索修远のBlog路漫漫其修远兮，吾将上下而求索。2024-08-12T14:30:59.785Zhttp://example.com/索修远Hexocentos7 yum报错：cannot find a valid baseurl for repo:base/7/x86_64的解决方案http://example.com/posts/e49a5a03.html2024-08-09T11:19:31.000Z2024-08-12T14:30:59.785Z
已解决centos7 yum报错：cannot find a valid baseurl for repo:base/7/x86_64的解决方案
报错说明
出现cannot find a valid baseurl for repo:base/7/x86_64错误通常是由于YUM仓库源无法找到或无法访问，导致YUM无法正常工作。这种情况常见于CentOS 7系统。解决这个问题需要检查几个方面，如网络连接、DNS设置和YUM仓库源配置。以下是详细的排查解决方法。
-- 查询用户的权限列表 SELECT DISTINCT p.permissin FROM tb_user u JOIN tb_role r ON JSON_CONTAINS(u.role, CAST(r.id AS CHAR)) JOIN tb_permission p ON JSON_CONTAINS(r.permissions, CAST(p.id AS CHAR)) WHERE u.id=用户id AND u.status=1;
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<select id="searchUserPermissions" parameterType="int" resultType="String"> SELECT DISTINCT p.permission_name FROM tb_user u JOIN tb_role r ON JSON_CONTAINS(u.role, CAST(r.id AS CHAR)) JOIN tb_permission p ON JSON_CONTAINS(r.permissions, CAST(p.id AS CHAR)) WHERE u.id=#{userId} AND u.status=1; </select>
@RestControllerAdvice @Slf4j public class ExceptionAdvice { @ResponseBody @ResponseStatus(HttpStatus.INTERNAL_SERVER_ERROR) @ExceptionHandler(Exception.class) // 捕获全局异常 public String validExceptionHandler(Exception e){
public class XssHttpServletRequestWrapper extends HttpServletRequestWrapper { public XssHttpServletRequestWrapper(HttpServletRequest request) { super(request); }
@Override public String getParameter(String name) { String value = super.getParameter(name); if (!StrUtil.hasEmpty(value)){ value=HtmlUtil.filter(value); } return value; }
@Override public String[] getParameterValues(String name) { String[] values = super.getParameterValues(name); if (values != null) { for (int i = 0; i < values.length; i++) { String value = values[i]; if (!StrUtil.hasEmpty(value)){ value=HtmlUtil.filter(value); } values[i]=value; } } return values; }
public class Main { public static void main(String[] args) {
Scanner sc=new Scanner(System.in); int n=sc.nextInt(); int[]arr=new int[n]; int[]sn=new int[n]; for (int i = 0; i < arr.length; i++) { arr[i]=sc.nextInt(); } sn[0]=arr[0]; for (int i = 1; i < arr.length; i++) { sn[i]=sn[i-1]+arr[i]; } for (int i = 0; i < sn.length; i++) { System.out.println(sn[i]); } // 求l 到r 的区间和 int l=sc.nextInt(); int r=sc.nextInt();
public class Main { public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
int n=sc.nextInt(); int m=sc.nextInt();
int[]arr=new int[n+100]; int[]b=new int[n+100];
for (int i = 1; i <= n; i++) { arr[i]=sc.nextInt(); //构建差分数组 b[i]=arr[i]-arr[i-1]; }
while(m-->0) {
int l=sc.nextInt(); int r=sc.nextInt(); int c=sc.nextInt(); //原数组是差分数组的前缀和数组，分别在差分数组的左右边界处分别加减c，会导致差分数组的前缀和数组除了边界值外的数保持不变 b[l]=b[l]+c; b[r+1]=b[r+1]-c;
}
for (int i = 1; i <=n; i++) { arr[i]=b[i]+arr[i-1]; System.out.print(arr[i]+" "); } } }
public class Main { static final int N= 10010; static int[][]arr=new int[N][N]; static int[][]b=new int[N][N]; public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
int n=sc.nextInt(); int m=sc.nextInt(); int q=sc.nextInt();
for (int i = 1; i <= n; i++) { for (int j = 1; j <=m ; j++) { arr[i][j]=sc.nextInt(); } }
for (int i = 1; i <= n; i++) { for (int j = 1; j <=m ; j++) {
constint N = 1e6 + 10; int n; int q[N]; voidquick_sort(int q[], int l, int r) { if (l >= r) return ; // 如果数组一个数都没有或者只有一个数，就不用排序、
//设置两个指针，以及分界点x int x = q[l], i = l - 1, j = r + 1; //这里两个指针我们设置在数组外，这样我们可以不管，直接让二者都往里面走
//开始循环 while (i < j) { //只有两个指针没有碰到才可以继续走 do ++i; while (q[i] < x); //当 q[i] < x就继续往下走 do --j; while (q[j] > x); //当q[j] > x就继续往下走，直到j指针指到一个<=x的数，说明应该放在左边
//判断i, j两个指针未碰到就可以交换 if (i < j) swap(q[i], q[j]); } //一次循环结束后，分界点x归位，接着继续分别递归左右两边 quick_sort(q, l, j); quick_sort(q, j + 1, r); } intmain() { scanf ("%d",&n); // 输入有几个数 for (int i = 0; i < n; ++i) { scanf ("%d",&q[i]); // 输入数组 }
// 快速排序 quick_sort(q, 0, n - 1);//范围0 - n-1 for (int i = 0; i < n; ++i) printf ("%d ",q[i]); printf ("\n"); return0; }
