Json/Apex:为 Json 数组上的节点赋值
Json/Apex: Assigning values to nodes on the Json Array
我有一个定义 JSON 结构的顶点代码。我想寻求有关如何使用 Apex 将值分配给 JSON 上的字符串字段的建议。 JSON 将有一个数组 (PackageData),其中包含应包含值
的字段
顶点代码:
public class Shipment{
public PackageData[] PackageData;
}
public class PackageData{
public Packaging Packaging;
public Dimensions Dimensions;
public PackageWeight PackageWeight;
}
public class Packaging{
public string Code;
}
public class Dimensions{
public UnitOfMeasurement UnitOfMeasurement;
public string Length;
public string Width;
public string Height;
}
public class UnitOfMeasurement{
public string Code;
}
public class PackageWeight{
public UOM UOM;
public string Weight;
}
public class UOM{
public string Code;
}
JSON:
{
"PackageData": [
{
"Packaging": {
"Code": ""
},
"Dimensions": {
"UnitOfMeasurement": {
"Code": ""
},
"Length": "",
"Width": "",
"Height": ""
},
"PackageWeight": {
"UOM": {
"Code": ""
},
"Weight": ""
}
}
]
}
JSON 总是字符串,所以有一个解析器将对象解析为 JSON String 和反之亦然
解析器完成它的工作并自动解析为 String,您不必为此担心
Json 对象:
ClassName objName = (ClassName) System.JSON.deserialize(jsonString, ClassName.class);
对象 Json:
String jsonString = System.JSON.serialize(objName);
我有一个定义 JSON 结构的顶点代码。我想寻求有关如何使用 Apex 将值分配给 JSON 上的字符串字段的建议。 JSON 将有一个数组 (PackageData),其中包含应包含值
的字段顶点代码:
public class Shipment{
public PackageData[] PackageData;
}
public class PackageData{
public Packaging Packaging;
public Dimensions Dimensions;
public PackageWeight PackageWeight;
}
public class Packaging{
public string Code;
}
public class Dimensions{
public UnitOfMeasurement UnitOfMeasurement;
public string Length;
public string Width;
public string Height;
}
public class UnitOfMeasurement{
public string Code;
}
public class PackageWeight{
public UOM UOM;
public string Weight;
}
public class UOM{
public string Code;
}
JSON:
{
"PackageData": [
{
"Packaging": {
"Code": ""
},
"Dimensions": {
"UnitOfMeasurement": {
"Code": ""
},
"Length": "",
"Width": "",
"Height": ""
},
"PackageWeight": {
"UOM": {
"Code": ""
},
"Weight": ""
}
}
]
}
JSON 总是字符串,所以有一个解析器将对象解析为 JSON String 和反之亦然
解析器完成它的工作并自动解析为 String,您不必为此担心
Json 对象:
ClassName objName = (ClassName) System.JSON.deserialize(jsonString, ClassName.class);
对象 Json:
String jsonString = System.JSON.serialize(objName);