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问题描述

我的要求是,使用WAITFOR条件,蚂蚁应该定期检查,如果字符串成功打造会显示在日志文件中。如果找到该字符串,那么应该执行特定动作。

My requirement is that, using waitfor condition, ant should periodically check if string "Build Successful" is displayed in log file. If the string is found, then particular action should be performed.

推荐答案

下面是你可以这样做的一种方式的例子:

Here's an example of one way you might do this:

<target name="wait-for">
    <waitfor maxwait="15" maxwaitunit="second" timeoutproperty="build.timeout">
        <resourcecontains resource="build.log" substring="Build Successful" />
    </waitfor>
    <antcall target="build-success" />
</target>

<target name="build-success" depends="build-fail" unless="build.timeout">
    <echo message="Success" />
</target>
<target name="build-fail" if="build.timeout">
    <echo message="Fail" />
</target>

使用 C>,除非取决于被用来做的if-else逻辑的需要。如果你只需要采取行动在成功的情况下的的失败,你可以稍微简化。

Use the resourcecontains condition to look for the string in the named resource - in this case the file 'build.log'.If it's not found in the allotted time, the build.timeout property is set. There are two targets, one that is to be runif the string is found, the other if not. The 'target' attributes if, unless, and depends are used to make the if-else logic need. If you only need to take an action in the case of success or failure, you can simplify slightly.

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08-28 06:28