parsing empty string to integer in java

Question

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 ..

and I get this error message .. can't parse '' to integer

Checkout my one-line code answer: stackoverflow.com/a/77739302/4621922 — Gbenga Adebowale
– Gbenga Adebowale, Commented Dec 31, 2023 at 17:37

Jon Skeet · Accepted Answer · 2011-10-15 08:20:02Z

2

Don't use == or != to compare strings in Java - it only compares the references, not the contents of the strings. Also, I doubt that toString would ever return null. I suspect you want:

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.)

In this particular case I've used length() > 0 to detect a non-empty string - but for more general equality, you'd want to use equals, so an alternative is:

if (x != null && !x.toString().trim().equals(""))

answered Oct 15, 2011 at 8:20

Jon Skeet

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Rosdi Kasim · Accepted Answer · 2011-10-15 08:29:42Z

0

Why not just,

int appointment.mSecNo = -1;

try {
    appointment.mSecNo = Integer.parseInt(jsonArray.getJSONObject(i).get("SECNO").toString());
}catch(Exception e) {
    log.error(" your error message ");
}

answered Oct 15, 2011 at 8:29

Rosdi Kasim

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Community · Accepted Answer · 2017-05-23 10:34:32Z

0

Don't compare Strings with ==

See: Java comparison with == of two strings is false? , Java String.equals versus ==

edited May 23, 2017 at 10:34

CommunityBot

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answered Oct 15, 2011 at 8:23

Dunes

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