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在controller中有两个action:
public class JapidTestController extends JapidController {
public static void view1() {
renderJapid();
}
public static void willBeTempalteInvoke() {
renderJapid();
}
}
view1 模板的代码如下:
can not be invoke?
`for(int i=0; i<10; i++){
`a JapidTestController.willBeTempalteInvoke()
`}
此时view1模板的编译结果为: @OverRide
protected void doLayout() {
//------
p("\n" +
"\n" +
"\n" +
"can not be invoke?\n");// line 1
for (int i = 0; i < 10; i++) {// line 5
------期望这里出现回调,可是没有--------------------------
}// line 7
p("\n" +
"\n");// line 7
}
The text was updated successfully, but these errors were encountered:
在项目中遇到一个问题:如果在模板中有一个循环,循环体中只做一件事:回调action。此时编译的结果不是所期望的,我将例子简化如下:
在controller中有两个action:
public class JapidTestController extends JapidController {
public static void view1() {
renderJapid();
}
public static void willBeTempalteInvoke() {
renderJapid();
}
}
view1 模板的代码如下:
can not be invoke? `for(int i=0; i<10; i++){ `a JapidTestController.willBeTempalteInvoke() `}此时view1模板的编译结果为:
@OverRide
protected void doLayout() {
//------
p("\n" +
"\n" +
"\n" +
"can not be invoke?\n");// line 1
for (int i = 0; i < 10; i++) {// line 5
------期望这里出现回调,可是没有--------------------------
}// line 7
p("\n" +
"\n");// line 7
The text was updated successfully, but these errors were encountered: