How to send a curl request in java

Question

I'm using a java WebService to send a curl request, this is my request :

curl -d "input = a b c d" 'localhost:8090/jobs?appName=Test40&classPath=fr.aid.cim.spark.JavaWord&sync=true&timeout=1000000000'

I don't know which library I should use and how I can write it in java. Can you please show me how to do it?

There is no such thing as a 'curl request'. Curl is a command line tool to do some form of an online request. — Gimby
– Gimby, Commented Jan 15, 2015 at 10:44
Ok thank you for your suggestions i will take a lokk. #Reimeus is not a duplication cause i saw it but i could'nt use it with localHost request !! — Amine CHERIFI
– Amine CHERIFI, Commented Jan 15, 2015 at 10:50

DmitryKanunnikoff · Accepted Answer · 2015-01-15 14:44:04Z

6

If you need execute console command (curl in this case):

Runtime runtime = Runtime.getRuntime();

try {
    Process process = runtime.exec("curl -d \"input = a b c d\" 'localhost:8090/jobs?appName=Test40&classPath=fr.aid.cim.spark.JavaWord&sync=true&timeout=1000000000'");
    int resultCode = process.waitFor();

    if (resultCode == 0) {
        // all is good
    } 
} catch (Throwable cause) {
    // process cause
}

UPDATE (according to your comment): Add folowing lines:

System.out.println("is: " + IOUtils.toString(process.getInputStream()));
System.out.println("es: " + IOUtils.toString(process.getErrorStream()));

What is the output? IOUtils can be found here.

edited Jan 15, 2015 at 14:44

answered Jan 15, 2015 at 10:51

DmitryKanunnikoff

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1 Comment

Amine CHERIFI Over a year ago

Thank you for your answer, but my resultCode is allways = 6 ??

ikettu · Accepted Answer · 2015-01-15 10:50:53Z

3

You should take look Apache HttpClient library. With HC Fluent component you can write something like:

Request.Post("http://localhost:8090/jobs?appName=Test40&classPath=fr.aid.cim.spark.JavaWord&sync=true&timeout=1000000000")
.execute().returnContent();

HC Fluent Javadocs

edited Jan 15, 2015 at 10:50

answered Jan 15, 2015 at 10:44

ikettu

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3 Comments

Amine CHERIFI Over a year ago

Thank you. But i got this error : java.lang.IllegalArgumentException: Illegal character in scheme name at index 4

ikettu Over a year ago

Pleace check the uri string that there is no extra characters etx.

ikettu Over a year ago

Did you get it working, or do you need further help?

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