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16/11/2009
11:25
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davidbemenderfer
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Forum Posts: 17
Member Since:
10/11/2009
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 Sorry for so many edits, I had trouble with the formatting.

What is the function jqGrid uses to do the actual POST command?  I've been trying to post my data after assembling it without success.

When using FireBug, my POST results are:

Actual: (string)

{'id' = 8}

Desired Post: (objects)

id  =  8

What am I doing wrong?  Is there a better built in way?

My data assemby code

$.each(rows,
function( intIndex, objValue ){

var rowdata = (jQuery(”#list”).getRowData(objValue));
if (strLaborID == “”) {
strLaborID = jQuery(”#list”).getCell(objValue,”labor_id”);
}
else
{
strLaborID = strLaborID + “,” + jQuery(”#list”).getCell(objValue,”labor_id”);
}
}
);

SubmitData(”POST”,”./JSON/partytime.JSON.Submit.asp”,”{'id'=” + strLaborID + “}”, “multipart/form-data” );

My SubmitData function (using sarissa.js to define the XMLHttpRequest() object)

function SubmitData(Method,Url,Params,ContentType){
    var http =  new XMLHttpRequest();
    if (Method == 'GET') {Url = url+”?”+Params};

http.open(Method, Url, true);

    if (Method == 'POST') {
     //Send the proper header information along with the request
     http.setRequestHeader(”Content-type”, ContentType);
     http.setRequestHeader(”Content-length”, Params.length);
     http.setRequestHeader(”Connection”, “close”);
    };

    http.onreadystatechange = function() {//Call a function when the state changes.
     if(http.readyState == 4 && http.status == 200) {
      //alert(http.responseText);
     }
    };
    if (Method == 'GET') {http.send(null)} else {if (Method == 'POST') {http.send(Params)}};
    return(http.readyState);

}

    

    

16/11/2009
13:09
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davidbemenderfer
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Forum Posts: 17
Member Since:
10/11/2009
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As usual, I answer my own question.  jQuery has a built in function.

         jQuery.post(
          './JSON/partytime.JSON.Submit.asp',
          { 'id': strLaborID },
          function() {
           //alert('Data has been sent to the server.');
          }
         );

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