我试图将我的数组列表中的值放到我的jsonObjectReturn中,但是我无法将其放入。它只返回0值。这是我的代码段:
List<String> values = new LinkedList<String>();
jsonObjectReturn = JSONFactoryUtil.createJSONObject();
for (int i = 4; i <= 27; i++) {
values.add(callableStatement.getString(i));
..//I'm getting the values here..
}
System.out.println("The list: " + values.toString());
此输出列表:[118708139,I,,LABRAMONTE,JESUS,CALUMPIANO,000,127,MAYON ST ,,, QUEZON CITY ,,,000,033,00003 ,,,,, A ,C,R,0]
for (String element: values){
jsonObjectReturn.put("tpTin", element);
jsonObjectReturn.put("tpClsfCode", element);
jsonObjectReturn.put("tpRegisteredName", element);
jsonObjectReturn.put("tpLname", element);
jsonObjectReturn.put("tpFname", element);
jsonObjectReturn.put("tpMname", element);
jsonObjectReturn.put("tpaBranchCode", element);
jsonObjectReturn.put("tpaSubstreet", element);
jsonObjectReturn.put("tpaStreet", element);
jsonObjectReturn.put("tpaBarangay", element);
jsonObjectReturn.put("tpaDistrict", element);
jsonObjectReturn.put("tpaCity", element);
jsonObjectReturn.put("tpaZipCode", element);
jsonObjectReturn.put("tplBranchCode", element);
jsonObjectReturn.put("tplRdoCode", element);
jsonObjectReturn.put("tplMunCode", element);
jsonObjectReturn.put("tplContactLname", element);
jsonObjectReturn.put("tplContactFname", element);
jsonObjectReturn.put("tplContactMname", element);
jsonObjectReturn.put("tplContactTelno", element);
jsonObjectReturn.put("tpStatus", element);
jsonObjectReturn.put("tpaStatusCode", element);
jsonObjectReturn.put("tpaRegisterFlag", element);
jsonObjectReturn.put("oTin", element);
System.out.println("test " + element);
}
System.out.println("The list2: " +jsonObjectReturn);
测试的输出是测试118708139
测试我
测试
测试LABRAMONTE
测试耶稣
测试CALUMPIANO
测试000
测试127
测试MAYON ST
测试
测试
测试QUEZON CITY
测试
测试000
测试033
测试00003
测试
测试
测试
测试
测试A
试验C
试验R
测试0
list2的输出是 list2:{&#34; tpaDistrict&#34;:&#34; 0&#34;,&#34; tpaBranchCode&#34;:&#34; 0&#34;,& #34; tpaStreet&#34;:&#34; 0&#34;&#34; tpaRegisterFlag&#34;:&#34; 0&#34;&#34; tpaStatusCode&#34;:&#34; 0&# 34;,&#34; tplContactFname&#34;:&#34; 0&#34;&#34; tpClsfCode&#34;:&#34; 0&#34;&#34; oTin&#34;:&# 34; 0&#34;&#34; tpFname&#34;:&#34; 0&#34;&#34; tpaCity&#34;:&#34; 0&#34;&#34; tplRdoCode&#34 ;:&#34; 0&#34;&#34; tpaBarangay&#34;:&#34; 0&#34;&#34; tpTin&#34;:&#34; 0&#34;&#34 ; tplContactMname&#34;:&#34; 0&#34;&#34; tpRegisteredName&#34;:&#34; 0&#34;&#34; tpMname&#34;:&#34; 0&#34; &#34; tpaZipCode&#34;:&#34; 0&#34;&#34; tplBranchCode&#34;:&#34; 0&#34;&#34; tpStatus&#34;:&#34; 0&#34;&#34; tpaSubstreet&#34;:&#34; 0&#34;&#34; tplContactTelno&#34;:&#34; 0&#34;&#34; tplMunCode&#34 ;: #&34; 0&#34;&#34; tplContactLname&#34;:&#34; 0&#34;&#34; tpLname&#34;:&#34; 0&#34;} < / p>
现在,我想要发生的是将&#34; test&#34;的值放在在我的jsonObjectReturn里面。我试着把它放在forloop中,但我只能得到&#34; 0&#34;作为价值观。感谢
答案 0 :(得分:0)
你的for..each
循环是罪魁祸首。它实际上最终会将你的JSON的最后一个元素(为零)设置为你所有的jsonObjectReturn
个键
因为订单是预先确定的(我想是因为您使用的是LinkedList
),所以只需使用它来设置jsonObjectReturn
值:
jsonObjectReturn.put("tpTin", values[0]);
jsonObjectReturn.put("tpClsfCode", values[1]);
// and so on..