js获取s:set值，来高手求解

Posted 2023-04-08

js获取 <s:set name='cur' value="1"/>
不要跟我说再页面执行完了查看源码，s:set在页面里是看不到源码的。

参考技术A 我也遇到了，稍微变通下，以下可行
<s:set name="buttonType" value="%'insertAction'"/>

<script type="text/javascript">
var buttonType = "<s:property value='#buttonType'/>";
</script> 参考技术B 你要去什么属性的值呢 ??
<s:property value="#cur.name"/> <br>
<s:property value="#cur.age"/> <br>
<script>
var name ="$属性名";
</script>追问

这个貌似可以在js里获取，但如何修改，s：set的值

追答

这个可以在后台去处理...

参考技术C document.getElementsByName('cur')[0].value

这个应该是你要的值追问

页面都看不到s:set执行完后的源码，显然这不行

追答

那个不是sturts2 的标签么

c++ 求解 编程高手

In statistics, the mode of a set of values is the value that occurs most often or with the greatest frequency. Write a function that accepts as arguments the following:
A) An array of integers
B) An integer that indicates the number of elements in the array
The function should determine the mode of the array. That is, it should determine which value in the array occurs most often. The mode is the value the function should return. If the array has no mode (none of the values occur more than once), the function should return -1. (Assume the array will always contain nonnegative values.)
Demonstrate your pointer prowess by using pointer notation instead of array notation in this function.

//设计上可能有点局限
#if 1
//In statistics, the mode of a set of values is the value that occurs most often or with the greatest frequency. Write a function that accepts as arguments the following:
// A) An array of integers
// B) An integer that indicates the number of elements in the array
// The function should determine the mode of the array. That is, it should determine which value in the array occurs most often. The mode is the value the function should return. If the array has no mode (none of the values occur more than once), the function should return -1. (Assume the array will always contain nonnegative values.)
// Demonstrate your pointer prowess by using pointer notation instead of array notation in this function.
#include<iostream>
#include<ctime>
const int ArSize = 10;
void createArr(int * arr,int size);/* 创建一个随机数组 */
void printArr(const int * arr,int size);/* 打印数组 */
int mode(const int * arr,int size,int * maxCount);
int main()

using namespace std;
int * arr = new int[ArSize];//申请空间
createArr(arr,ArSize);
printArr(arr,ArSize);
int maxCount=0;
int modei = mode(arr,ArSize,&maxCount);
cout<<"the value:"<<modei<<" occurs most often,total "<<maxCount<<" times\n";
delete [] arr;//释放空间
return 0;

void createArr( int * arr,int size )

int i;
srand((unsigned int)time(0));//真随机数
for (i=0;i<size;i++)

*(arr+i) = rand()%size;


void printArr(const int * arr,int size )

printf("arr = [");
for (int i=0;i<size-1;i++)

printf("%2d,",*(arr+i));
if ((i+1)%10==0)

printf("\n");

if((i+1)%10==0&&(i+1)/10>0)

printf(" ");


printf("%d]\n",arr[size-1]);


int mode(const int * arr,int size, int * maxCount)

int * temp = new int[size];//假设100个值各出现一次,用索引表示arr中的值,用temp元素值表示出现的次数
for (int k=0;k<size;k++)

temp[k]=0;//数组初始化

for (int i=0;i<size;i++)

temp[arr[i]]++;

printArr(temp,ArSize);
//#define TEST
#ifdef TEST
using namespace std;
int finds = 0;
for (int u=0;u<size;u++)

cout<<u<<" find "<<temp[u]<<endl;
finds+=temp[u];

cout<<"finds===="<<finds<<endl;
#endif
//然后找到数组中最大的值
*maxCount=temp[0];
int value = -1;
value = temp[0]>1?0:-1;
for (int j=0;j<size;j++)

if (temp[j]>(*maxCount)&&temp[j]>1)

(*maxCount) = temp[j];
value = j;


delete temp;
return value;


#endif追问

void reset_timerr ()

driven_times=0;

类中 初始化这个 对象 可以这样写么

参考技术A #include<iostream>
using namespace std;
int Mode(int n,int *p)

int position,count=1,countx=1;
int *next;
int *first=p;
for(int i=0;i<n-1;i++)

p=first+i;
next=p+1;
for(int j=i+1;j<n;j++)


if(*p==*next)
countx++;
if(countx>count)

count=countx;
position=i;

next++;

countx=1;

if(1==count)
return -1;
else
return p[position];

int main()

int n;
cin>>n;
int *a=new int[n];
for(int i=0;i<n;i++)
cin>>a[i];
cout<<Mode(n,a)<<endl;
return 0;


//看下对不对。追问

能给下你QQ么

追答

我QQ就是我名字咯。

参考技术B #include <cstdlib>
#include <iostream>

using namespace std;

int modeFinder(int*,int size);

int main(int argc, char *argv[])

int sizeOfArray = 1;
int mode = -1;
cin>>sizeOfArray;
int array[sizeOfArray];
for(int i=0;i!=sizeOfArray;i++)
cin>>array[i];
mode = modeFinder(array,sizeOfArray);
if(mode == -1)
cout<<"mode of the array is not existed"<<endl;
else
cout<<mode<<endl;
while(getchar()=='\n');
return 0;


int modeFinder(int* ptr,int size)

int freq = 0;
int mode = -1;
for(int i=0; i!=size;i++)

int tmp_freq = 0;
for(int j=0;j!=size;j++)

if(*(ptr+i) == *(ptr+j) && i!=j)
tmp_freq++;

if(tmp_freq > freq)

freq = tmp_freq;
mode = *(ptr+i);


return mode;


the array name is exactly the pointer to the array, that is, var array is the pointer of array[n];
you can use both array[i] and *(array + i) to access ith value in array[n];
hope my answer helps :-) 参考技术C 汗。代码看懂了，E文看不懂。 参考技术D 学习了