time-period-from" id="datepicker-from" class="datepicker"></p>

I also tried using $.post() instead of $.ajax(), but I ran into the same issue:

$.post("include/rent-fetch-pick-up-point.php", {name: selectedDateTo}, function(data) {

  $("#pick-up-point-container").text(data)

});

You do realize you're sending the data in action and not in time-period-from? — Charlotte Dunois
– Charlotte Dunois, Commented Apr 18, 2016 at 18:00
Your posting action, not time-period-from... try: $_POST['action'] — M. Eriksson
– M. Eriksson, Commented Apr 18, 2016 at 18:01
Little Bobby says your script is at risk for SQL Injection Attacks. Learn about prepared statements for MySQLi. Even escaping the string is not safe! — Jay Blanchard
– Jay Blanchard, Commented Apr 18, 2016 at 18:02
@CharlotteDunois oh, that's what's wrong! I didn't quite understand how or where I was supposed to get the POST data from, but things got a little clearer now. Thanks so much. — user4495602
– user4495602, Commented Apr 18, 2016 at 18:08

Barmar · Accepted Answer · 2016-04-18 18:09:13Z

2

The keys of $_POST come from the keys of the object you pass to the data: option, not the names of the form fields where the values originally came from. Since you used:

data: { action: selectedDateFrom }

the value will be in $_POST['action'], not $_POST['time-period-from']. So you need to use:

if (isset($_POST['action']))

and:

$dateFrom = $_POST['action'];

or you could change the Javascript to:

data: { "time-period-from": selectedDateFrom }

answered Apr 18, 2016 at 18:09

Barmar

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1 Comment

user4495602 Over a year ago

I got it working now. Thanks for explaining it thoroughly.

tapos ghosh · Accepted Answer · 2016-04-18 18:05:09Z

0

I think your selectedDateFrom variable is array that cause your post info can't you get properly .

data: {action: $('#selectedDateFrom').serializeArray()}

then you get your form data properly

answered Apr 18, 2016 at 18:05

tapos ghosh

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5 Comments

Barmar Over a year ago

Why do you think it's an array?

tapos ghosh Over a year ago

you use selectedDateFrom.length that cause i think it is array or string

Barmar Over a year ago

My guess is it's a string, probably $("#selectedDateFrom").val().

Barmar Over a year ago

I'm not sure why you would use .serializeArray() on a single element.

tapos ghosh Over a year ago

$("#datepicker-from").val() this process you can easily get this data

A.Sharma · Accepted Answer · 2016-04-18 18:08:04Z

0

You aren't grabbing the right variable on the PHP side:

if (isset($_POST['action'])) {
  require '../include/connection.php';

  $dateFrom = $_POST['action'];
  $sql = "SELECT * FROM order WHERE $dateFrom BETWEEN date_from AND date_to";
  $result = mysqli_query($connection, $sql);
  $numRows = mysqli_num_rows($result);
  echo $sql; // For testing purposes
}

answered Apr 18, 2016 at 18:08

A.Sharma

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