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我有应用程序，我在 sqlite 表中插入数据我希望如果数据已经插入，那么它不应该插入相同的记录这里是我插入数据的代码。

这是用于插入数据的带有插入语句的代码

       - (void) addCoffee:(NSString *)dbPath {

        if(addStmt == nil) {
        const char *sql = "insert into library(userID,contentAddedDateTime,contentType,contentTitle,contentSource,contentDescription,categoryTitle,subCategoryTitle,organizationCode,userName) Values(?,?,?,?,?,?,?,?,?,?)";

       if (sqlite3_open([dbPath UTF8String], &database) == SQLITE_OK) 


       if(sqlite3_prepare_v2(database, sql, -1, &addStmt, NULL) != SQLITE_OK)
        NSAssert1(0, @"Error while creating add statement. '%s'", sqlite3_errmsg(database));


    }



  sqlite3_bind_text(addStmt, 1, [userID UTF8String], -1, SQLITE_TRANSIENT);

  sqlite3_bind_text(addStmt, 2, [contentAddedDateTime UTF8String], -1, SQLITE_TRANSIENT);

  sqlite3_bind_text(addStmt, 3, [contentType UTF8String], -1, SQLITE_TRANSIENT);

  sqlite3_bind_text(addStmt, 4, [contentTitle UTF8String], -1, SQLITE_TRANSIENT);

  sqlite3_bind_text(addStmt, 5, [contentSource UTF8String], -1, SQLITE_TRANSIENT);

      sqlite3_bind_text(addStmt, 6, [contentDescription UTF8String], -1, SQLITE_TRANSIENT);


      sqlite3_bind_text(addStmt, 7, [categoryTitle UTF8String], -1, SQLITE_TRANSIENT);

      sqlite3_bind_text(addStmt, 8, [subCategoryTitle UTF8String], -1, SQLITE_TRANSIENT);

      sqlite3_bind_text(addStmt, 9, [organizationCode UTF8String], -1, SQLITE_TRANSIENT);
      sqlite3_bind_text(addStmt, 10, [userName UTF8String], -1, SQLITE_TRANSIENT);



if(SQLITE_DONE != sqlite3_step(addStmt))
    NSAssert1(0, @"Error while inserting data. '%s'", sqlite3_errmsg(database));
else
    //SQLite provides a method to get the last primary key inserted by using sqlite3_last_insert_rowid
    coffeeID = sqlite3_last_insert_rowid(database);

//Reset the add statement.
sqlite3_reset(addStmt);


   }
4

2 回答 2

0

您应该使用唯一列编写选择查询。让我们说你的情况下的用户ID

const char *sql = "select * from library where userID=%@";

如果记录存在，则输入 else 跳过

于 2013-07-30T07:36:33.743 回答
0 
- (int) GetCount :(NSString *)UserID
{
[self createEditableCopyOfDatabaseIfNeeded];
[self initializeDatabase];

int count = 0;


NSArray *paths = NSSearchPathForDirectoriesInDomains(NSDocumentDirectory, NSUserDomainMask, YES);
NSString *documentsDirectory = [paths objectAtIndex:0];

NSString *path = [documentsDirectory stringByAppendingPathComponent:@"dbfav2table.sqlite"];
if (sqlite3_open([path UTF8String], &database) == SQLITE_OK)
{
    sqlite3_stmt *statement;

    NSString *sql_str = [NSString stringWithFormat:@"SELECT count(*) FROM tablename WHERE userID='%@'", UserID];

    const char *sqlStatement = (char *)[sql_str UTF8String];
     NSLog(@"%@", [NSString stringWithUTF8String:sqlStatement]);

    if( sqlite3_prepare_v2(database, sqlStatement, -1, &statement, NULL) == SQLITE_OK )
    {

        while( sqlite3_step(statement) == SQLITE_ROW )
        {
            count = sqlite3_column_int(statement, 0);
            NSLog(@"count for %@ is %i",UserID,count);
        }

    }
    else

 {
        NSLog( @" Error is:  %s", sqlite3_errmsg(database) );

 }

    // Finalize and close database.
    sqlite3_finalize(statement);
    sqlite3_close(database);
}

return count;
}

如果返回 count = 1，则表示已经存在具有相同 userId 的值，否则不存在该值。希望能帮助到你

于 2013-07-30T07:36:43.853 回答