首页 \ 问答 \ 从UITextfield中给出的输入中搜索iPhone中的SQLite3数据库(Searching SQLite3 DB in iPhone from input given in UITextfield)

从UITextfield中给出的输入中搜索iPhone中的SQLite3数据库(Searching SQLite3 DB in iPhone from input given in UITextfield)

我是iOS开发的新手。 我正在尝试构建在sqllite3 DB中搜索给定搜索词的应用程序。 我遇到了将参数绑定到该sql语句的问题。以下是我从数据库中读取数据的代码。

-(void) readPlayersFromDatabase 
{
NSString * strTemp=[[NSString alloc]init];
strTemp=@"ben";
sqlite3 *database;
players = [[NSMutableArray alloc] init];
if(sqlite3_open([dbPath UTF8String], &database) == SQLITE_OK) 
{
    char *sqlStatement = "SELECT * FROM Player WHERE LENGTH (lastname)>0 AND LENGTH (firstname)>0 AND ( lastname LIKE '%?%' OR firstname LIKE'%?%' OR Height LIKE '%?%' OR Weight LIKE '%?%') ORDER BY lastname,firstname";

sqlite3_stmt *compiledStatement;
if(sqlite3_prepare_v2(database, sqlStatement, -1, &compiledStatement, NULL) == SQLITE_OK) 
{
    sqlite3_bind_text(compiledStatement, 1, [strTemp UTF8String],-1, SQLITE_TRANSIENT );
    sqlite3_bind_text(compiledStatement, 2, [strTemp UTF8String],-1, SQLITE_TRANSIENT );
    sqlite3_bind_text(compiledStatement, 3, [strTemp UTF8String],-1, SQLITE_TRANSIENT );
    sqlite3_bind_text(compiledStatement, 4, [strTemp UTF8String],-1, SQLITE_TRANSIENT );
    while(sqlite3_step(compiledStatement) == SQLITE_ROW) 
            {
            NSString *playerId = [NSString stringWithUTF8String:(char *)sqlite3_column_text(compiledStatement, 0)];
        NSString *playerName = [NSString stringWithUTF8String:(char *)sqlite3_column_text(compiledStatement, 1)];
        Player *temp = [[Player alloc] initWithID:playerId name:playerName];
            [players addObject:temp];
                [temp release];
    }
}
sqlite3_finalize(compiledStatement);
}
sqlite3_close(database);

}

我到底得到的只是玩家的名字? 以他们的名义。 有人可以帮我从这里出去吗。 你能告诉我如何在上面的代码UITextfield输入连接到strTemp吗?

谢谢。


I am fairly new to iOS development. I am trying to build application which searches given search term in sqllite3 DB. I am having trouble with binding parameter to that sql statement.Following is my code to read data from database.

-(void) readPlayersFromDatabase 
{
NSString * strTemp=[[NSString alloc]init];
strTemp=@"ben";
sqlite3 *database;
players = [[NSMutableArray alloc] init];
if(sqlite3_open([dbPath UTF8String], &database) == SQLITE_OK) 
{
    char *sqlStatement = "SELECT * FROM Player WHERE LENGTH (lastname)>0 AND LENGTH (firstname)>0 AND ( lastname LIKE '%?%' OR firstname LIKE'%?%' OR Height LIKE '%?%' OR Weight LIKE '%?%') ORDER BY lastname,firstname";

sqlite3_stmt *compiledStatement;
if(sqlite3_prepare_v2(database, sqlStatement, -1, &compiledStatement, NULL) == SQLITE_OK) 
{
    sqlite3_bind_text(compiledStatement, 1, [strTemp UTF8String],-1, SQLITE_TRANSIENT );
    sqlite3_bind_text(compiledStatement, 2, [strTemp UTF8String],-1, SQLITE_TRANSIENT );
    sqlite3_bind_text(compiledStatement, 3, [strTemp UTF8String],-1, SQLITE_TRANSIENT );
    sqlite3_bind_text(compiledStatement, 4, [strTemp UTF8String],-1, SQLITE_TRANSIENT );
    while(sqlite3_step(compiledStatement) == SQLITE_ROW) 
            {
            NSString *playerId = [NSString stringWithUTF8String:(char *)sqlite3_column_text(compiledStatement, 0)];
        NSString *playerName = [NSString stringWithUTF8String:(char *)sqlite3_column_text(compiledStatement, 1)];
        Player *temp = [[Player alloc] initWithID:playerId name:playerName];
            [players addObject:temp];
                [temp release];
    }
}
sqlite3_finalize(compiledStatement);
}
sqlite3_close(database);

}

All I am getting at the end is name of player which has ? in their name. Can anyone help me out here. Can you also tell me how can I connect UITextfield input to the strTemp in above code ?

Thanks.


原文:https://stackoverflow.com/questions/6386546
更新时间:2023-08-12 18:08

最满意答案

最后,对数的实际用途。

记住这一点

log(X^Y) = log(X) * Y

所以

log(X) = log(X^Y) / Y

所以

X = exp(log(X^Y) / Y)

在这种情况下,X是每个步骤的比例。 Y是步数。 X ^ Y是您想要的总比例。 例如,

Steps = 10;
TotalScale = 0.5;
ScaleEachTime = exp(log(TotalScale) / Steps);

给出(点......因为我在例子中四舍五入)

              = exp(log(0.5) / 10)
              = exp(-0.693... / 10)
              = exp(-0.0693...)
              = 0.9330....

因此,按0.9330缩放10次......总得分为0.5


At last, a practical use for logarithms.

Remember that

log(X^Y) = log(X) * Y

So

log(X) = log(X^Y) / Y

So

X = exp(log(X^Y) / Y)

In this case, X is the scale to make at each step. Y is the number of steps. X^Y is the total scale you want to make. For example,

Steps = 10;
TotalScale = 0.5;
ScaleEachTime = exp(log(TotalScale) / Steps);

gives (dots... because I am rounding off in the example)

              = exp(log(0.5) / 10)
              = exp(-0.693... / 10)
              = exp(-0.0693...)
              = 0.9330....

So scaling 10 times by 0.9330... gives a total scale of 0.5

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