Question

我有一个以编码形式传入URL的参数。所以，如果JSON是

{"size":{"value":"5\"x10\" Kraft (250 Envelopes) Value Pack"}, "color":{"value":""},"packSize":{"value":""},"style":{"value":""}}

然后以编码形式：

%7B%22size%22%3A%7B%22value%22%3A%225%5C%22x10%5C%22%20Kraft%20(250%20Envelopes)%20Value%20Pack%22%7D%2C%20%22color%22%3A%7B%22value%22%3A%22%22%7D%2C%22packSize%22%3A%7B%22value%22%3A%22%22%7D%2C%22style%22%3A%7B%22value%22%3A%22%22%7D%7D

当我使用我的Java代码时，我会这样：

{"size":{"value":"5\"x10\" Kraft (250 Envelopes) Value Pack"}, "color":{"value":""},"packSize":{"value":""},"style":{"value":""}}

我有一个VariantContainer类，我想分配这些值。

所以我尝试了两种方法，但是我得到了例外。

方法1：

try {
   # variant is value I get from URL 
   JsonElement variantRequestJson = new JsonParser().parse(variant);


   variantContainer.setSize(variantApiRequestJson.getAsJsonObject().
                    get("size").getAsJsonObject().get("value").toString());

 # Similarly other variations, but its giving null pointer exception
 } catch (Exception ex) {
            logger.error(ex);
}

方法2：

try {
    variantContainer = gson.fromJson(variant, VariantContainer.class);
    } catch (Exception ex) {
            logger.error(ex);
  }

它给出例外：

com.google.gson.JsonSyntaxException: java.lang.IllegalStateException: Expected a string but was BEGIN_OBJECT at line 1 column 10

让我知道，遗漏了什么。我怎样才能以最佳方式获得它？

VariantContainer类

import com.google.gson.Gson;

public class VariantContainer {
private String size;
private String color;
private String packSize;
private String style;

public String getSize() {
    return size;
}

public void setSize(String size) {
    this.size = size;
}

public String getColor() {
    return color;
}

public void setColor(String color) {
    this.color = color;
}

public String getStyle() {
    return style;
}

public void setStyle(String style) {
    this.style = style;
}

public String getPackSize() {
    return packSize;
}

public void setPackSize(String packSize) {
    this.packSize = packSize;
}
@Override
public String toString() {
    Gson gson = new Gson();
    return gson.toJson(this);
}
}

在URL中编码Json给出异常

0 个答案: