Question

我的表格和行如下所示;

Table : Visit
Column : VisitDate
Value : 634993344000000000

VisitDate列包含Int64值（例如DateTime.Now.Ticks）

当我从数据库中读取此内容时，SQLite无法检测列是否为Int64值。我的代码是;

switch (sqlite3_column_type(statement,i))
      {
         case SQLITE_INTEGER:
                intValue  = (int)sqlite3_column_int(statement, i);
                colNameChar = (const char *)sqlite3_column_name(statement, i);
                colName =  [NSString stringWithCString:colNameChar encoding:NSUTF8StringEncoding];
                colValue = [NSNumber numberWithInteger: intValue];
                break;
        case SQLITE_FLOAT:
                dblValue = (double)sqlite3_column_double(statement, i);
                colNameChar = (const char *)sqlite3_column_name(statement, i);
                colName =  [NSString stringWithCString:colNameChar encoding:NSUTF8StringEncoding];
                colValue  = [NSNumber numberWithDouble: dblValue];
                break;
         case SQLITE_TEXT:
                 strValue = (const char *)sqlite3_column_text(statement, i);
                 colNameChar = (const char *)sqlite3_column_name(statement, i);
                 colName =  [NSString stringWithCString:colNameChar encoding:NSUTF8StringEncoding];
                 colValue = [NSString stringWithCString:strValue encoding:NSUTF8StringEncoding];
                        break;
          case SQLITE_BLOB:
                  strValue = (const char *)sqlite3_column_value(statement, i);
                  colNameChar = (const char *)sqlite3_column_name(statement, i);
                  colName =  [NSString stringWithCString:colNameChar encoding:NSUTF8StringEncoding];
                  colValue = [NSString stringWithCString:strValue encoding:NSUTF8StringEncoding];
                  break;
           case SQLITE_NULL:
                  strValue = nil;
                  colNameChar = (const char *)sqlite3_column_name(statement, i);
                  colName =  [NSString stringWithCString:colNameChar encoding:NSUTF8StringEncoding];
                  colValue = [[NSObject alloc] init];
                  break;
            default:
                   strValue = (const char *)sqlite3_column_value(statement, i);
                   colNameChar = (const char *)sqlite3_column_name(statement, i);
                   colName =  [NSString stringWithCString:colNameChar encoding:NSUTF8StringEncoding];
                   colValue = [NSString stringWithCString:strValue encoding:NSUTF8StringEncoding];
                    break;
       }

调试器输入SQLITE_INTEGER的大小写并产生一个垃圾值（类似于-2323423）。我怎样才能得到正确的价值？

Answer 1

在SQLite中，所有整数值都是64位值。通常只使用sqlite3_column_int，因为大多数应用程序已经知道它们的值确实适合32位（或者使用来自其他应用程序的代码复制 - 表示这个假设）。

如果要检查SQLITE_INTEGER值是否适合32位或64位，则必须先调用sqlite3_column_int64并检查结果。

如何从sqlite数据库中读取Int64值？

1 个答案: