Question

CREATE TABLE #tmpTbl (m VARCHAR(100))  
INSERT INTO #tmpTbl VALUES
 (',[Undergraduate1]')    
,(',[Undergraduate10]')   
,(',[Undergraduate11]')   
;
GO

select LEFT(m, PATINDEX('%[0-9]%', m)-1) as a,
         SUBSTRING(m, PATINDEX('%[0-9]%', m), LEN(m)) as b from #tmpTbl


drop table #tmpTbl

如果给出上面的tmptable和select语句，结果如下：

           a            |      b
-------------------------------------------------------
    ,[Undergraduate     |      1]
    ,[Undergraduate     |      10]
    ,[Undergraduate     |      11]

但我希望它是这样的。

           a            |      b
-------------------------------------------------------
    ,[Undergraduate     |      1
    ,[Undergraduate     |      10
    ,[Undergraduate     |      11

我怎样才能实现这一目标？我尝试了很多与PATINDEX，LEFT，RIGHT，SUBSTRING，LEN的组合。但是不能在B栏中找到正确的

Answer 1

您可以使用替换删除]。狡猾，但它实现了你想要的东西

select LEFT(m, PATINDEX('%[0-9]%', m)-1) as a,
     REPLACE(SUBSTRING(m, PATINDEX('%[0-9]%', m), LEN(m)),']','') as b from #tmpTbl

替代方法：反转字符串，子串以删除第一个字符，反向返回

select LEFT(m, PATINDEX('%[0-9]%', m)-1) as a,
         REVERSE(SUBSTRING(REVERSE(SUBSTRING(m, PATINDEX('%[0-9]%', m), LEN(m))),2,LEN(M))) as b from #tmpTbl

Answer 2

您可以使用REPLACE将]替换为''

select LEFT(m, PATINDEX('%[0-9]%', m)-1) as a,
         REPLACE(SUBSTRING(m, PATINDEX('%[0-9]%', m), LEN(m)), ']', '') as b from #tmpTbl

Answer 3

我倾向于将stuff()用于此目的：

select replace(stuff(m, 1, patindex(m, '%[0-9]%'), ''), ']', '')

Answer 4

这是一种替代方法，可以删除任何文本，只留下数字。

 static class Node{
    int data;
    Node next;
    Node(int num){
        data=num;
        next=null;
    }
}

// Function for merging two sorted linked list

public void merge(Linkedlist list1,Linkedlist list2){       
    Linkedlist l=new Linkedlist();

    Node new_node1=new Node(0); 
    Node new_node=new_node1;   //This  line of code is my doubt!
    while(list1.head!=null || list2.head!=null){    
        if(list1.head!=null && list2.head!=null){
        if(list1.head.data<=list2.head.data ){
            new_node.next=list1.head; // what difference it makes by using new_node.next instead of new_node1
            list1.head=list1.head.next;         

        }
        else{
            new_node.next=list2.head;
            list2.head=list2.head.next; 
            }           
        }
        else if(list1.head==null){           
                new_node.next=list2.head;
                list2.head=list2.head.next; 
            }
        else if(list2.head==null){          
                new_node.next=list1.head;
                list1.head=list1.head.next;                 
        }
        new_node=new_node.next;
        //new_node1=new_node1.next;
    }
    l.printlist(new_node1);

}

从字符串中选择正确的字母

4 个答案: