How to bind a LIKE wildcard in SQLite3 - sql

Question

I'd like to run the SQL query: SELECT COUNT(id) AS nbr FROM user WHERE name LIKE '%John%', and get the number of records matched by the keyword John.
So I've written the following code :
unsigned int nbr;
char *sql, *like;
sqlite3 *db;
sqlite3_stmt *stmt;
/* Database connection etc. */
like = _make_string("%%%s%%", keyword); /* keyword: John */
printf("Like clause: %s\n", like);
sql = _make_string("SELECT COUNT(id) AS nbr FROM user WHERE name LIKE ?");
sqlite3_prepare_v2(db, sql, -1, &stmt, NULL);
sqlite3_bind_text(stmt, 1, like, -1, SQLITE_STATIC);
if (sqlite3_step(stmt) != SQLITE_ROW) {
/* error, SQL execution failed */
}
nbr = sqlite3_column_int64(stmt, 1);
printf("Number of record: %u\n", nbr);
/* Free strings, finalize statement, disconnect db etc. */
For reason of simplicity, some return value tests are ignored in the snippet. _make_string(const char *fmt, ...) is a printf-like function which can form strings, and it did return %John% to me.
However, my code always returns 0 record. I have tried the SQL in sqlite3 command line shell, but it has found 2 records.
Any idea please ?
I've hard coded the LIKE clause and removed the sqlite3_bind_text function, and it doesn't change the result.
sql = _make_string("SELECT COUNT(id) AS nbr FROM client WHERE name LIKE '%John%'");
sqlite3_prepare_v2(db, sql, -1, &stmt, NULL);
if (sqlite3_step(stmt) != SQLITE_ROW) {
/* error */
}
nbr = sqlite3_column_int64(stmt, 1);
printf("Number of record: %u\n", nbr);
The program always returns 0 records.

Answer 1

Well, I've found the answer in the mailing list archive of SQLite Why bind indexes start from 1 and column indexes start from 0?
In fact, the bind index starts from 1, but the column index starts from 0. So the code should be modified as the following
nbr = sqlite3_column_int64(stmt, 0);
printf("Number of record: %u.\n", nbr);
Now the program returns 2 (records) !