Java:如果结果为null，则如何分配到变量

Question

这有效：

String address1 = HtmlUtil.escape(rs.getString("address1"));

...但是我不想显示" null"。所以我尝试了这一点，但不起作用：

String address1 = HtmlUtil.escape(rs.getString("address1") || "");

是使用此问题的最佳选择：？

String address1 = HtmlUtil.escape(rs.getString("address1"));
if (address1 == null) address1 = "";

Answer 1

使用三元运算符，您可以用一行代码分配一个值：

String address1 = (rs.getString("address1") != null) ?
                  HtmlUtil.escape(rs.getString("address1")) : "";

Answer 2

String address1 = HtmlUtil.escape(rs.getString("address1"));
if (address1 == null || address1.equals("null")) {
    address1 = "";
}

Answer 3

String address1 = HtmlUtil.escape(rs.getString("address1")) == null ? "" : HtmlUtil.escape(rs.getString("address1")); 

如果您想要一个丑陋的单线： - ）

Answer 4

解决方案：

 String address1 = HtmlUtil.escape(rs.getString("address1"));
 if (address1 == null) address1 = "";

Answer 5

使用一个温度变量仅调用HtmlUtil.escape一次。然后检查null并分配给address1：

String tempAddress =  HtmlUtil.escape(rs.getString("address1"));
String address1 = tempAddress != null ? tempAddress : "";

Answer 6

尝试类似的东西

<c:when test="${empty rs.getString("address1")}">
  ...
</c:when>
<c:otherwise>
   .....
</c:otherwise>

Answer 7

您可以尝试以下方法：

String address1 = "";
if(HtmlUtil.escape(rs.getString("address1") != null && !(HtmlUtil.escape(rs.getString("address1").equalsIgnoreCase("null"))){
address1 = HtmlUtil.escape(rs.getString("address1"));
}