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Use Intermedia Text in oracle8i or Text in oracle9i to create an index of type on the clob field, and then use contain as the query condition during the query. ** As follows: --Create table tb clob(
userid number primary key,text clob);- Create index search idx
on tb_clob(text)
indextype is ;- Query select count(*).
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Syntax: select * from table where
dbms field name (clod type),'Query criteria',1,1) >0
create table products(
productid number(10) not null,name varchar2(255),description clob);
Query statement: select , from products t
where dbms_, 'aaa', 1, 1) >0;
In Oracle, you can use the instr function to determine whether a string contains a specified character. Its syntax is:
where sourcestring represents the source string; deststring represents the substring you want to find in the source string; start indicates the start position of the lookup, and this parameter is optional, which is default.
1;appearposition means that you want to find out the first occurrence of deststring from the source character, this parameter is also optional, the default is 1; If the value of start is negative, that.
It stands for searching from right to left.
The return value is: the location of the found string.
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To perform a fuzzy query on a clob in Oracle, you need to convert the content of the clob to a character type before you can use the fuzzy query.
Example: 1. Enter the following data in the table:
create table test
id int,str clob);
insert into test values (1,'Dongdong is the bad guy');
insert into test values (2,'Physics is the bad guy');
insert into test values (3,'Little frogs are bad guys');
insert into test values (4,'Badkano is a good person');
commit;
2. Now query the content of "bad guy" in the str field of the test table, use the following statement:
select * from test where to_char(str) like '%Bad guys%';
3. Query results (three records were found, but the content of the clob cannot be directly displayed).
4. If you want to display the content of clob, you can use the to char function to convert it, and you can see that the content you are checking is the row containing "bad guys".
select to_char(str) from test where to_char(str) like '%Bad guys%';
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