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How to Extract a Substring From a String in PostgreSQL/MySQL

Problem:

How to Extract a Substring From a String in PostgreSQL/MySQL

Example 1:

In the emails table, there is an email column. You'd like to display the first seven characters of each email.

The table looks like this:

email
jake99@gmail.com
tamarablack@zoho.com
notine@yahoo.fr
jessica1995@onet.pl

Solution 1:

SELECT
  email,
  SUBSTRING(email, 1, 7) AS substring
FROM emails;

Another syntax:

SELECT
  email,
  SUBSTRING(email FROM 1 FOR 7) AS substring
FROM emails;

The result is:

emailsubstring
jake99@gmail.comjake99@
tamarablack@zoho.comtamarab
notine@yahoo.frnotine@
jessica1995@onet.pljessica

Discussion:

Use the SUBSTRING() function. The first argument is the string or the column name. The second argument is the index of the character at which the substring should begin. The third argument is the length of the substring.

Watch out! Unlike in some other programming languages, the indexes start at 1, not 0. This means the first character has index 1, the second character has index 2, etc.

SUBSTRING(email, 1, 7) will return the substrings of the values in the email column that start at the beginning of the strings (first character) and go for seven characters. The other notation, SUBSTRING(email FROM 1 FOR 7), does exactly the same. The argument after the FROM is the starting index, and the argument after the FOR is the substring length.

The third argument of the SUBSTRING() function is optional. If you omit it, you'll get the substring that starts at the index in the second argument and goes all the way up to the end of the string. SUBSTRING(email, 1) will return the whole string, just as will SUBSTRING(email FROM 1).

Example 2:

You'd like to display the substring between indexes 2 and 6 (inclusive).

Solution 2:

SELECT
  email,
  SUBSTRING(email, 2, 5) AS substring
FROM emails;

Another syntax:

SELECT
  email,
  SUBSTRING(email FROM POSITION('@' IN email)) AS substring
FROM emails;

The result is:

emailsubstring
jake99@gmail.com@gmail.com
tamarablack@zoho.com@zoho.com
notine@yahoo.fr@yahoo.fr
jessica1995@onet.pl@onet.pl

Discussion:

You use the SUBSTRING() function like in the previous examples. This time, you're looking for a specific character whose position can vary from row to row. To find the index of the specific character, you can use the POSITION(character IN column) function, where character is the specific character at which you'd like to start the substring (here, @) . The argument column is the column from which you'd like to retrieve the substring; it can also be a literal string.

If you want the substring to go all the way to the end of the original string, the third argument in the SUBSTRING() function (or the FOR argument) is not needed. Otherwise, it should be the length of the substring, or you can calculate it using the POSITION() function. You may also want to retrieve a substring that doesn't end at the end of the string but at some specific character, e.g., before '.'. Here's an example:

SELECT
  email,
  SUBSTRING(email, POSITION('@' IN email), POSITION('.' IN email) - POSITION('@' IN email)) AS substring
FROM emails;

Another syntax:

SELECT
  email,
  SUBSTRING(email FROM POSITION('@' IN email) FOR POSITION('.' IN email) - POSITION('@' IN email)) AS substring
FROM emails;

The result of this query is:

emailsubstring
jake99@gmail.com@gmail
tamarablack@zoho.com@zoho
notine@yahoo.fr@yahoo
jessica1995@onet.pl@onet

The part POSITION('.' IN email) - POSITION('@' IN email) simply calculates the length of the substring.

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