how to convert maven pom.xml into build.gradle file?
up vote
1
down vote
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I have a maven pom.xml file I need to convert into a build.gradle file. How do I do this? For example, I am defining a resources file in the pom.xml file:
<resources>
<resource>
<directory>src/main/resources/</directory>
<includes>
<include>config.properties</include>
</includes>
</resource>
</resources>
xml maven gradle
asked Nov 19 at 20:49
user840930
1,366144163
add a comment |
up vote
1
down vote
favorite
I have a maven pom.xml file I need to convert into a build.gradle file. How do I do this? For example, I am defining a resources file in the pom.xml file:
<resources>
<resource>
<directory>src/main/resources/</directory>
<includes>
<include>config.properties</include>
</includes>
</resource>
</resources>
xml maven gradle
asked Nov 19 at 20:49
user840930
1,366144163
1
did you check guides.gradle.org/migrating-from-maven ?
– M.Ricciuti
Nov 19 at 20:53
Cool! Thank you for the link! I will try it!
– user840930
Nov 19 at 21:09
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have a maven pom.xml file I need to convert into a build.gradle file. How do I do this? For example, I am defining a resources file in the pom.xml file:
<resources>
<resource>
<directory>src/main/resources/</directory>
<includes>
<include>config.properties</include>
</includes>
</resource>
</resources>
xml maven gradle
asked Nov 19 at 20:49
user840930
1,366144163
I have a maven pom.xml file I need to convert into a build.gradle file. How do I do this? For example, I am defining a resources file in the pom.xml file:
<resources>
<resource>
<directory>src/main/resources/</directory>
<includes>
<include>config.properties</include>
</includes>
</resource>
</resources>
xml maven gradle
xml maven gradle
asked Nov 19 at 20:49
user840930
1,366144163
asked Nov 19 at 20:49
user840930
1,366144163
asked Nov 19 at 20:49
user840930
1,366144163
asked Nov 19 at 20:49
user840930
1,366144163
asked Nov 19 at 20:49
user840930
1,366144163
1,366144163
1
did you check guides.gradle.org/migrating-from-maven ?
– M.Ricciuti
Nov 19 at 20:53
Cool! Thank you for the link! I will try it!
– user840930
Nov 19 at 21:09
add a comment |
1
did you check guides.gradle.org/migrating-from-maven ?
– M.Ricciuti
Nov 19 at 20:53
Cool! Thank you for the link! I will try it!
– user840930
Nov 19 at 21:09
1
1
did you check guides.gradle.org/migrating-from-maven ?
– M.Ricciuti
Nov 19 at 20:53
did you check guides.gradle.org/migrating-from-maven ?
– M.Ricciuti
Nov 19 at 20:53
Cool! Thank you for the link! I will try it!
– user840930
Nov 19 at 21:09
Cool! Thank you for the link! I will try it!
– user840930
Nov 19 at 21:09
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
When you do gradle init in a directory with pom.xml it will try to convert some configuration parts automagically for you. Though it's not complete but you'll get the most basic parts converted: multi-module setup, repositories and dependencies.
Generally, src/main/resources is already recognized as "resources" directory by both Maven and Gradle, so you don't need to configure it separately unless the config is not standard. E.g. you can enable filtering by:
<resource>
<directory>src/main/resources</directory>
<filtering>true</filtering>
</resource>
In Gradle you'll need to configure processResources (processMainResources) task:
processResources {
expand project.properties
}
However, it's not the only way and you may find filter more useful.
answered Nov 21 at 6:55
madhead
14k1382117
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
When you do gradle init in a directory with pom.xml it will try to convert some configuration parts automagically for you. Though it's not complete but you'll get the most basic parts converted: multi-module setup, repositories and dependencies.
Generally, src/main/resources is already recognized as "resources" directory by both Maven and Gradle, so you don't need to configure it separately unless the config is not standard. E.g. you can enable filtering by:
<resource>
<directory>src/main/resources</directory>
<filtering>true</filtering>
</resource>
In Gradle you'll need to configure processResources (processMainResources) task:
processResources {
expand project.properties
}
However, it's not the only way and you may find filter more useful.
answered Nov 21 at 6:55
madhead
14k1382117
add a comment |
up vote
0
down vote
accepted
When you do gradle init in a directory with pom.xml it will try to convert some configuration parts automagically for you. Though it's not complete but you'll get the most basic parts converted: multi-module setup, repositories and dependencies.
Generally, src/main/resources is already recognized as "resources" directory by both Maven and Gradle, so you don't need to configure it separately unless the config is not standard. E.g. you can enable filtering by:
<resource>
<directory>src/main/resources</directory>
<filtering>true</filtering>
</resource>
In Gradle you'll need to configure processResources (processMainResources) task:
processResources {
expand project.properties
}
However, it's not the only way and you may find filter more useful.
answered Nov 21 at 6:55
madhead
14k1382117
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
When you do gradle init in a directory with pom.xml it will try to convert some configuration parts automagically for you. Though it's not complete but you'll get the most basic parts converted: multi-module setup, repositories and dependencies.
Generally, src/main/resources is already recognized as "resources" directory by both Maven and Gradle, so you don't need to configure it separately unless the config is not standard. E.g. you can enable filtering by:
<resource>
<directory>src/main/resources</directory>
<filtering>true</filtering>
</resource>
In Gradle you'll need to configure processResources (processMainResources) task:
processResources {
expand project.properties
}
However, it's not the only way and you may find filter more useful.
answered Nov 21 at 6:55
madhead
14k1382117
When you do gradle init in a directory with pom.xml it will try to convert some configuration parts automagically for you. Though it's not complete but you'll get the most basic parts converted: multi-module setup, repositories and dependencies.
Generally, src/main/resources is already recognized as "resources" directory by both Maven and Gradle, so you don't need to configure it separately unless the config is not standard. E.g. you can enable filtering by:
<resource>
<directory>src/main/resources</directory>
<filtering>true</filtering>
</resource>
In Gradle you'll need to configure processResources (processMainResources) task:
processResources {
expand project.properties
}
However, it's not the only way and you may find filter more useful.
answered Nov 21 at 6:55
madhead
14k1382117
answered Nov 21 at 6:55
madhead
14k1382117
answered Nov 21 at 6:55
madhead
14k1382117
answered Nov 21 at 6:55
madhead
14k1382117
14k1382117
add a comment |
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1
did you check guides.gradle.org/migrating-from-maven ?
– M.Ricciuti
Nov 19 at 20:53
Cool! Thank you for the link! I will try it!
– user840930
Nov 19 at 21:09