Question

如果我有一个包含完整名称的列的表，例如：

fullname
------------
Joe Bloggs
Peter Smith
Mary Jones and Liz Stone

如何使用SQL重试全名列中每个条目的名字和姓氏。我并不担心我的例子中的第3个条目中的第二个名字，即Liz Stone。

所以基本上要检索

Firstname
---------
Joe
Peter
Mary

Lastname  
--------
Bloggs 
Smith
Jones

Answer 1

这是一个SQL Server 2016之前的方法，它使用基本的字符串函数来隔离名字和姓氏。

SELECT SUBSTRING(fullname, 1, CHARINDEX(' ', fullname) - 1) AS Firstname,     
       SUBSTRING(fullname,
                 CHARINDEX(' ', fullname) + 1,
                 LEN(fullname) - CHARINDEX(' ', fullname)) AS Lastname
FROM yourTable

请注意，此解决方案假定fullname列仅包含单个名字和单个姓氏（即没有中间名，缩写等）。

Answer 2

这是一个滑坡，没有简单的答案。也就是说，请考虑以下内容

private Uri uri;
@Override
protected void onActivityResult(int requestCode, int resultCode, Intent data) {
    switch (requestCode) {
        case FILE_SELECT_CODE:
            if (resultCode == RESULT_OK) {
                // Get the Uri of the selected file
                uri = data.getData();
            }
            break;
    }
    super.onActivityResult(requestCode, resultCode, data);
}

public void sendAttachment() {

        File filePhoto = new File(uri.getPath());
        Boolean fileIsPublic = false;

        QBContent.uploadFileTask(filePhoto, fileIsPublic, null, new QBProgressCallback() {
            @Override
            public void onProgressUpdate(int i) {
                // i - progress in percentages
                Log.e(TAG, "onProgressUpdate: " + i);
            }
        }).performAsync(new QBEntityCallback<QBFile>() {
            @Override
            public void onSuccess(QBFile qbFile, Bundle bundle) {

                Integer id = qbFile.getId();

                // create a message
                QBChatMessage chatMessage = new QBChatMessage();
                chatMessage.setProperty("save_to_history", "1"); // Save a message to history

                // attach a photo
                QBAttachment attachment = new QBAttachment("photo");
                attachment.setId(qbFile.getId().toString());
                chatMessage.setBody("");
                chatMessage.addAttachment(attachment);

                // send a message
                try {
                    mQBChatDialog.sendMessage(chatMessage);
                    mtvEmpty.setVisibility(View.INVISIBLE);
                } catch (SmackException.NotConnectedException e) {
                    e.printStackTrace();
                }
            }

            @Override
            public void onError(QBResponseException errors) {
                Log.e("DATA", TextUtils.join(",", errors.getErrors()));
            }
        });
}

返回

Declare @YourTable table (FullName varchar(50))
Insert Into @YourTable values
('Joe Bloggs'),
('Peter Smith'),
('Betty Jane Martinez'),
('Mary Jones and Liz Stone')


Select A.*
      ,FirstName = Pos1+case when Pos3 is not null then ' '+Pos2 else '' end
      ,LastName  = case when Pos3 is null then Pos2 else Pos3 end
 From @YourTable A
 Cross Apply (
    Select Pos1 = xDim.value('/x[1]','varchar(max)')
          ,Pos2 = xDim.value('/x[2]','varchar(max)')
          ,Pos3 = xDim.value('/x[3]','varchar(max)')
          ,Pos4 = xDim.value('/x[4]','varchar(max)')
          ,Pos5 = xDim.value('/x[5]','varchar(max)')
          ,Pos6 = xDim.value('/x[6]','varchar(max)')
     From  (Select Cast('<x>' + replace((Select substring(FullName,1,charindex(' and ',FullName+' and ')-1) as [*] For XML Path('')),' ','</x><x>')+'</x>' as xml) as xDim) as A 
 ) B

如果它对视觉有帮助，CROSS APPLY会生成

Answer 3

SELECT CASE
           WHEN CHARINDEX(' ', FullName) > 0
           THEN SUBSTRING(FullName, 1, LEN(FullName) - CHARINDEX(' ', REVERSE(FullName)))
           ELSE ''
       END AS FirstName,
       CASE
           WHEN CHARINDEX(' ', FullName) > 0
           THEN REVERSE(SUBSTRING(REVERSE(FullName), 
                       1, 
                       CHARINDEX(' ', REVERSE(FullName)) - 1))
           ELSE FullName
       END AS LastName
FROM(VALUES('Mary Anne Bloggs'), ('Joe Bloggs'), ('Bloggs')) AS T(FullName);

此版本检查全名中是否有空格可分割。如果没有，则将名字设置为空字符串，并将全名放入姓氏中。另外，当有多个空格时，使用反向可在最后一个空格上分割

Answer 4

BigQuery：标准Sql

substr（name，1，STRPOS（name，''）-1）作为名字，

substr（name，STRPOS（name，''）+ 1，length（name））作为姓氏

Answer 5

这是这个问题中最容易和最短的，没有任何假设。您还可以使用rtrim(ltrim('firstname lastname'))进一步增强此功能。

如果字符串前面有空格，

Select 
    substring('Firstname Lastname',1,CHARINDEX(' ', 'Firstname Lastname')) as firstname, 
    substring('Firstname Lastname',CHARINDEX(' ', 'Firstname Lastname'),LEN('Firstname Lastname')) as Lastname

Answer 6

我使用此查询来检索first和lastname

SELECT SUBSTRING(FULLNAME, 1, CASE WHEN CHARINDEX(' ', FULLNAME)>0 THEN CHARINDEX(' ', FULLNAME) - 1 ELSE LEN(FULLNAME) END ) AS Firstname,
  REVERSE(SUBSTRING(REVERSE(FULLNAME), 1, CASE WHEN CHARINDEX(' ', REVERSE(FULLNAME))>0 THEN CHARINDEX(' ', REVERSE(FULLNAME)) - 1 ELSE LEN(REVERSE(FULLNAME)) END ) )AS Firstname FROM  HRMDESFO.EMPLOID

Resuls

Answer 7

select  passemail,substring(passemail,1,instr(passemail,'@') - 1) as name ,
substring(passemail,instr(passemail,'@') + 1,length(passemail)) from passenger

Answer 8

用于获取名字

SELECT SUBSTR(FULLNAME,1,(LOCATE(' ',FULLNAME)))  AS FIRSTTNAME from EmployeeDetails;

姓氏

SELECT SUBSTR(FULLNAME,(LOCATE(' ',FULLNAME)))  AS LASTNAME from EmployeeDetails;

SO

SELECT  SUBSTR(FULLNAME,1,(LOCATE(' ',FULLNAME)))  AS FIRSTTNAME, SUBSTR(FULLNAME,(LOCATE(' ',FULLNAME)))  AS LASTNAME from EmployeeDetails;

Answer 9

SELECT 
  LEFT(column_name,  POSITION(' ' IN column_name)-1)                     AS first_name,
  RIGHT(column_name, LENGTH(column_name) - POSITION(' ' IN column_name)) AS last_name
FROM table_name

Answer 10

SELECT SUBSTR(NAME,1,(LOCATE(NAME, ' '))) AS FIRSTTNAME, SUBSTR(NAME,(LOCATE(NAME, ' ')+1)) AS LASTNAME 从 yourTABLE;

Answer 11

你可以使用，

 STRING_SPLIT (string , separator)

sql - 如何将fullname拆分为名字和姓氏

11 个答案: