Efficient way to replace strings in java

Question

I have a string template like this:

http://server/{x}/{y}/{z}/{t}/{a}.json

And I have the values:

int x=1,y=2,z=3,t=4,a=5;

I want to know which is the efficent way to replace the {x} to the value of x, and so does y,z,t,z?

Do you know the exact order of the parameters in advanced, or the template could also be something like http://server/{z}/{x}/{y}.json? — Elist
– Elist, Commented May 13, 2013 at 7:28
The order is not sure, since the url may be something like this http://server/{y}.json?x={x}&y={y} — hguser
– hguser, Commented May 13, 2013 at 7:29
I read all the fllowing answsers, and it seems that all of them will require the order of the placeholders? — hguser
– hguser, Commented May 13, 2013 at 7:34
That's right. I think there's no built in way in Java to do it, and you should manually iterate over the string and replace the '{?}' with the accurate value. — Elist
– Elist, Commented May 13, 2013 at 7:36
@hguser No. You can use http://server/{1}.json?x={0}&y={1}. — Eng.Fouad
– Eng.Fouad, Commented May 13, 2013 at 7:36

jlordo · Accepted Answer · 2013-05-13 07:17:45Z

16

String template = "http://server/%s/%s/%s/%s/%s.json";
String output = String.format(template, x, y, z, t, a);

answered May 13, 2013 at 7:17

jlordo

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Eng.Fouad · Accepted Answer · 2013-05-13 07:22:33Z

4

Another way to do it (C# way ;)):

MessageFormat mFormat = new MessageFormat("http://server/{0}/{1}/{2}/{3}/{4}.json");
Object[] params = {x, y, z, t, a};
System.out.println(mFormat.format(params));

OUTPUT:

http://server/1/2/3/4/5.json

answered May 13, 2013 at 7:22

Eng.Fouad

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2 Comments

hguser Over a year ago

While int x=27315, then the formatted string is something like .../27,315/....... What is the problem?

Eng.Fouad Over a year ago

You could use Object[] params = {Integer.toString(x)}; or Object[] params = {"" + x};.

Rahul · Accepted Answer · 2013-05-13 07:30:12Z

4

Use MessageFormat.java

MessageFormat messageFormat = new MessageFormat("http://server/{0}/{1}/{2}/{3}/{4}.json");
Object[] args = {x,y,z,t,a};
String result = messageFormat.format(args);

edited May 13, 2013 at 7:30

answered May 13, 2013 at 7:21

Rahul

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user207421 · Accepted Answer · 2013-05-13 07:21:41Z

3

http://server/{x}/{y}/{z}/{t}/{a}.json

If you can change that to http://server/{0}/{1}/{2}/{3}/{4}.json you can use MessageFormat:

String s = MessageFormat.format("http://server/{0}/{1}/{2}/{3}/{4}.json", x, y, z, t, a);

answered May 13, 2013 at 7:21

user207421

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Cjxcz Odjcayrwl · Accepted Answer · 2013-05-13 07:45:32Z

1

To replace the placeholders exact how you use them in example, you can use StrinUtils.replaceEach

org.apache.commons.lang.StringUtils.replaceEach(
"http://server/{x}/{y}/{z}/{t}/{a}.json",
new String[]{"{x}","{y}","{z}","{t}","{a}"},
new String[]{"1","2","3","4","5"});

However, MessageFormat will be more efficient, but requiring to replace x with 0, y with 1 etc.

If you change your format to

"http://server/${x}/${y}/${z}/${t}/${a}.json",

you can consider using Velocity, which has parser specialized on finding ${ and } occurences.

The most efficient way would be to write own parser searching for next { occurence, than }, than replacing the placeholder.

answered May 13, 2013 at 7:45

Cjxcz Odjcayrwl

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Chakit Patel · Accepted Answer · 2013-05-13 08:28:08Z

0

This Could be probable answer.

    String url= "http://server/{x}/{y}/{z}/{t}/{a}.json";
    int x=1,y=2,z=3,t=4,a=5;

    url = url.replaceAll("\\{x\\}", String.valueOf(x));
    url = url.replaceAll("\\{y\\}", String.valueOf(y));
    url = url.replaceAll("\\{z\\}", String.valueOf(z));
    url = url.replaceAll("\\{t\\}", String.valueOf(t));
    url = url.replaceAll("\\{a\\}", String.valueOf(a));

    System.out.println("url after : "+ url);

answered May 13, 2013 at 8:28

Chakit Patel

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