带有JSON数据的jqGrid将表呈现为空
我正在尝试创建一个jqgrid，但表是空的。 表格呈现，但数据未显示。
我从php调用回来的数据是：
{ "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"]} ]} javascript配置如下所示：
$("#"+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 });
HTML看起来像：
我不确定我是否需要定义jsonReader，因为我已经尝试保持默认值。 如果php代码有帮助，我也可以发布它。
我得到了它的工作！
dataType字段应为数据类型 。 这是区分大小写的。
当您在grid.locale-en.js之前包含脚本jquery.jqGrid.min.js时，也会出现此问题。 如果控制器的方法调用有任何问题，请检查此项。
从jqGrid 3.6迁移到jqGrid 3.7.2时遇到了同样的问题。 问题是我的JSON没有正确地双引号（根据JSON规范的要求）。 jqGrid 3.6容忍我的无效JSON，但jqGrid 3.7更严格。
请参阅此处： http ： //simonwillison.net/2006/Oct/11/json/
无效 ：
{ 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"]} ]}
有效期 ：
{ "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"]} ]}
我也得到它的工作：数据类型是正确的拼写 – 它在示例中显示，但它与库中的其他所有不一致所以它很容易出错
我非常厌倦追逐这个稀疏的文档，我真的觉得JSON，在JavaScript中使用是正确和恰当的，实际上已经得到了简短的覆盖，支持XML。 Python和JavaScript一起通过JSON，是一个非常强大的组合，但它与这个特定的库不断斗争。
有替代方案的任何人：
1>正确支持jQuery UI主题（包括圆角！）（ http://datatables.net对主题有更好的支持）
2>允许调整列的大小（ http://datatables.net不支持开箱即用）
3>允许子网格（ http://datatables.net让你通过一个事件做你想做的任何事情）
请告诉我。 我花了更多的时间在我的界面的这一部分而不是整个其余的部分组合，并且它花费了所有的时间来寻找工作示例和“尝试的东西”，这只是让人讨厌。
小号
这可能是一个较旧的post，但我会发布我的成功只是为了帮助别人。
您的JSON需要采用以下格式：
{ "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" ] } ]
}
我在Zend写了这个模型，所以如果你愿意，你可以使用它。 按照您的要求操纵它。
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; }
伙计们只想帮助你。 我得到了以下工作：
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"] } ]};
//标记下面重要的一行。 数据类型“jsonstring”为我而不是“json”工作。
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'}],
问候，
在我的情况下，问题是由以下PHP代码行引起的（它来自jqGrid演示）：
$responce->page = $page;
这里有什么问题：我访问对象$responce属性页面而不先创建它。 这导致Apache显示以下错误消息：
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
最后，错误消息曾经发送到脚本中的json reader。
我通过创建空对象来解决问题：
$responce = new stdClass();
我不认为你的ID是正确的类型，我认为它应该是一个int。
对于给定的json，您实际上不需要jsonreader设置。 你列出的是默认值，加上你的json中没有子网格。
试试这个：
{ "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"]} ]}
我正在使用WAMP 2.4，我对这个问题很疯狂，我尝试了很多东西，比如安装以前版本的PHP和5.2，我在Windows XP中尝试过，还有很多jqGrid选项。 非常感谢Oleg终于和Mariusz找到了唯一的一句话：
$responce = new stdClass();
在使用$ responce之前可以解决所有问题，现在我的网格工作得很棒!!!
谢谢我的朋友们。