Dart将JSON解析为表
我正在试图弄清楚如何将JSON提要解析为Dart中的表。 我不确定我是否应该使用Map，但如果我这样做，我不确定如何处理提取数据。 我希望有一个单元格来保存assetID，然后另一个单元格用于保存域名，另一个用于保存IP地址。 如果您需要更多信息，请与我们联系。
镖：
void loadData() { var url = "http://localhost:8080/***selectAll"; //call the web server asynchronously var request = HttpRequest.getString(url).then(onDataLoaded); } void onDataLoaded(String response) { TableElement table = querySelector("#domainTableSelector"); //table.children.add(); var jsonString = response; // print(jsonString); List list = new List(); list.add(jsonString); for (var x = 0; x < list.length; x++) { //create new table row TableRowElement row = table.insertRow(x+1); TableCellElement cell = row.insertCell(x); cell.text = list.toString(); // print(list); } // Iterator itr= list.iterator(); // // print("Displaying List Elements,"); // // while(itr.hasNext()) // { // print(itr.next()); // } } JSON
[{"serviceResponseValue":[{"assetId":"8a41250446b89b5f0146b04d49910023","oplock":0,"longitude":115.86,"domainName":"free-aus-trip.au","latitude":-31.95,"ipAddress":"6.0.0.6"},{"assetId":"8a49859246918966014691b1aac9000c","oplock":0,"longitude":-65.30,"domainName":null,"latitude":-24.18,"ipAddress":"4.0.0.4"},{"assetId":"8a49859246876566014691b1437512e4","oplock":0,"longitude":77.60,"domainName":"allmovies.cn","latitude":12.97,"ipAddress":"14.0.0.14"},{"assetId":"8a49850446b04b5f0146b04d49910000","oplock":0,"longitude":112.47,"domainName":"getrichez.cn","latitude":32.98,"ipAddress":"5.0.0.5"},{"assetId":"8a498592469189660146919b7a210006","oplock":0,"longitude":-37.61,"domainName":"googles.com","latitude":55.75,"ipAddress":null},{"assetId":"8a42250876b89b5f0876b04d49910763","oplock":0,"longitude":-68.90,"domainName":"lolcatzfun.net","latitude":-22.48,"ipAddress":"8.0.0.8"},{"assetId":"8a498592469189660146919f8d700008","oplock":0,"longitude":113.50,"domainName":"ccn.com","latitude":52.03,"ipAddress":null},{"assetId":"8a45250446b89b5f0876b04d49910187","oplock":0,"longitude":115.84,"domainName":"free-aus-trip.au","latitude":-31.86,"ipAddress":"7.0.0.7"},{"assetId":"8a49859246918966014691aeda76000a","oplock":0,"longitude":3.38,"domainName":"cashnow.net","latitude":6.52,"ipAddress":"2.0.0.2"},{"assetId":"8a49859246918966014691ae19df0009","oplock":0,"longitude":7.48,"domainName":"free-money.tv","latitude":9.07,"ipAddress":"222.222.222.222"},{"assetId":"8a498592469189660146919e09900007","oplock":0,"longitude":30.34,"domainName":"facebok.com","latitude":59.93,"ipAddress":"111.111.111.222"},{"assetId":"8a49859246918966014691b14375000b","oplock":0,"longitude":116.41,"domainName":null,"latitude":39.90,"ipAddress":"0.0.0.111"}],"messages":{"messages":[]}}]
要使用JSON，你在dart中有一件好事： dart：convert => JSON
这里有一些使用示例：
var encoded = JSON.encode([1, 2, { "a": null }]); var decoded = JSON.decode('["foo", { "bar": 499 }]');
我做了一些事情，但我不确定这是否完全适合你的需要
import 'dart:html'; import 'dart:convert'; void loadData() { var url = "http://localhost:8080/***selectAll"; //call the web server asynchronously var request = HttpRequest.getString(url).then(onDataLoaded); } void onDataLoaded(String response) { TableElement table = querySelector("#domainTableSelector"); //table.children.add(); var jsonString = response; // print(jsonString); var jsonObject = JSON.decode(jsonString); for (var x = 0; x < jsonObject.length; x++) { //create new table row TableRowElement row = table.insertRow(x+1); for (var d in jsonObject[x]["serviceResponseValue"]) { TableCellElement cell = row.insertCell(x); cell.text = d["assetId"]; cell = row.insertCell(x); cell.text = d["domainName"]; cell = row.insertCell(x); cell.text = d["ipAddress"]; print(d["assetId"]); print(d["domainName"]); print(d["ipAddress"]); } // print(list); } }
您使用将JSON解析为Dart数据结构
import 'dart:convert' show JSON; var decoded = JSON.decode(jsonString);
另请参阅如何在Dart中处理JSON （Alexandres回答）
然后decoded为Dart数据结构（列表，映射，值），您可以在调试器中迭代或调查。 如果您需要进一步的帮助，请添加评论。