我想知道是否可以编写一个春季批处理作业，该作业的步骤只有一个编写器。我在 spring 批处理文档中找不到任何关于给定步骤本质上必要的文档。
我希望做类似的事情：
public class MyBatchConfiguration {
@Bean
public ItemWriter<myInfo> myWriter() {
return new MyWriter();
}
@Bean
public Step myStep(StepBuilderFactory stepBuilderFactory,
ItemWriter<? super Object> myWriter,
PlatformTransactionManager transactionManager) {
return stepBuilderFactory.get("myStep")
.chunk(1)
.writer(myWriter).
transactionManager(transactionManager).
build();
}
@Bean
public Job myBatch(JobBuilderFactory jobs, Step myStep, JobExecutionListener listener) {
return jobs.get("myBatch")
.incrementer(new RunIdIncrementer())
.flow(exceptionReporterStep)
.end()
.listener(listener)
.build();
}
}
出于某种原因，嵌入的 CSS 或 HTML 中都没有选择我的样式标签。字体系列更改和border-top: 5px solid black;行都没有生效。有小费吗？
<!DOCTYPE html>
<html>
<head>
<title>
My Resume
</title>
<style type=“text/css”>
#ob {
border-top: 5px solid black;
}
</style>
</head>
<body>
<div>
<h1 style=“font-family: sans-serif”>
<strong>Corey Schnedl </strong>
</h1>
<h4 style=“font-family: Sans-Serif”>
<strong> XXXXXXXXXXX / XXXXXXXX / XXXXX / XXXXX </strong>
<br>
<strong>XXXXXXXXXXXXX</strong>
</h4>
</div>
<div>
<h3 style=“font-family:Arial”>
<strong>Objective</strong>
</h3>
<p>
To obtain an entry-level position as a Java software developer.
</div>
</body>
</html>
monster* monster1 = new monster("Frankenstein", "The Ugly One", BackYard);
Player* player1 = new Player("Corey", "The Chosen one", Atrium);
Player* player2 = new Player("Darth Vader", "The Evil One", Atrium);
vector<Agent*> agents;
agents.push_back(monster1);
agents.push_back(player1);
agents.push_back(player2);
while (true)
{
vector<Agent*>::iterator it;
for (it = agents.begin(); it < agents.end(); it++) {
it->act(); // Error here
if (it->act() == false) // Error here
return 0;
}
...
}
我收到一个错误说:
成员引用基类型'Agent*'不是结构或联合.
我真的不明白为什么这不能导航向量.