how to fix java.lang.NumberFormatException: Invalid int: ""

Question

public class VitalCompare implements Comparator<VitalReportsDetails> {
    @Override
    public int compare(VitalReportsDetails vitalReportsDetails, VitalReportsDetails t1) {
        int n1 = Integer.parseInt((vitalReportsDetails.getValue().equals("") ? "0" : vitalReportsDetails.getValue()));

        int n2 = Integer.parseInt((t1.getValue().equals("")? "0" : t1.getValue()));
        if (n1 >= n2) {
            return 1;
        }
        return -1;

    }

}

calling like this :

int min=Integer.parseInt(Collections.min(listOfData, new VitalCompare()).getValue());

Logcat

  8-07 11:17:49.604 27972-27972/com.cognistrength.caregiver E/AndroidRuntime: FATAL EXCEPTION: main
                                                                             Process: com.cognistrength.caregiver, PID: 27972
                                                                             java.lang.NumberFormatException: Invalid int: ""
                                                                                 at java.lang.Integer.invalidInt(Integer.java:138)
                                                                                 at java.lang.Integer.parseInt(Integer.java:358)
                                                                                 at java.lang.Integer.parseInt(Integer.java:334)
                                                                                 at com.cognistrength.caregiver.adapters.VitalGraphAdapter.onBindViewHolder(VitalGraphAdapter.java:123)
                                                                                 at com.cognistrength.caregiver.adapters.VitalGraphAdapter

Just throwing this out there... why is `vitalReportsDetails.getValue()` a `String` and not an `Integer`? — Joe C, Aug 07 '17 at 05:34
The comparator is used to sort the data, it does not change them.. — , Aug 07 '17 at 05:34
Also, your `Comparator` needs to return `0` when the two values are the same. Yours does not. — Joe C, Aug 07 '17 at 05:34
Start by making `getValue()` return an `int` and not a `String`. Then your fix becomes trivial. — Joe C, Aug 07 '17 at 05:38
as @IntelliJ Amiya said, it has no value that can be convert to int. for example if the value is "Ali", how can you expect to change that to integer!? — Mehran Zamani, Aug 07 '17 at 05:41
Post the full exception, the text of the exception for me (Oracle java 8) is java.lang.NumberFormatException: For input string: "", if it is really saying Invalid int: "" then you have a different error than I have seen or a different version of the JVM. You are handling the single invalid string empty string, any non-numeric string will produce the same error, 'Ali' for example as Mehran suggested. If the string has nothing but unprintable characters then the exception message could show an empty string. — Steve Bauer, Aug 07 '17 at 05:46

score 3 · Answer 1 · answered Aug 07 '17 at 05:50

You get "" because that value is the min value in that list (your comparator said that). You can handle that by simply call String temp = Collections.min(listOfData, new VitalCompare()).getValue(); int min = Integer.parseInt(temp.equals("") ? "0" : temp);

score 1 · Answer 2 · edited Aug 07 '17 at 05:39

1

Maybe you can use:

vitalReportsDetails.getValue().equals("") || vitalReportsDetails.isEmpty() ? "0" : vitalReportsDetails.getValue()

edited Aug 07 '17 at 05:39

ΦXocę 웃 Пepeúpa ツ

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answered Aug 07 '17 at 05:37

Andres Lopez

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IntelliJ Amiya · Answer 3 · 2017-08-07T05:48:30.163

1

java.lang.NumberFormatException: Invalid int: ""

NumberFormatException

Thrown to indicate that the application has attempted to convert a string to one of the numeric types, but that the string does not have the appropriate format.

     int n1 = Integer.parseInt((vitalReportsDetails.getValue().equals("") ? "0" : vitalReportsDetails.getValue()));
     int n2 = Integer.parseInt((t1.getValue().equals("")? "0" : t1.getValue()));

Problem coming from n1 & n2 . Debug both .

The statement [Either n1 or n2] would throw NumberFormatException because it generate String Which cannot be parsed to int.

edited Aug 07 '17 at 05:48

answered Aug 07 '17 at 05:37

IntelliJ Amiya

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I am converting String to Integer – Adevelopment Aug 07 '17 at 05:40
@Adevelopment for test case DEBUG please – IntelliJ Amiya Aug 07 '17 at 05:42
@Adevelopment You are Getting `SPACE` – IntelliJ Amiya Aug 07 '17 at 05:54
i have applied isEmpty and trim also then also same exception is coming – Adevelopment Aug 07 '17 at 06:13

Omar Dhanish · Answer 4 · 2017-08-07T06:28:18.257

0

Same problem happened to me , this happens when string is in the form of Float or Double or like other form other than numbers this error throws if it is in form of double or float try like below method

    float floatstring = FLoat.parseFloat("your_string");
    //then
    int int1= (int) Math.round(floatstring );

for Double

double doublestring = Double.parseDouble("100.22");
        //then
        int int2= (int) Math.round(doublestring );

edited Aug 07 '17 at 06:28

answered Aug 07 '17 at 05:45

Omar Dhanish

885
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i am getting String that we have to COnvert in int – Adevelopment Aug 07 '17 at 05:58
Yes , First convert your string to float or double and then convert it to int . See the above code . – Omar Dhanish Aug 07 '17 at 06:08
I have edited the answer recheck now – Omar Dhanish Aug 07 '17 at 06:13
still same issue i have changed according ur logic – Adevelopment Aug 07 '17 at 06:17
@ Omar Danisha are u there can u check my code ? – Adevelopment Aug 07 '17 at 06:38
ok , actually it should work , how is your string received ? , if the string is in like "hello" it can't parse .... – Omar Dhanish Aug 07 '17 at 06:58
Let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/151240/discussion-between-adevelopment-and-omar-danisha). – Adevelopment Aug 07 '17 at 07:17

how to fix java.lang.NumberFormatException: Invalid int: ""

4 Answers4