如何使用PHP中的参数实例化/调用java类? [英] How to instantiate/call java class with parameter from PHP?
问题描述
例如,我有java类JRXlsExporter,如果我从java实例化java我的代码是:
For example, i have java class JRXlsExporter, if i instantiated that java from java my code is :
JRXlsExporter myObject= new JRXlsExporter();
从PHP变为:
and from PHP become:
$myObject = new Java("net.sf.jasperreports.engine.export.JRXlsExporter");
它有效,但如果我有带参数的java类,例如:
It works, but if i have java class with parameter, for example :
JRXlsExporter myObject= new JRXlsExporter(param1,param2);
如何从PHP实例化/调用该java类?
How to instantiate/call that java class from PHP ?
推荐答案
这是示例如何 java.util.Date
对象实例化,带有一些构造函数参数:
Here is an example of how java.util.Date
object is instantiated, with some constructor arguments:
$date = new Java("java.util.Date", 70, 9, 4);
此外,它说:
新的Java(java.util.Date,70,9,4)调用使用java.util.Date(int year,int month)创建java.util.Date类的新实例,int day)构造函数。
The new Java("java.util.Date", 70, 9, 4) call creates a new instance of the java.util.Date class using the java.util.Date(int year, int month, int day) constructor.
所以,你可以试试这个:
So, you could maybe try this:
$myObject = new Java("net.sf.jasperreports.engine.export.JRXlsExporter", param1, param2);
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