为什么在IOS的sqlite中插入数据时数据库被锁定?
为什么在sqlite中插入数据时数据库被锁定?我已经打开和关闭了我所有的数据库代码?我有以下代码:为什么在IOS的sqlite中插入数据时数据库被锁定?,ios,database,sqlite,Ios,Database,Sqlite,为什么在sqlite中插入数据时数据库被锁定?我已经打开和关闭了我所有的数据库代码?我有以下代码: -(void)insertDataIn_tbl_selectItem_data: (NSString *)empID ProdId: (NSString *)prodId ProdName: (NSString *)prodName GenName: (NSString *)genName ComputeType: (NSString *)computeType UOM: (NSString *
-(void)insertDataIn_tbl_selectItem_data: (NSString *)empID ProdId: (NSString *)prodId ProdName: (NSString *)prodName GenName: (NSString *)genName ComputeType: (NSString *)computeType UOM: (NSString *)uom ListPrice: (NSString *)listPrice UOMQty: (NSString *)uomqty{
const char *query = "insert into tbl_selectItem_data (femployeeid,fproductid,fname,fgeneric_name,fcompute_type,fuom,flist_price,fuomqty) values (?,?, ?, ?, ?, ?, ?, ?)";
sqlite3_stmt *stmt;
if (sqlite3_open([sqLiteDb UTF8String], &(_database)) == SQLITE_OK) {
if (sqlite3_prepare_v2(_database, query, -1, &stmt, nil) == SQLITE_OK) {
sqlite3_bind_text(stmt, 1, [empID UTF8String], -1, SQLITE_TRANSIENT);
sqlite3_bind_text(stmt, 2, [prodId UTF8String], -1, SQLITE_TRANSIENT);
sqlite3_bind_text(stmt, 3, [prodName UTF8String], -1, SQLITE_TRANSIENT);
sqlite3_bind_text(stmt, 4, [genName UTF8String], -1, SQLITE_TRANSIENT);
sqlite3_bind_text(stmt, 5, [computeType UTF8String], -1, SQLITE_TRANSIENT);
sqlite3_bind_text(stmt, 6, [uom UTF8String], -1, SQLITE_TRANSIENT);
sqlite3_bind_text(stmt, 7, [listPrice UTF8String], -1, SQLITE_TRANSIENT);
sqlite3_bind_text(stmt, 8, [uomqty UTF8String], -1, SQLITE_TRANSIENT);
if(sqlite3_step(stmt) == SQLITE_DONE){
NSLog(@"Insert Successful");
sqlite3_finalize(stmt);
}else{
NSLog(@"insertDataIn_tbl_selectItem_data error: %s", sqlite3_errmsg(_database));
}
}
}
sqlite3_close(_database);
}
首先。。。即使有错误,也尝试“完成”您的语句…因此您的方法:
if(sqlite3_step(stmt) == SQLITE_DONE){
NSLog(@"Insert Successful");
sqlite3_finalize(stmt);
}else{
NSLog(@"insertDataIn_tbl_selectItem_data error: %s", sqlite3_errmsg(_database));
}
}
}
sqlite3_close(_database);
最好是:
if(sqlite3_step(stmt) == SQLITE_DONE){
NSLog(@"Insert Successful");
}else{
NSLog(@"insertDataIn_tbl_selectItem_data error: %s", sqlite3_errmsg(_database));
}
}
sqlite3_finalize(stmt);
}
sqlite3_close(_database);
第二
您可以尝试使用sqlite3\u open\u v2
来代替sqlite3\u open
更多信息:
因此……取而代之的是:
if (sqlite3_open([sqLiteDb UTF8String], &(_database)) == SQLITE_OK) {...}
将是:
if(sqlite3\u open\u v2([sqLiteDb UTF8String],&(\u database),SQLITE\u open\u READWRITE,NULL)==SQLITE\u OK){……}
您是否在循环中调用此方法?不,我只调用了一次该方法。当我单击“添加”按钮时,您可以打印sqlite3\u errmsg
,并在其中调用此方法。'我已经打印了该slite3\u errmsg,它返回我一个数据库值已锁定检查是否打开数据库进行其他操作?是否读取了其他操作?