How to use the SQLITE3 LIKE statement
I have problems running a dynamic LIKE statement in my project: this query works like a charm and returns all items with a 't' in their name:
const char *sql = "select * from bbc_ipad_v1_node where name LIKE '%%t%%'";
When I try to do this dynamically I do not get errors, but just an empty result. It seems that the value is null. I try to bind a string value 's' which outputs a correct value
NSLog(@"bbc_ : search menu items from db based on: %@",s);
const char *sql = "select * from bbc_ipad_v1_node where name LIKE '%%?%%'";
sqlite3_stmt *statement;
if (sqlite3_prepare_v2(database, sql, -1, &statement, NULL) == SQLITE_OK) {
sqlite3_bind_text(statement, 1, [s UTF8String],-1,SQLITE_TRANSIENT);
How should I bind this value instead of using:
const char开发者_JS百科 *sql = "select * from bbc_ipad_v1_node where name LIKE '%%?%%'";
Just found a great explanation at http://www.innerexception.com/2008/10/using-like-statement-in-sqlite-3-from.html
I changed following lines:
const char *sql = "select * from bbc_ipad_v1_node where name LIKE ?001";
and
NSString *searchInput = [NSString stringWithFormat:@"%@%%", s];
sqlite3_bind_text(statement, 1, [searchInput UTF8String],-1,SQLITE_TRANSIENT);
You can actually just say
const char *sql = "select * from bbc_ipad_v1_node where name LIKE '%t%'";
(Note the single-%
s)
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