How to sort PHP loop without page refresh using JQUERY

Question

I am trying to send variables using ajax to php page so when I click a button, the sent variable will update to the page without refreshing it.

I have a button called #toshow and when clicked, it will send ?show=okay as a post.

$('#toshow' ).click( function( e ) {
  $.post('?show=okay')
  .done(function(data) {
  });
});

and on my php

if(isset($_POST['show']) && $_POST['show']) {
  echo $_POST['show'];
}

I click the button, but it doesn't show the text 'okay'. What else do I need to add to make this work?

Thank you!

You need to fill in the code in the .done() callback to take the data and place it wherever it's needed. — VLAZ
– VLAZ, Commented Jan 26, 2019 at 13:35

Martin Heralecký · Accepted Answer · 2019-01-26 13:45:18Z

1

Whatever you print with echo in PHP wil be available (if no error occurs) in the JavaScript data variable in the callback of .done() function. So you need to implement this callback to display the returned text where you need it to be displayed.

Example:

.done(function(data) {
    document.write(data);
});

Also, the way you send the show parameter to the server (as a part of URL) is a GET method, not POST. If you want to use the POST method, pass the data as the second argument of the $.post() function:

$.post('', {
    show: 'okay'
})
    .done(...)

answered Jan 26, 2019 at 13:45

Martin Heralecký

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Comments

SilvioQ · Accepted Answer · 2019-01-26 13:45:27Z

0

You are sending "show" variable in query part of URL, then, you must retrieve it from $_GET global variable ... check ...

$('#toshow' ).click( function( e ) {
  $.post('', {"show": "okay"})
  .done(function(data) {
      console.log(data);
  });
});

https://api.jquery.com/jquery.post/#jQuery-post-url-data-success-dataType

answered Jan 26, 2019 at 13:45

SilvioQ

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