First I am conface that I am Newbie to php, I am using jquery(knockout js) at client side & PHP at server side. my code.

Client side: I am using knockout js(Javascript). to call my PHP service.

My Code:

self.VMSaveEditUserMode = function () {
   try {
      var params = { "ClientData": [controllerVM_.ClientID(), controllerVM_.VMList[0].ClientName(), controllerVM_.VMList[0].ShortName(), controllerVM_.VMList[0].Address(), controllerVM_.VMList[0].CreatedBy(), controllerVM_.VMList[0].CityName(), controllerVM_.VMList[0].PostalCode(), controllerVM_.VMList[0].ContactEmail(), controllerVM_.VMList[0].ContactPhone(), controllerVM_.VMList[0].IsCorporate()] };

      $.ajax({
         type: "POST",
         url: URL + "index.php/phpService/SaveClient/" + controllerVM_.TokenKey(),
         data: JSON.stringify(ko.toJS(params)),
         contentType: "application/json",
         async: true,
         dataType: 'json',
         cache: false,
         success: function (response) {
         },
         error: function (ErrorResponse) {
            if (ErrorResponse.statusText == "OK") {
            }
            else {
               alert("ErrorMsg:" + ErrorResponse.statusText);
            }
         }
      });
   }
   catch (error) {
      alert("Catch:" + error);
   }
}

Server Side My Code, I am using this PHP code to connect with DB.

PHP Code:

public function SaveClient($userToken)
    {   
       $value = json_decode($Clientdata);
       echo $value->ClientData[0];
    }

*My Question *:

• I am not clear on how to POST data in PHP ? I tried with $_POST[''] method as well as many more.
• I am using eclipse as a php framework. so, not able to debug it when i post the data.Normally mode i am able to debug my code.but not from remotely.for that i made changes on php.ini file also.

How to get Response of Post Data on php code ?

How to debug via remote post ?

My Request sample:

suppose i use:

For, data: params, only at that time my request format is.

ClientData%5B%5D=4&ClientData%5B%5D=kamlesh&ClientData%5B%5D=KAM&ClientData%5B%5D=Junagadh&ClientData%5B%5D=me&ClientData%5B%5D=SANTA+ROSA&ClientData%5B%5D=76220&ClientData%5B%5D=kamlesh.vadiyatar%40gmail&ClientData%5B%5D=9998305904&ClientData%5B%5D=false

For, data: JSON.stringify(ko.toJS(params)),

{"ClientData":["4","kamlesh","KAM","Junagadh","me","SANTA ROSA","76220","[email protected]","9998305904",false]}

First I am conface that I am Newbie to php, I am using jquery(knockout js) at client side & PHP at server side. my code.

Client side: I am using knockout js(Javascript). to call my PHP service.

My Code:

self.VMSaveEditUserMode = function () {
   try {
      var params = { "ClientData": [controllerVM_.ClientID(), controllerVM_.VMList[0].ClientName(), controllerVM_.VMList[0].ShortName(), controllerVM_.VMList[0].Address(), controllerVM_.VMList[0].CreatedBy(), controllerVM_.VMList[0].CityName(), controllerVM_.VMList[0].PostalCode(), controllerVM_.VMList[0].ContactEmail(), controllerVM_.VMList[0].ContactPhone(), controllerVM_.VMList[0].IsCorporate()] };

      $.ajax({
         type: "POST",
         url: URL + "index.php/phpService/SaveClient/" + controllerVM_.TokenKey(),
         data: JSON.stringify(ko.toJS(params)),
         contentType: "application/json",
         async: true,
         dataType: 'json',
         cache: false,
         success: function (response) {
         },
         error: function (ErrorResponse) {
            if (ErrorResponse.statusText == "OK") {
            }
            else {
               alert("ErrorMsg:" + ErrorResponse.statusText);
            }
         }
      });
   }
   catch (error) {
      alert("Catch:" + error);
   }
}

Server Side My Code, I am using this PHP code to connect with DB.

PHP Code:

public function SaveClient($userToken)
    {   
       $value = json_decode($Clientdata);
       echo $value->ClientData[0];
    }

*My Question *:

• I am not clear on how to POST data in PHP ? I tried with $_POST[''] method as well as many more.
• I am using eclipse as a php framework. so, not able to debug it when i post the data.Normally mode i am able to debug my code.but not from remotely.for that i made changes on php.ini file also.

How to get Response of Post Data on php code ?

How to debug via remote post ?

My Request sample:

suppose i use:

For, data: params, only at that time my request format is.

ClientData%5B%5D=4&ClientData%5B%5D=kamlesh&ClientData%5B%5D=KAM&ClientData%5B%5D=Junagadh&ClientData%5B%5D=me&ClientData%5B%5D=SANTA+ROSA&ClientData%5B%5D=76220&ClientData%5B%5D=kamlesh.vadiyatar%40gmail.&ClientData%5B%5D=9998305904&ClientData%5B%5D=false

For, data: JSON.stringify(ko.toJS(params)),

{"ClientData":["4","kamlesh","KAM","Junagadh","me","SANTA ROSA","76220","[email protected]","9998305904",false]}

• javascript
• php
• sql-server
• knockout.js

• try replacing data: JSON.stringify(ko.toJS(params)), with data: params, – Dave Chen Commented May 8, 2013 at 6:13
• For debugging this I'd reent FireBug + FirePHP extensions – k102 Commented May 8, 2013 at 6:13
• @k102, sorry but i am not familiar with firePHp...and i dont think so, here firebug should help me. as per my knowledge. – Jignesh.Raj Commented May 8, 2013 at 6:15

4 Answers 4

If I understand correctly you need to create a PHP service which is able to receive REST-like requests from client.

In order to do thad you need to access raw POST data. In PHP its being done like this:

$ClientData = file_get_contents('php://input');

You can read more about php://input in the wrappers documentation.

Of course from the client's side the data need to be sent using the POST method and as raw data, i.e. as a string. You can obtain a string from object using JSON.stringify() which you already do.

If you pass an object, it will be converted to string internally by jQuery using query-string format. More on that in the jQuery documentation for $.ajax (the most importatnt options being data and processData).

Just pass the ajax data param as an object, don't convert it into JSON. Then in PHP use $_POST directly.

Use firebug or chrome dev tools to analyze the ajax request and see which data is sent

Use this simple jquery function to acplish your task

$.ajax({           
            type: "POST",  
            url:"scripts/dummy.php",  
            data:"tbl="+table,
            dataType:"json", //if you want to get back response in json

            beforeSend: function()
            {                   

            },
            success: function(resp)
            {               

            }, 

            plete: function()
            {

            },

            error: function(e)
            {  
            alert('Error: ' + e);  
            }  

    }); //end Ajax

in PHP use:

if(isset($_POST['ClientData'])){
   $client_data = $_POST['ClientData']
}

now $client_data variable should contain the array.

For debugging purpose you can use php's built-in print_r() function. It's pretty handy.

here's is an example:

//make sure it's post request
if(isset($_POST)){
   //now print the array nicely
   echo "<pre>";
   print_r($_POST);
   echo "</pre>";
}