Question

您好，以下是代码，我试图将数据插入数据库但我无法插入表中。

-(BOOL) saveData:(NSString *)registerNumber name:(NSString *)name department:(NSString *)department year:(NSString *)year
{
    const char *dbpath=[databasePath UTF8String];
    if(sqlite3_open(dbpath, &database) == SQLITE_OK)
    {

        NSString *insertSQL=[NSString stringWithFormat:@"insert into studentsDetail (regno,name,department,year)values(\"%d\",\"%@\",\"%@\",\"%@\")",[registerNumber integerValue],name,department,year];
        const char *insert_stmt=[insertSQL UTF8String];


        sqlite3_prepare_v2(database, insert_stmt,-1, &statement,  NULL);
        if(sqlite3_step(statement) == SQLITE_DONE)
        {
            return YES;

        }
        else
        {
            return NO;
        }
        sqlite3_reset(statement);
    }
    return NO;
}

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Answer 1

而不是将这些值添加到NSString。尝试使用sqlite3_bind。它可能有用。

const char *sql = "INSERT INTO studentsDetail (regno,name,department,year) VALUES(?,?,?,?)";
if(sqlite3_prepare_v2(cruddb, sql, 1, &statement, NULL)) {
   sqlite3_bind_int(stmt, 1, [registerNumber integerValue]);
   sqlite3_bind_text(stmt, 2, [name UTF8String], -1, SQLITE_TRANSIENT);
   sqlite3_bind_text(stmt, 3, [department UTF8String], -1, SQLITE_TRANSIENT);
   sqlite3_bind_text(stmt, 4, [year UTF8String], -1, SQLITE_TRANSIENT);
   sqlite3_step(statement);
   sqlite3_finalize(statement);
}

Answer 2

在 -

中尝试单引号

 NSString *insertSQL=[NSString stringWithFormat:@"insert into studentsDetail (regno,name,department,year)values(\'%d\',\'%@\',\'%@\',\'%@\')",[registerNumber integerValue],name,department,year];

这在类似情况下对我有用。您可以使用sqlite3_Bind_Blob数据类型，因为它将按原样存储数据。

将值插入sql数据库的表中

2 个答案: