Send data to DB and get back via AJAX

Question

The following code works properly and I am using it 'cause I need to update some notes in DB without refreshing a page. But what have to do when I need in the same event get some data back?

index.php where change should occur

<?php
$s = $GET['...']
$p = $GET['...']
print '
 <form id="notesForm">
  <textarea name="c">'.$note.'</textarea>
  <input type="text" value="LAST_CHANGE_NEED_GIVE_BACK_FROM_DB">
  <input type="hidden" name="s" value="'.$s.'">
  <input type="hidden" name="p" value="'.$p.'">
  <input type="submit" value="save note">
  <div id="response"></div>
</form>
';
?>

script which is posting data to live_notes.php and probably should be changed because of the goal

$(document).ready(function(){
    $('#notesForm').submit(function(){
        event.preventDefault();
        $('#response').html("<b>saving...</b>");
        $.ajax({
            type: 'POST',
            url: 'notes/live_notes.php',
            data: $(this).serialize()
        })
        .done(function(data){
            $('#response').html(data);
        })
        .fail(function() {
            alert("bad luck");
        });
        return false;
    });
});

notes/live_notes.php

<?php
$s = $_POST['s'];
$p = $_POST['p'];
$c = $_POST['c'];

// connecting DB

mysql_query("
UPDATE `poznamky`
SET
    last_change = now(),
    page = '$p',
    content = '$c'
WHERE
    page = '$p';
");

any idea?

Actually, need to reload some data from db in the same event and really don't know how to do it...:/ — falcon
– falcon, Commented Mar 28, 2014 at 6:20
you want to perform update then select some data from db and return to ajax? — William Buttlicker
– William Buttlicker, Commented Mar 28, 2014 at 6:21

Awlad Liton · Accepted Answer · 2014-03-28 06:22:27Z

1

<?php
$s = $_POST['s'];
$p = $_POST['p'];
$c = $_POST['c'];

// connecting DB

$blnSuccess = mysql_query("
UPDATE `poznamky`
SET
    last_change = now(),
    page = '$p',
    content = '$c'
WHERE
    page = '$p';
");

if($blnSuccess){
echo "success";
}
else {
echo "not success!";
}

in your jquery :

.done(function(data){
            $('#response').html(data); 
}

data have success or not success! which return as ajax response.

Update
If you want to return some data from database then:

$result = mysql_query("
   select * 
      from `poznamky`
");

while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) {
    echo $row['content'];
}

Then it will return all content field from poznamky table into data in ajax.

answered Mar 28, 2014 at 6:22

Awlad Liton

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3 Comments

falcon Over a year ago

Thank you for ur time and response Awlad :)

falcon Over a year ago

Well, you said "which return as ajax response, but how can I require specific column instead of echoing (not) success. Isn't Sarah's response something what I need?? Thanks

falcon Over a year ago

Ops... Im sorry.. .] Have a nice day

Shahid Karimi · Accepted Answer · 2014-03-28 09:04:29Z

1

<?php
$s = $_POST['s'];
$p = $_POST['p'];
$c = $_POST['c'];

// connecting DB

$OK = mysql_query("
UPDATE `poznamky`
SET
    last_change = now(),
    page = '$p',
    content = '$c'
WHERE
    page = '$p';
");

$id = mysql_insert_id();
$res = mysql_query("select * from poznamky where id = $id");


$data = mysql_fetch_array($res);

//Print raw or JSON of $data like json_encode($data) etc

edited Mar 28, 2014 at 9:04

answered Mar 28, 2014 at 6:23

Shahid Karimi

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1 Comment

falcon Over a year ago

I would like to, but have not enough reputation .)

William Buttlicker · Accepted Answer · 2014-03-28 06:25:44Z

0

If you want to return something back then in your notes.php you can do something like this:

$s = $_POST['s'];
$p = $_POST['p'];
$c = $_POST['c'];

// connecting DB

mysql_query("
UPDATE `poznamky`
SET
    last_change = now(),
    page = '$p',
    content = '$c'
WHERE
    page = '$p';
");

$select = "//use your select query and loop here and get all data in this variable";

exit($select);

and in jquery:

.done(function(data){
           //data has the response. You can do whatever you want to do with it.
}

answered Mar 28, 2014 at 6:25

William Buttlicker

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