我在我的项目中运行动态 LIKE 语句时遇到问题:这个查询就像一个魅力,并返回所有名称中带有“t”的项目:
const char *sql = "select * from bbc_ipad_v1_node where name LIKE '%%t%%'";
当我尝试动态执行此操作时,我没有收到错误,而只是一个空结果。似乎该值为空。我尝试绑定一个输出正确值的字符串值's'
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);
我应该如何绑定这个值而不是使用:
const char *sql = "select * from bbc_ipad_v1_node where name LIKE '%%?%%'";