java 在java中将空字符串解析为整数
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parsing empty string to integer in java
提问by Adham
if(jsonArray.getJSONObject(i).get("SECNO").toString()!=null && jsonArray.getJSONObject(i).get("SECNO").toString().trim()!="")
appointment.mSecNo =Integer.parseInt(jsonArray.getJSONObject(i).get("SECNO").toString());
else
appointment.mSecNo = -1;
In the previouse lines, when the value of jsonArray.getJSONObject(i).get("SECNO").toString()equales to '' it doesn't be caught by the if statement ..
在前几行中，当值jsonArray.getJSONObject(i).get("SECNO").toString()等于 '' 时，它不会被 if 语句捕获。
and I get this error message .. can't parse '' to integer
我收到此错误消息.. can't parse '' to integer
回答by Jon Skeet
Don't use ==or !=to compare strings in Java - it only compares the references, not the contentsof the strings. Also, I doubt that toStringwould ever return null. I suspect you want:
不要在 Java 中使用==或!=比较字符串——它只比较引用，而不是字符串的内容。另外，我怀疑这toString会永远返回 null。我怀疑你想要：
Foo x = jsonArray.getJSONObject(i).get("SECNO");
if (x != null && x.toString().trim().length() > 0)
(I don't know what the type of jsonArray.getJSONObject(i).get("SECNO")would be, hence the Foo.)
（我不知道jsonArray.getJSONObject(i).get("SECNO")会是什么类型，因此是Foo.）
In this particular case I've used length() > 0to detect a non-empty string - but for more general equality, you'd want to use equals, so an alternative is:
在这种特殊情况下，我曾经length() > 0检测过一个非空字符串 - 但对于更一般的相等性，您需要使用equals，所以另一种方法是：
if (x != null && !x.toString().trim().equals(""))
回答by Dunes
Don't compare Strings with ==
不要将字符串与 ==
See: Java comparison with == of two strings is false?, Java String.equals versus ==
回答by Rosdi Kasim
Why not just,
为什么不只是，
int appointment.mSecNo = -1;
try {
appointment.mSecNo = Integer.parseInt(jsonArray.getJSONObject(i).get("SECNO").toString());
}catch(Exception e) {
log.error(" your error message ");
}
