jqGrid：formatter：リモートjson urlに対して "actions"が機能しません

Question

jqGridフォーマッタを使用しようとしました：リモートjson urlの "actions"が列名と実際に返されたデータの数が一致しないためact列は列名に余分なものです）。私はオンラインで検索しましたが、すべての例がjsonがキー/値として使用されているローカルデータ（ページ上）用です。url：somepage.phpはcsvの種類の列（キーなしの値） 。また、すべての例がidを表示している間、私は隠しフィールドとしてIDを使用します。もう1つの非表示列も使用します。 は、ここに私のcolmodelです：

url: "manager_json", 
       editurl: "manager_edit", 
       datatype: "json", 
       //data: mydata, 
       width:1000, 
       //jsonReader: { 
       //    repeatitems : false, 
       //   }, 
       colNames: ["","Id","Uid","Custom Id","Image Name","Coord. X","Coord. Y","Gender","Progress","Status","Created","Updated"], 
       colModel: [{"name":"act","template":"actions","formatoptions":{"editformbutton":true}},{"name":"id","align":"center","width":33,"editable":false,"hidden":true},{"name":"uid","align":"center","width":33,"editable":true,"hidden":true,"editrules":{"edithidden":false,"required":false}},{"name":"ta_id","align":"center","width":100,"editrules":{"required":true}},{"name":"image","width":150,"template":"text","editrules":{"required":true}},{"name":"x","width":100,"align":"center","template":"integer","editrules":{"required":true}},{"name":"y","width":100,"align":"center","template":"integer","editrules":{"required":true}},{"name":"gender","width":100,"align":"center","formatter":"select","stype":"select","editrules":{"required":true},"edittype":"select","editoptions":{"value":"f:Female;m:Male","defaultValue":"m"},"searchoptions":{"sopt":["eq","ne"],"value":":Any;f:Female;m:Male"}},{"name":"progress","width":120,"align":"center","formatter":"select","stype":"select","editrules":{"required":true},"editable":false,"edittype":"select","editoptions":{"value":"-:New;pf:Process Failed;ps:Process Scheduled;pss:Process Success;p:Processing...;s:Staging;tf:Test Failed;ts:Test Scheduled;tss:Test Success;t:Testing...","defaultValue":"m"},"searchoptions":{"sopt":["eq","ne"],"value":":Any;-:New;pf:Process Failed;ps:Process Scheduled;pss:Process Success;p:Processing...;s:Staging;tf:Test Failed;ts:Test Scheduled;tss:Test Success;t:Testing..."}},{"name":"status","width":100,"align":"center","formatter":"select","stype":"select","edittype":"select","editoptions":{"value":"a:Active;n:New","defaultValue":"m"},"searchoptions":{"sopt":["eq","ne"],"value":":Any;a:Active;n:New"}},{"name":"date_created","width":150,"template":"text","editable":false},{"name":"last_updated","width":150,"template":"text","editable":false}] 

そして、ここで私のサーバー（DB）からjqgridの仕様に従って、私の実際のデータ：

{"page":1,"total":1,"records":7,"rows":[{"id":32,"cell":[32,"889daf31ff3e49544f52850258439600","2uu","2ok",2,2,"m","-","a","2017-02-10 18:57:05","2017-02-10 23:37:12"]},{"id":30,"cell":[30,"","11","11",1,1,"m","-","a","2017-02-10 18:01:52","2017-02-10 18:01:52"]},{"id":29,"cell":[29,"aaa","ww222111uu","11",1,1,"m","-","a","2017-02-10 18:00:36","2017-02-10 23:37:08"]},{"id":27,"cell":[27,"","11","1",1,1,"m","-","a","2017-02-10 17:57:41","2017-02-10 17:57:41"]},{"id":25,"cell":[25,"","4tt","4img",4,4,"f","-","n","2017-02-10 17:50:21","2017-02-11 00:26:03"]},{"id":24,"cell":[24,"","1","1",1,1,"m","-","a","2017-02-10 17:49:38","2017-02-10 17:49:38"]},{"id":22,"cell":[22,"","bbb","imam 222",2,22,"f","p","n","2017-02-08 20:14:55","2017-02-10 13:27:57"]}]} 

あなたはセル行がcolNamesとcolModel未満の列を持って見ることができるように。また、idとuidの列は表示されません。私は、DBから余分な偽の欄に追加しましたが、それは私が事前4.13.7 jqGridを使用しています

を助けなかった

Answer 1

私はあなたが

{ 
    "id": 32, 
    "uid": "889daf31ff3e49544f52850258439600", 
    "ta_id": "2uu", 
    "image": "2ok", 
    "x": 2, 
    "y": 2, 
    "gender": "m", 
    "progress": "-", 
    "status": "a", 
    "date_created": "2017-02-10 18:57:05", 
    "last_updated": "2017-02-10 23:37:12" 
} 

代わりの

のようなオブジェクトを返すようにお勧めします

{ 
    "id": 32, 
    "cell": [ 
     32, "889daf31ff3e49544f52850258439600", "2uu", "2ok", 2, 2, "m", "-", "a", 
     "2017-02-10 18:57:05", "2017-02-10 23:37:12" 
    ] 
} 

をサーバーの応答として使用します。それはあなたのコードを簡素化します。

あなたがデータの現在の形式を処理する必要があります場合は、使用状況によってオプション

jsonReader: { 
    repeatitems: false 
} 

をrepeatitems: falseモードを強制するとのどの要素からjqGridを知らせるためにcolModelのjsonmapプロパティを使用する必要があります項目1は列データを読み取る必要があります。あなたは

var columnOrder = { 
     id: 0, 
     uid: 1, 
     ta_id: 2, 
     image: 3, 
     x: 4, 
     y: 5, 
     gender: 6, 
     progress: 7, 
     status: 8, 
     date_created: 9, 
     last_updated: 10 
    }; 

どのようなオブジェクトを定義することができ、コードをより良いメンテナンス性にするために

jsonmap: function (item) { 
    return item.cell[1]; 
} 

：jsonmapはどちらか

jsonmap: "cell.1" 

または関数形式のような文字列の形であることができcellのインデックスを列名で提供し、jsonmapのように

jsonmap: function (item) { 
    return item.cell[columnOrder.uid]; 
} 

デモhttps://jsfiddle.net/OlegKi/ozzgnaeh/2/がアプローチを示しています。