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我怎樣才能創建一個JSON與逗號分隔的許多小數據集?生成一組數據到Json而不是一個大的混合Json
而不是由雙大括號括起來的一個大Json?
我確實收到一個Json和php,我使用foreach來循環它,使得很多數據處理在裏面。
然後生成一個新的Json,只是爲了避免在客戶端的數據處理,這將通過angularjs ng-repeat進行處理。
所有JSON數據被混合到一個大的JSON集(雙大括號內)
我的目標是分成小數據集。
我可以使用NrType屬性。在這個腳本中,NrType接收最後一個屬性,只有最後一個屬性可用。
//The php script
$arr = json_decode($returnedJson); //The original json to be pre-processed
$processedData = "[]";
$processedJson = json_decode($processedData,true);
foreach($arr as $key=>$value) {
foreach($value as $vkey=>$vvalue) {
if($value[$i]->NrType == 1) {
$VlMIni = $value[$i]->QttyInitial;
$VlMSub = $value[$i]->QttyPeriod + $value[$i]->QttyRealAfter;
$VlMRec = $value[$i]->RealValue;
$VlMTotal += $VlMesRece;
//much more data processing going on here ...
} elseif($value[$i]->NrType == 2) {
.
.
.
//and much more data processing going on here ...
}
}
//simple data atribution here
$processedJson['labelDesIni'] = 'Instruments';
$processedJson['labelValueMIni'] = $lblVlMIni;
$processedJson['labelValuePIni'] = $lblVlPlIni;
$processedJson['labelValueAIni'] = $lblVlAIni;
$processedJson['labelValuePAIni'] = $lblVlPAIni;
$processedJson['labelValuePercInic'] = $lblVlPercInic;
$processedJson['labelValuePerc2Inic'] = $lblVlPerc2Inic;
//much more data atribution ...
echo json_encode($processedJson); //the new hgenerated Json
生成的JSON:
{
labelDesI: "Inspection",
labelValueMI: "2357",
labelValuePlI: "3914066",
labelValueAI: "1389406",
labelValuePAI: "2431425",
labelValuePercI: 57.143691456656,
labelValuePerc2I: 35.497766261478,
labelDesR: "Instruments",
labelValueMR: "734.54",
labelValuePR: "819.14",
labelValueAR: "660.05",
labelValuePAR: "877.94",
labelValuePercR: 80.087,
labelValuePerc2R: 44.739,
labelDesAcfi: "Fiscalização",
labelValueMAcfi: "343",
labelValuePlAcfi: "29907",
labelValueAAcfi: "16718",
labelValuePAAcfi: "16493",
labelValuePercAcfi: 101.36421512157,
labelValuePerc2Acfi: 55.899956531916,
labelDesT: "Totals",
labelValueMT: 365.59,
labelValuePlT: 547.62,
labelValueAnT: 909.63,
labelValuePAnT: 957.63,
labelValuePercT: 22949,
labelValuePerc2T: 25065
}
所述期望的格式將是這樣:
{
label: "Inspection",
labelValue1: "2357",
labelValue2: "3914066",
labelValue3: "1389406",
labelValue4: "2431425",
labelValue5: 57.1456656,
labelValue6: 35.4961478
},
{
labelDesR: "Instruments",
labelValue1: "734.54",
labelValue2: "819.14",
labelValue3: "660.05",
labelValue4: "877.94",
labelValue5: 80.087,
labelValue6: 44.739
},
{
labelDesT: "Totals",
labelValue1: 365.59,
labelValue2: 547.62,
labelValue3: 909.63,
labelValue4: 957.63,
labelValue5: 22949,
labelValue6: 25065
}
Thank's預先
我嘗試直接添加一個索引,如下所示:** $ processedJson [11] ['label1'] \t = $ label1; **並生成劃分部分,我需要謝謝@eggmaters –
甚至更好。別客氣。如果需要,您可以將其作爲接受的答案進行檢查。 – eggmatters
謝謝!你的回答對我很有幫助。 –