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Strings in JPQL and Criteria Queries

States') is evaluated to 13. LENGTH('China') is evaluated to 5. LOCATE - Locating Substrings The LOCATE(str, substr [, start]) function searches a substring and returns its position. For example: LOCATE ... ). Therefore, the position of the first character is 1 . Zero ( 0 ) is returned if the substring

GROUP BY and HAVING clauses

(no Aggregates) The following query groups all the countries by their first letter: SELECT SUBSTRING (c.name, 1, 1) FROM Country c GROUP BY SUBSTRING (c.name, 1, 1); The FROM clause defines iteration ... (which does not use aggregates) is equivalent to the following query: SELECT DISTINCT SUBSTRING (c.name, 1

CriteriaBuilder.replace(x,substring,replacement) - JPA Method

; Expression  x,    String  substring ,     Expression  replacement ) Create an expression replacing every occurrence of a substring within a string. Parameters: x - string expression substring - the literal substring to replace replacement - the replacement

CriteriaBuilder.replace(x,substring,replacement) - JPA Method

; Expression  x,     Expression   substring ,     Expression  replacement ) Create an expression replacing every occurrence of a substring within a string. Parameters: x - string expression substring - the literal substring to replace replacement - the replacement

CriteriaBuilder.replace(x,substring,replacement) - JPA Method

; Expression  x,    String  substring ,    String replacement ) Create an expression replacing every occurrence of a substring within a string. Parameters: x - string expression substring - the literal substring to replace replacement - the replacement string

CriteriaBuilder.replace(x,substring,replacement) - JPA Method

; Expression  x,     Expression   substring ,    String replacement ) Create an expression replacing every occurrence of a substring within a string. Parameters: x - string expression substring - the literal substring to replace replacement - the replacement

CriteriaBuilder.substring(x,from,len) - JPA Method

JPA Method in jakarta.persistence.criteria.CriteriaBuilder Expression substring (    ... ) Create an expression for substring extraction. Extracts a substring of given length starting ... expression len - length expression Return: expression corresponding to substring extraction Since: JPA 2.0

CriteriaBuilder.substring(x,from) - JPA Method

JPA Method in jakarta.persistence.criteria.CriteriaBuilder Expression substring (    Expression  x,     Expression  from ) Create an expression for substring extraction. Extracts a substring starting at the specified position through to end of the string. First

CriteriaBuilder.substring(x,from) - JPA Method

JPA Method in jakarta.persistence.criteria.CriteriaBuilder Expression substring (    Expression  x,    int from ) Create an expression for substring extraction. Extracts a substring starting at the specified position through to end of the string. First position

CriteriaBuilder.substring(x,from,len) - JPA Method

JPA Method in jakarta.persistence.criteria.CriteriaBuilder Expression substring (    ... an expression for substring extraction. Extracts a substring of given length starting at the specified ... : expression corresponding to substring extraction Since: JPA 2.0

substring works with string but not with char

i want to select all entities whose name starts with a specific letter, i can use this query with a string parameter select f from Foo f where substring (f.name, 1, 1) = ?1 but when i use a char, the query does not return anything. i think it should also be possible to use char/Character types

FROM clause (JPQL / Criteria API)

, SUBSTRING , SUM, THEN,TRAILING, TRIM, TRUE, TYPE, UNKNOWN, UPDATE, UPPER, VALUE, WHEN, WHERE. JPQL

JPA Query Expressions (JPQL / Criteria)

, substring , ... ). Data expressions ( currentDate , currentTime , currentTimestamp ). Collection

JPA Criteria API Queries

, neg , sqrt ). String expressions ( like , length , locate , lower , upper , concat , substring

InterfaceRef jakarta.persistence.criteria.CriteriaBuilder

for the leftmost substring of a string, Parameters: x - string expression len - length of the substring to return Return: expression for the leftmost substring Since: JPA 2.0 Expression left ( Expression  x, int len) Create an expression for the leftmost substring of a string, Parameters: x

CriteriaBuilder.right(x,len) - JPA Method

JPA Method in jakarta.persistence.criteria.CriteriaBuilder Expression right (    Expression  x,    int len ) Create an expression for the rightmost substring of a string, Parameters: x - string expression len - length of the substring to return Return: expression for the rightmost substring Since: JPA 2.0

CriteriaBuilder.left(x,len) - JPA Method

JPA Method in jakarta.persistence.criteria.CriteriaBuilder Expression left (    Expression  x,    int len ) Create an expression for the leftmost substring of a string, Parameters: x - string expression len - length of the substring to return Return: expression for the leftmost substring Since: JPA 2.0

CriteriaBuilder.right(x,len) - JPA Method

JPA Method in jakarta.persistence.criteria.CriteriaBuilder Expression right (    Expression  x,     Expression  len ) Create an expression for the rightmost substring of a string, Parameters: x - string expression len - length of the substring to return Return: expression for the rightmost substring Since: JPA 2.0

CriteriaBuilder.left(x,len) - JPA Method

JPA Method in jakarta.persistence.criteria.CriteriaBuilder Expression left (    Expression  x,     Expression  len ) Create an expression for the leftmost substring of a string, Parameters: x - string expression len - length of the substring to return Return: expression for the leftmost substring Since: JPA 2.0

order by string with numbers

  The sql that works is ORDER BY CAST( SUBSTR (username FROM 4) AS UNSIGNED) ASC Using JPQL SELECT u FROM Users u ORDER BY SUBSTRING (u.username,4)*1 ASC If you have other alternative for JPA

Partial Indexing

I have a question about partial indexing. For example, a field of String type ttt, will be distinguished by the first 5 characters, then I can make a index of substr (ttt,1,5) in MySQL. With ObjectDB, how can I write the annotation for this purpose? TIA gzdillon Lai Yang Currently ObjectDB supports

combined index not used

a unique substring from the query plan XML, so specifying the index name should work. support Support ... ? If it is, then any unique substring from that query plan that doesn't depend on parameters should work

com.objectdb.o.InternalException: null

ParseException {         String dateOfBirthStr = personalNumber. substring (0