Write data to a String

Question

This is probably very easy to solve, but I am stuck on this for a while now. I got a while loop which I use to write data. I want to write the data from the while loop to a String.

public void dumpPart(Part p) throws Exception {
    InputStream is = p.getInputStream();
    if (!(is instanceof BufferedInputStream)) {
        is = new BufferedInputStream(is);
    }
    int c;
    System.out.println("Message: ");
    while ((c = is.read()) != -1) {
        System.out.write(c);  //I want to write this data to a String.

    }
    sendmail.VerstuurEmail(mpMessage, kenmerk);
}

Solved:

public void dumpPart(Part p) throws Exception {
    InputStream is = p.getInputStream();
    if (!(is instanceof BufferedInputStream)) {
        is = new BufferedInputStream(is);
    }
    int c;
     final StringWriter sw = new StringWriter();
    System.out.println("Message: ");
    while ((c = is.read()) != -1) {
        sw.write(c);
    }
    mpMessage = sw.toString();;
    sendmail.VerstuurEmail(mpMessage, kenmerk);
}

Thanks for your help.

@PeterLawrey I want to store the data in just a String, sorry. — Jef
– Jef, Commented Sep 17, 2012 at 11:19

Alexandre Dupriez · Accepted Answer · 2012-09-17 11:20:03Z

6

You can consider a java.io.StringWriter (as of JDK 1.4+) :

 System.out.println("Message: ");

 final StringWriter sw = new StringWriter();

 int c;
 while ((c = is.read()) != -1) {
    sw.write(c);
 }

 String data = sw.toString();

answered Sep 17, 2012 at 11:20

Alexandre Dupriez

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Comments

Peter Lawrey · Accepted Answer · 2012-09-17 11:19:21Z

5

I would use IOUtils.toString(inputStream) or something like it.

answered Sep 17, 2012 at 11:19

Peter Lawrey

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1 Comment

Herm Over a year ago

this is what I'd prefer if you don't need the while-loop for other reasons. Just convert it with: String sPart = IOUtils.toString(is);

Dan D. · Accepted Answer · 2012-09-17 11:25:32Z

2

Instead of the System.out call just initialize a StringBuffer before the loop and append to it:

StringBuffer s = new StringBuffer();
while ((c = is.read()) != -1) {
  s.append((char) c);
}

edited Sep 17, 2012 at 11:25

answered Sep 17, 2012 at 11:18

Dan D.

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3 Comments

Peter Lawrey Over a year ago

c is an int so it may not compile.

Peter Lawrey Over a year ago

I wouldn't use a StringBuffer if I can use a StringBuilder vanillajava.blogspot.co.uk/2012/08/…

Dan D. Over a year ago

Right. Just that the StringBuffer is compatible with the pre-1.5 era.

Community · Accepted Answer · 2017-05-23 11:55:56Z

1

Its best to do it using the StringBuilder Object rather than StringBuffer ( Difference between StringBuilder and StringBuffer )

public void dumpPart(Part p) throws Exception {
    InputStream is = p.getInputStream();
    if (!(is instanceof BufferedInputStream)) {
        is = new BufferedInputStream(is);
    }
    int c;
    StringBuilder sb = new StringBuilder();
    System.out.println("Message: ");
    while ((c = is.read()) != -1) {
        sb.append(c);

    }
    String result= sb.toString();
    sendmail.VerstuurEmail(mpMessage, kenmerk);
}

edited May 23, 2017 at 11:55

CommunityBot

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answered Sep 17, 2012 at 11:20

MaVRoSCy

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Comments

Henrique · Accepted Answer · 2012-09-17 11:24:57Z

1

One possibility is:

        int c;
    System.out.println("Message: ");
    StringWriter sw = new StringWriter();
    while ((c = is.read()) != -1) {
        sw.write(c);
    }
    System.out.println(sw.toString());

answered Sep 17, 2012 at 11:24

Henrique

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Comments

Santosh · Accepted Answer · 2012-09-17 11:28:52Z

1

One more way :

StringBuffer buffer = new StringBuffer();
byte[] b = new byte[4096];
for (int n; (n = is.read(b)) != -1;) {
  buffer.append(new String(b, 0, n));
}
String str = buffer.toString();

answered Sep 17, 2012 at 11:28

Santosh

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