jqGrid with JSON data renders table as empty

Question

I'm trying to create a jqgrid, but the table is empty. The table renders, but the data doesn't show.

The data I'm getting back from the php call is:

{
"page":"1",
"total":1,
"records":"10",
"rows":[
{"id":"2:1","cell":["1","image","Chief Scout","Highest Award test","0"]},
{"id":"2:2","cell":["2","image","Link Badge","When you are invested as a Scout, you may be eligible to receive a Link Badge. (See page 45)","0"]},
{"id":"2:3","cell":["3","image","Pioneer Scout","Upon completion of requirements, the youth is invested as a Pioneer Scout","0"]},
{"id":"2:4","cell":["4","image","Voyageur Scout Award","Voyageur Scout Award is the right after Pioneer Scout.","0"]},
{"id":"2:5","cell":["5","image","Voyageur Citizenship","Learning about and caring for your community.","0"]},
{"id":"2:6","cell":["6","image","Fish and Wildlife","Demonstrate your knowledge and involvement in fish and wildlife management.","0"]},
{"id":"2:7","cell":["7","image","Photography","To recognize photography knowledge and skills","0"]},
{"id":"2:8","cell":["8","image","Recycling","Demonstrate your knowledge and involvement in Recycling","0"]},
{"id":"2:10","cell":["10","image","Voyageur Leadership ","Show leadership ability","0"]},
{"id":"2:11","cell":["11","image","World Conservation","World Conservation Badge","0"]}
]}

The javascript configuration looks like so:

$("#"+tableId).jqGrid ({
    url:'getAwards.php?id='+classId,
    dataType : 'json',
    mtype:'POST',
    colNames:['Id','Badge','Name','Description',''],
    colModel : [
        {name:'awardId', width:30, sortable:true, align:'center'},
        {name:'badge', width:40, sortable:false, align:'center'},
        {name:'name', width:180, sortable:true, align:'left'},
        {name:'description', width:380, sortable:true, align:'left'},
        {name:'selected', width:0, sortable:false, align:'center'}
        ],
    sortname: "awardId",
    sortorder: "asc",
    pager: $('#'+tableId+'_pager'),
    rowNum:15,
    rowList:[15,30,50],
    caption: 'Awards',
    viewrecords:true,
    imgpath: 'scripts/jqGrid/themes/green/images',
    jsonReader : { 
        root: "rows", 
        page: "page", 
        total: "total", 
        records: "records", 
        repeatitems: true, 
        cell: "cell", 
        id: "id",
        userdata: "userdata", 
        subgrid: {root:"rows", repeatitems: true, cell:"cell" } 
    },
    width: 700,
    height: 200
});

The HTML looks like:

<table class="awardsList" id="awardsList2" class="scroll" name="awardsList" />
<div id="awardsList2_pager" class="scroll"></div>

I'm not sure that I needed to define jsonReader, since I've tried to keep to the default. If the php code will help, I can post it too.

Answer 1

I got it to work!

The dataType field should be datatype. It's case sensitive.

Answer 2

The problem also occures when you include script jquery.jqGrid.min.js before then grid.locale-en.js. Check this if there are any problems with controller's method call.

Answer 3

I experienced the same problem when migrating from jqGrid 3.6 to jqGrid 3.7.2. The problem was my JSON was not properly double-quoted (as required by JSON spec). jqGrid 3.6 tolerated my invalid JSON but jqGrid 3.7 is stricter.

Refer here: http://simonwillison.net/2006/Oct/11/json/

Invalid:

{
page:"1",
total:1,
records:"10",
rows:[
    {"id":"2:1","cell":["1","image","Chief Scout","Highest Award test","0"]},
    {"id":"2:2","cell":["2","image","Link Badge","When you are invested as a Scout, you may be eligible to receive a Link Badge. (See page 45)","0"]},
    {"id":"2:3","cell":["3","image","Pioneer Scout","Upon completion of requirements, the youth is invested as a Pioneer Scout","0"]}
]}

Valid:

{
"page":"1",
"total":1,
"records":"10",
"rows":[
    {"id":"2:1","cell":["1","image","Chief Scout","Highest Award test","0"]},
    {"id":"2:2","cell":["2","image","Link Badge","When you are invested as a Scout, you may be eligible to receive a Link Badge. (See page 45)","0"]},
    {"id":"2:3","cell":["3","image","Pioneer Scout","Upon completion of requirements, the youth is invested as a Pioneer Scout","0"]}
]}

Answer 4

I also got it to work: datatype is the correct spelling -- it's shown that way in the example but it is inconsistent with everything else in the library so it was easy to get wrong

I'm getting very tired chasing this sparse documentation around and I really feel like JSON, which is right and proper to be using in JavaScript, has really been given short coverage in favor of XML. Python and JavaScript together, through JSON, is a really strong combination, but it's a constant struggle with this particular library.

Anyone with an alternative that:

1> Properly supports jQuery UI themes (including rounded corners!) (http://datatables.net has much nicer support for themes)

2> Allows resizing of columns (http://datatables.net doesn't support this out of the box)

3> Allows sub-grids (http://datatables.net lets you do whatever you want here, through an event)

please let me know. I'm spending more time on this one part of my interface than on the whole rest of it combined and it's all the time spent searching for working examples and "trying things" which is just getting annoying.

S

Answer 5

This might be a older post but I will post my success just to help others.

Your JSON needs to be in this format:

{
"rows": [
    {
        "id": 1,
        "cell": [
            1,
           "lname",
            "fname",
            "mi",
            phone,
            "cell1",
            "cell2",
            "address",
            "email"
        ]
    },
    {
        "id": 2,
        "cell": [
            2,
            "lname",
            "fname",
            "mi",
            phone,
            "cell1",
            "cell2",
            "address",
            "email"
        ]
    }
]

}

and I wrote this model in Zend so you can use it if you feel like it. Manipulate it how you want.

public function fetchall ($sid, $sord)
{
    $select = $this->getDbTable()->select(Zend_Db_Table::SELECT_WITH_FROM_PART);
    $select->setIntegrityCheck(false)
           ->join('Subdiv', 'Subdiv.SID = Contacts.SID', array("RepLastName" => "LastName", 
                                                                "Subdivision" => "Subdivision",
                                                                "RepFirstName" => "FirstName"))
           ->order($sid . " ". $sord);

    $resultset = $this->getDbTable()->fetchAll($select);
    $i=0;
    foreach ($resultset as $row) {
        $entry  = new Application_Model_Contacts();

        $entry->setId($row->id);
        $entry->setLastName($row->LastName);
        $entry->setFirstName1($row->FirstName1);
        $entry->setFirstName2($row->FirstName2);
        $entry->setHomePhone($row->HomePhone);
        $entry->setCell1($row->Cell1);
        $entry->setCell2($row->Cell2);
        $entry->setAddress($row->Address);
        $entry->setSubdivision($row->Subdivision);
        $entry->setRepName($row->RepFirstName . " " . $row->RepLastName);
        $entry->setEmail1($row->Email1); 
        $entry->setEmail2($row->Email2);

        $response['rows'][$i]['id'] = $entry->getId(); //id
        $response['rows'][$i]['cell'] = array (
                                                $entry->getId(),
                                                $entry->getLastName(),
                                                $entry->getFirstName1(),
                                                $entry->getFirstName2(),
                                                $entry->getHomePhone(),
                                                $entry->getCell1(),
                                                $entry->getCell2(),
                                                $entry->getAddress(),
                                                $entry->getSubdivision(),
                                                $entry->getRepName(),
                                                $entry->getEmail1(),
                                                $entry->getEmail2()
                                            );
        $i++;

    }
    return $response;
}

Answer 6

Guys just want to help you in this. I got following worked:

JSON

var mydata1 = { "page": "1", "total": 1, "records": "4","rows": [{ "id": 1, "cell": ["1", "cell11", "values1" ] },
    { "id": 2, "cell": ["2", "cell21", "values1"] },
    { "id": 3, "cell": ["3", "cell21", "values1"] },
    { "id": 4, "cell": ["4", "cell21", "values1"] }
]};

//Mark below important line. datatype "jsonstring" worked for me instead of "json".

datatype: "jsonstring",

contentType: "application/json; charset=utf-8",

datastr: mydata1,

colNames: ['Id1', 'Name1', 'Values1'],

colModel: [
      { name: 'id1', index: 'id1', width: 55 },
      { name: 'name1', index: 'name1', width: 80, align: 'right', sorttype: 'string' },
      { name: 'values1', index: 'values1', width: 80, align: 'right', sorttype: 'string'}],

Regards,

Answer 7

In my case, the problem was caused by the following line of PHP code (which was taken from jqGrid demo):

$responce->page = $page;

What is wrong here is that: I am accessing property page of object $responce without creating it first. This caused Apache to display the following error message:

Strict Standards: Creating default object from empty value in /home/mariusz/public_html/rezerwacja/apps/frontend/modules/service/actions/actions.class.php on line 35

And finally the error message used to be send to json reader within the script.

I fixed the problem by creating empty object:

$responce = new stdClass();

Answer 8

I don't think your ID is the correct type, I think it should be an int.

For the given json you really don't need the jsonreader settings. What you have listed is the defaults anyway, plus you don't have a subgrid in your json.

Try this:

{
"page":"1",
"total":1,
"records":"10",
"rows":[
{"id":1 ,"cell":["1","image","Chief Scout","Highest Award test","0"]},
{"id":2,"cell":["2","image","Link Badge","When you are invested as a Scout, you maybe eligible to receive a Link Badge. (See page 45)","0"]},
{"id":3,"cell":["3","image","Pioneer Scout","Upon completion of requirements, the youth is invested as a Pioneer Scout","0"]},
{"id":4,"cell":["4","image","Voyageur Scout Award","Voyageur Scout Award is the right after Pioneer Scout.","0"]},
{"id":5,"cell":["5","image","Voyageur Citizenship","Learning about and caring for your community.","0"]},
{"id":6,"cell":["6","image","Fish and Wildlife","Demonstrate your knowledge and involvement in fish and wildlife management.","0"]},
{"id":7,"cell":["7","image","Photography","To recognize photography knowledge and skills","0"]},
{"id":8,"cell":["8","image","Recycling","Demonstrate your knowledge and involvement in Recycling","0"]},
{"id":9,"cell":["10","image","Voyageur Leadership ","Show leadership ability","0"]},
{"id":10,"cell":["11","image","World Conservation","World Conservation Badge","0"]}
]}

Answer 9

I was working with WAMP 2.4, I was being crazy with this problem, I tried lot of things, like install previous versions of PHP and like 5.2, een I tried in Windows XP, and lots of jqGrid options. Well thank to Oleg finally and Mariusz I find the only line:

$responce = new stdClass(); 

Before the use of $responce could solve all, and now my grid is works Great!!!

Thanks my friends.