SQL Oracle：从字符串中删除前 4 个字符
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Oracle: remove first 4 characters from a string
提问by Gamidron
So I want to remove the first 4 characters from a string in oracle. Those characters can be different every time. In my case I need to take away the first 4 characters of an IBAN and put them at the end of the string. I got the part of putting them to the end of the string but I can't get the first 4 characters to be removed. Every solution I find on the internet removes specified characters, not characters from a certain position in the string (1 to 4). I used the code below to get the first 4 characters to the end of the string and wanted to try something similar for removing them at the front but without success.
所以我想从 oracle 中的字符串中删除前 4 个字符。这些角色每次都可能不同。就我而言，我需要去掉 IBAN 的前 4 个字符并将它们放在字符串的末尾。我得到了将它们放在字符串末尾的部分，但我无法删除前 4 个字符。我在互联网上找到的每个解决方案都会删除指定的字符，而不是字符串中某个位置（1 到 4）的字符。我使用下面的代码将前 4 个字符移到字符串的末尾，并想尝试类似的方法将它们从前面删除，但没有成功。
SELECT SUBSTR(iban_nummer, 1, 4) INTO iban_substring FROM dual;
iban_nummer := iban_nummer || iban_substring;
回答by Guntram Blohm supports Monica
See the docs:
查看文档：
substring_length ... When you do not specify a value for this argument, then the function returns all characters to the end of string. When you specify a value that is less than 1, the function returns NA.
substring_length ... 如果您没有为此参数指定值，则该函数将返回到字符串末尾的所有字符。当您指定一个小于 1 的值时，该函数返回 NA。
So iban_nummer := substr(iban_nummer, 5) || substr(iban_nummer, 1,4)should work. The first part selects all characters beginning from the 5th, the second character numbers 1..4.
所以iban_nummer := substr(iban_nummer, 5) || substr(iban_nummer, 1,4)应该工作。第一部分选择从第 5 个开始的所有字符，第二个字符编号为 1..4。
回答by Sai
update table_name set col_name=substr(col_name,5);
回答by ElvisZhu
try regexp, like:
尝试regexp，例如：
SELECT regexp_replace(t.iban_nummer,'(.{4})(.*)','') FROM t;
