需求：lr需要在一串数字中随机位置插入一个新数字的实现方式

Action()
{
char s1[20],s2[4],s3[20];

int n=atoi(lr_eval_string("{rand}"));//n从1开始取值      //int n=2;//n从1开始取值

int i=0,l,y,m=0,k=0; int j=n-1;

lr_output_message("s1原始：%s", s1);

sscanf(lr_eval_string("{num}"),"%s",s1);//作用：将num参数的值，赋值给s1字符串，也可以用字符串连接函数

sscanf(lr_eval_string("{rand}"),"%s",s2);

lr_output_message("s1新值：%s", s1);

while(s1[j]!='\0')//改if为while
{
s3[m]=s1[j];
m++;
j++;
}
s3[m] = '\0';//加入字符串结束标识
while(s2[i]!='\0')//改if为while
{
s1[n-1+i]=s2[i];
i++;
}

l=n+i-1;
while(s3[k]!='\0')//改if为while
{
s1[l]=s3[k];
k++;
l++;
}
s1[l] = '\0';//加入字符串结束标识

lr_output_message("s1最终：%d", atoi(s1));

return 0;
}

int Action()
{ //char s[10]="222222"; //长度为6，i的取值只能为0-6
//int i = 3;
char s[10];
int i = atoi(lr_eval_string("{rand}"));
char *ss,*p,*ch; int j;

sscanf(lr_eval_string("{num}"),"%s",s);

lr_output_message("%s",s);

p = s;
ss = (char *)malloc(20);
ch = ss;

for(j = 0;j <strlen(s)+2;j++)
{
if(i == j)
{
*ss = 'A';
ss++;
}
else
{
*ss = *p;
ss++;
p++;

}
}

ss++;

*ss ='\0' ;

lr_output_message("%s",ch);

}