How to post complex XML using rest assured
Using rest-assured we can easily perform GET, POST and other methods. In the example below we are sending a POST to an API that returns a JSON response.
@Test
public void reserveARide()
{
given().
header("Authorization", "abcdefgh-123456").
param("rideId", "gffgr-3423-gsdgh").
param("guestCount", 2).
when().
post("http://someWebsite/reserveRide").
then().
contentType(ContentType.JSON).
body("result.message", equalTo("success"));
}
But I need to create POST request with complex XML body.
Body example:
<?xml version="1.0" encoding="UTF-8"?>
<request protocol="3.0" version="xxx" session="xxx">
<info1 param1="xxx" version="xxx" size="xxx" notes="xxx"/>
<info2 param1="xxx" version="xxx" size="xxx" notes="xxx"/>
</request>
How can I do this?
Thank you in advance
java xml post rest-assured
asked Feb 5 '16 at 14:21
Art PolArt Pol
2612
add a comment |
Using rest-assured we can easily perform GET, POST and other methods. In the example below we are sending a POST to an API that returns a JSON response.
@Test
public void reserveARide()
{
given().
header("Authorization", "abcdefgh-123456").
param("rideId", "gffgr-3423-gsdgh").
param("guestCount", 2).
when().
post("http://someWebsite/reserveRide").
then().
contentType(ContentType.JSON).
body("result.message", equalTo("success"));
}
But I need to create POST request with complex XML body.
Body example:
<?xml version="1.0" encoding="UTF-8"?>
<request protocol="3.0" version="xxx" session="xxx">
<info1 param1="xxx" version="xxx" size="xxx" notes="xxx"/>
<info2 param1="xxx" version="xxx" size="xxx" notes="xxx"/>
</request>
How can I do this?
Thank you in advance
java xml post rest-assured
asked Feb 5 '16 at 14:21
Art PolArt Pol
2612
add a comment |
Using rest-assured we can easily perform GET, POST and other methods. In the example below we are sending a POST to an API that returns a JSON response.
@Test
public void reserveARide()
{
given().
header("Authorization", "abcdefgh-123456").
param("rideId", "gffgr-3423-gsdgh").
param("guestCount", 2).
when().
post("http://someWebsite/reserveRide").
then().
contentType(ContentType.JSON).
body("result.message", equalTo("success"));
}
But I need to create POST request with complex XML body.
Body example:
<?xml version="1.0" encoding="UTF-8"?>
<request protocol="3.0" version="xxx" session="xxx">
<info1 param1="xxx" version="xxx" size="xxx" notes="xxx"/>
<info2 param1="xxx" version="xxx" size="xxx" notes="xxx"/>
</request>
How can I do this?
Thank you in advance
java xml post rest-assured
asked Feb 5 '16 at 14:21
Art PolArt Pol
2612
Using rest-assured we can easily perform GET, POST and other methods. In the example below we are sending a POST to an API that returns a JSON response.
@Test
public void reserveARide()
{
given().
header("Authorization", "abcdefgh-123456").
param("rideId", "gffgr-3423-gsdgh").
param("guestCount", 2).
when().
post("http://someWebsite/reserveRide").
then().
contentType(ContentType.JSON).
body("result.message", equalTo("success"));
}
But I need to create POST request with complex XML body.
Body example:
<?xml version="1.0" encoding="UTF-8"?>
<request protocol="3.0" version="xxx" session="xxx">
<info1 param1="xxx" version="xxx" size="xxx" notes="xxx"/>
<info2 param1="xxx" version="xxx" size="xxx" notes="xxx"/>
</request>
How can I do this?
Thank you in advance
java xml post rest-assured
java xml post rest-assured
asked Feb 5 '16 at 14:21
Art PolArt Pol
2612
asked Feb 5 '16 at 14:21
Art PolArt Pol
2612
asked Feb 5 '16 at 14:21
Art PolArt Pol
2612
asked Feb 5 '16 at 14:21
Art PolArt Pol
2612
asked Feb 5 '16 at 14:21
Art PolArt Pol
2612
2612
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
I keep my bodies in the resources directory, and read them into a string using the following method:
public static String generateStringFromResource(String path) throws IOException {
return new String(Files.readAllBytes(Paths.get(path)));
}
Then in my request I can say
String myRequest = generateStringFromResource("path/to/xml.xml")
given()
.contentType("application/xml")
.body(myRequest)
.when()
.put("my.url/endpoint/")
.then()
statusCode(200)
answered Mar 23 '16 at 16:12
Luke D. SmithLuke D. Smith
1015
add a comment |
I believe you can simply do this:
given().
contentType("application/xml").
body(yourbody).
...
...
You can also send serializable objects see:
https://github.com/jayway/rest-assured/wiki/Usage#serialization
answered Mar 23 '16 at 13:59
Ranil WijeyratneRanil Wijeyratne
206210
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
I keep my bodies in the resources directory, and read them into a string using the following method:
public static String generateStringFromResource(String path) throws IOException {
return new String(Files.readAllBytes(Paths.get(path)));
}
Then in my request I can say
String myRequest = generateStringFromResource("path/to/xml.xml")
given()
.contentType("application/xml")
.body(myRequest)
.when()
.put("my.url/endpoint/")
.then()
statusCode(200)
answered Mar 23 '16 at 16:12
Luke D. SmithLuke D. Smith
1015
add a comment |
I keep my bodies in the resources directory, and read them into a string using the following method:
public static String generateStringFromResource(String path) throws IOException {
return new String(Files.readAllBytes(Paths.get(path)));
}
Then in my request I can say
String myRequest = generateStringFromResource("path/to/xml.xml")
given()
.contentType("application/xml")
.body(myRequest)
.when()
.put("my.url/endpoint/")
.then()
statusCode(200)
answered Mar 23 '16 at 16:12
Luke D. SmithLuke D. Smith
1015
add a comment |
I keep my bodies in the resources directory, and read them into a string using the following method:
public static String generateStringFromResource(String path) throws IOException {
return new String(Files.readAllBytes(Paths.get(path)));
}
Then in my request I can say
String myRequest = generateStringFromResource("path/to/xml.xml")
given()
.contentType("application/xml")
.body(myRequest)
.when()
.put("my.url/endpoint/")
.then()
statusCode(200)
answered Mar 23 '16 at 16:12
Luke D. SmithLuke D. Smith
1015
I keep my bodies in the resources directory, and read them into a string using the following method:
public static String generateStringFromResource(String path) throws IOException {
return new String(Files.readAllBytes(Paths.get(path)));
}
Then in my request I can say
String myRequest = generateStringFromResource("path/to/xml.xml")
given()
.contentType("application/xml")
.body(myRequest)
.when()
.put("my.url/endpoint/")
.then()
statusCode(200)
answered Mar 23 '16 at 16:12
Luke D. SmithLuke D. Smith
1015
edited Nov 13 '18 at 8:47
answered Mar 23 '16 at 16:12
Luke D. SmithLuke D. Smith
1015
answered Mar 23 '16 at 16:12
Luke D. SmithLuke D. Smith
1015
answered Mar 23 '16 at 16:12
Luke D. SmithLuke D. Smith
1015
1015
add a comment |
add a comment |
I believe you can simply do this:
given().
contentType("application/xml").
body(yourbody).
...
...
You can also send serializable objects see:
https://github.com/jayway/rest-assured/wiki/Usage#serialization
answered Mar 23 '16 at 13:59
Ranil WijeyratneRanil Wijeyratne
206210
add a comment |
I believe you can simply do this:
given().
contentType("application/xml").
body(yourbody).
...
...
You can also send serializable objects see:
https://github.com/jayway/rest-assured/wiki/Usage#serialization
answered Mar 23 '16 at 13:59
Ranil WijeyratneRanil Wijeyratne
206210
add a comment |
I believe you can simply do this:
given().
contentType("application/xml").
body(yourbody).
...
...
You can also send serializable objects see:
https://github.com/jayway/rest-assured/wiki/Usage#serialization
answered Mar 23 '16 at 13:59
Ranil WijeyratneRanil Wijeyratne
206210
I believe you can simply do this:
given().
contentType("application/xml").
body(yourbody).
...
...
You can also send serializable objects see:
https://github.com/jayway/rest-assured/wiki/Usage#serialization
answered Mar 23 '16 at 13:59
Ranil WijeyratneRanil Wijeyratne
206210
answered Mar 23 '16 at 13:59
Ranil WijeyratneRanil Wijeyratne
206210
answered Mar 23 '16 at 13:59
Ranil WijeyratneRanil Wijeyratne
206210
answered Mar 23 '16 at 13:59
Ranil WijeyratneRanil Wijeyratne
206210
206210
add a comment |
add a comment |
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