Deserialize Response

Question

I am try to post a data from a from I have this code on my html.

<javascript>
$(document).ready(function(){
$("#btnsubmit").click(function(e){
    e.preventDefault();
    var testData = $("#test").serialize();
    $.ajax({
        type: "POST",
        url: "ajaxSurvey.php",             
        data: {survey:testData}          
    });
});
})
</javascript>

and my form

<form id="test" name="test" method="POST">
  <input name="surveyperiod" id="surveyperiod" type="date">
  <input name="deadline" id="deadline" type="date"  >
  <input type="submit" id="btnsubmit"name="btnsubmit" value="   Update    ">
</form>

and my php page

if(isset($_POST['survey']){
    $myDate = $_POST['survey'];

    mysql_query('INSERT INTO (surveyperiod,deadline) VALUES (????????)');

}

Now how can I deserialize $myDate which is look like below

surveyperiod=2014-02-25&deadline=2014-02-18

mysql_* functions have been deprecated. Please use PDO or mysqli instead! — Patrick Geyer, Feb 20 '14 at 08:54
This ir pretty the same questions with good answeres: http://stackoverflow.com/questions/1792603/how-do-i-php-unserialize-a-jquery-serialized-form — lukasz, Feb 20 '14 at 08:55
it should be corrected from script itself, not from PHP dcript — Nisam, Feb 20 '14 at 08:57

score 0 · Answer 1 · answered Feb 20 '14 at 08:54

You do not need to deserialize, you can do that in different way by putting hidden field like;

HTML:

<form id="test" name="test" method="POST">
  <input name="survey" type="hidden" value="true"/>
  <input name="surveyperiod" id="surveyperiod" type="date">
  <input name="deadline" id="deadline" type="date"  >
  <input type="submit" id="btnsubmit"name="btnsubmit" value="   Update    ">
</form>

PHP:

if(isset($_POST['survey']){
    $myDate = $_POST['survey'];

    mysql_query('INSERT INTO (surveyperiod,deadline) VALUES ($_POST["surveyperiod"], $_POST["deadline"])');

}

JS:

<javascript>
$(document).ready(function(){
$("#btnsubmit").click(function(e){
    e.preventDefault();
    var testData = $("#test").serialize();
    $.ajax({
        type: "POST",
        url: "ajaxSurvey.php",             
        data: testData          
    });
});
})
</javascript>

By doing this, you dont neede to post data like {survey:testData}. Simply add,

<input name="survey" type="hidden" value="true"/>

html and check hidden field on php side. If a field with name survey exists, then run your code

jogesh_pi · Accepted Answer · 2014-02-20T09:17:44.583

0

Tricky method is use parse_str()

// Access as Variable
if( isset($_POST['survey']) ){

    // surveyperiod=2014-02-25&deadline=2014-02-18
    parse_str($_POST['survey']);

    $S_Period = $surveyperiod;
    $S_Deadline = $deadline;

   // do whatever you want
   mysql_query('INSERT INTO (surveyperiod,deadline) VALUES ( "'.$S_Period.'", "'.$S_Deadline.'" )');

}

Explanation:

In you Ajax Request you send the data via POST method( data: {survey:testData} ) and assigned a POST variable that is survey and this POST variables contains the data string surveyperiod=2014-02-25&deadline=2014-02-18 as you assigned in your javascript testData. Now what we have to do is, Now we have to parse string into variables and parse_str() inbuilt function do it for you. That's it :)

edited Feb 20 '14 at 09:17

answered Feb 20 '14 at 08:54

jogesh_pi

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jogesh_pi, it's so tricky that I can't understand :). Thank you so much. This solves my problem. – jules Feb 20 '14 at 09:11
@jules I explain it, now hope you understand it :) – jogesh_pi Feb 20 '14 at 09:18

Deserialize Response

2 Answers2

Explanation: