Warning: file_get_contents(/data/phpspider/zhask/data//catemap/9/java/368.json): failed to open stream: No such file or directory in /data/phpspider/zhask/libs/function.php on line 167

Warning: Invalid argument supplied for foreach() in /data/phpspider/zhask/libs/tag.function.php on line 1116

Notice: Undefined index: in /data/phpspider/zhask/libs/function.php on line 180

Warning: array_chunk() expects parameter 1 to be array, null given in /data/phpspider/zhask/libs/function.php on line 181
Java 如何从xml中提取参数?_Java_Xml_Xpath - Fatal编程技术网

Java 如何从xml中提取参数?

Java 如何从xml中提取参数?,java,xml,xpath,Java,Xml,Xpath,如何提取价值总主题 <?xml version="1.0"?> <config> <command param = "Total topics" > Check title </command> </config> 请帮我拿param XPathExpression expr = xpath.compile("//command[@param]/text()"); Object result = expr.evaluate(doc

如何提取价值
总主题

<?xml version="1.0"?>
<config>
<command param = "Total topics" > Check title
</command> 
</config>
请帮我拿param

XPathExpression  expr = xpath.compile("//command[@param]/text()");
Object result = expr.evaluate(doc, XPathConstants.NODESET);

我没有编译它,但要选择一个属性,您需要在选择节点后使用
@attributeName

如何使用
@
作为属性?试一试

XPathExpression  expr = xpath.compile("//command/@param");

这里仅作评论,有许多可用于Java的XML处理解决方案。您不仅限于XPath。简单XML和SAX以及其他许多方法都是可选的。
XPathExpression  expr = xpath.compile("//command/@param");