CODEWITHSUNDEEP
phpjquery
Azad Chouhan
Azad Chouhan

Reputation: 79

Unable to get the value from jquery post in php

I have written this code in choose-item.php 

function myFunction() {
    var values =[];
    $('#sortable2').each(function(){// id of ul
  var li = $(this).find('li')//get each li in ul
   var res = li.text().split("::");
    var myJsonString = JSON.stringify(res);
  $.post("choose-items.php", {"result": myJsonString});

console.log(myJsonString)//get text of each li*/
})

}

in the same page I have added this php script 

if(!empty($_POST['result']))
 {
      $res = $_POST['result'];
 echo "alert('$res')";
 }

But I am not getting the result in my php variable.

any body can help? Thanks

Upvotes: 1

Views: 63

Answers (2)

Jaquarh
Jaquarh

Reputation: 6673

jQuery could to be like this:

[...]
    $.post("choose-items.php", {"result": myJsonString})
        .done(function (data)
        { alert(data); }); // Your alert can be inside the function
[...]

PHP could to be like this:

if(isset($_POST['result']) && !empty($_POST['result'])):
    $res = $_POST['result'];
    echo $res; // This is the data you're passing back
endif;

However, your response will not be accurate since you're not sending it to a generic handler page followed then by your response you wanted to send ($res).

The response the request will receive is anything that the handler puts out onto the screen, so in your case, your response will be your normal page.

TL;DR & IMO: You should create a new PHP file for handling this so your code is laid out and so your response doesn't get mixed up with page content.

Upvotes: 1

shivgre
shivgre

Reputation: 1173

If you are using wrong variable names in $_POST to reference which actually do not exists then you will not be able to get the value of desired variable :

//$_POST['result[]'] do not exists but $_POST['result'] is what you passed in $.post
if(!empty($_POST['result[]']))
 {
      $res = $_POST['result'];
      echo "alert('$res')";
 }

You should correct the if condition to :

 if(!empty($_POST['result']))
 {
      $res = $_POST['result'];
      echo "alert('$res')";
 }

Also as you are using the same page to return your ajax response so you should move your process logic for ajax at top of the page with a conditional check so that it only executes when ajax is fired and not on page load. Code should be like this

// Conditional statements to allow ease of HTML
if(!empty($_POST['result'])):
    $res = $_POST['result'];
    echo "alert('$res')";
    exit;
else:
    // your normal page content
endif;

You should do a print_r at top of you script to know what is actually getting posted, or you can inspect the console ajax request for same

echo "<pre>"
print_r($_POST);
echo "</pre>"

Upvotes: 1

Related Questions