Mst137
Mst137

Reputation: 135

Unable to read from SQLite database

Im trying to read from a database row-by-row. I am getting a warning that states "assignment makes integer from pointer without a cast [enabled by default" for all the rows containing the sqlite3_column_text() function. However no such error occurs on the row in which I try to get an integer, although this line still only returns 0 from the database.

rc handling is removed for compactness.

    typedef struct {
   char InstrumentType[32];
   char Protocol[16];
   int RegNum;
   char RW[4];
   char RegisterType[32];
   char Signed[8];
   char Inverted[8];
   char DataType[16];
} Register;

typedef struct {
       char InstrumentType[32];
       int Bus;
       int Address;
       int Port;
       int Sampling;
       char Sync[2];
       Register Reg[];
       } Instrument;

int main(void) {
    sqlite3 *db;
    Register Register[8];
    sqlite3_stmt* SQL;
    char SQL2[128];
    int i;
    int MPTotalReg;

    int rc;

    rc = sqlite3_open("/root/BBBTest/Instruments.db", &db);

     for (i=0; i<4; i++){

     rc = sqlite3_prepare_v2(db, "SELECT * from instrumentsdefinition WHERE Type='Magprobe' AND Register=?;", -1, &SQL, 0);

     rc = sqlite3_bind_int(SQL, 1, i);

       *Register[i].InstrumentType=(const char*) sqlite3_column_text(SQL,2);
       *Register[i].Protocol=(const char*) sqlite3_column_text(SQL,3);
       Register[i].RegNum= sqlite3_column_int(SQL,5);
       *Register[i].RW=(const char*) sqlite3_column_text(SQL,5);
       *Register[i].RegisterType=(const char*) sqlite3_column_text(SQL,6);
       *Register[i].Signed=(const char*) sqlite3_column_text(SQL,7);
       *Register[i].Inverted=(const char*) sqlite3_column_text(SQL,8);
       *Register[i].DataType=(const char*) sqlite3_column_text(SQL,9);
       printf("Magprobe %d\n", i);
       printf("Instrument Type: ");
       printf("%s\n", Register[i].InstrumentType);
       printf("Protocol: ");
       printf("%s\n", Register[i].Protocol);
       printf("Register: ");
       printf("%d\n", Register[i].RegNum);
       printf("R/W?: ");
       printf("%s\n", Register[i].RW);
       printf("Register Type: ");
       printf("%s\n", Register[i].RegisterType);
       printf("Signed?: ");
       printf("%s\n", Register[i].Signed);
       printf("Inverted?: ");
       printf("%s\n", Register[i].Inverted);
       printf("Data Type: ");
       printf("%s\n", Register[i].DataType);
     }
       sqlite3_finalize(SQL);
       sqlite3_close(db);

    return EXIT_SUCCESS;
}

I know the error is something related to char arrays and chars however I've tried numerous methods to fix this. Also the fact that the int returned is always 0 leads me to believe there is an error with my prepare or bind statement also.

Upvotes: 0

Views: 267

Answers (1)

LPs
LPs

Reputation: 16223

Using you cannot assign an array to another, you have to copy source array content to destination array, eg using strcpy

unsigned char *value = sqlite3_column_text(SQL,2);
if ((value != NULL) && (strlen(value) < sizeof(Register[i].InstrumentType)))
{       
    strcpy(Register[i].InstrumentType, value);
}

In your specific case you can also use pointers to do that refactoring your struct.

Register[i].InstrumentType=(const char*) sqlite3_column_text(SQL,2);
if (Register[i].InstrumentType != NULL)
{
   //...
}

I wrote your specific case due to the fact that as soon as you call sqlite3_finalize, the return values from sqlite3_column_text become invalid.


Another thing is the query execution, after bind you must call

rc = sqlite3_step(stmt);

to retrieve result of query execution. You should loop until returned value is == SQLITE_ROW

 for (i=0; i<4; i++)
 {
    rc = sqlite3_prepare_v2(db, "SELECT * from instrumentsdefinition WHERE Type='Magprobe' AND Register=?;", -1, &SQL, 0);

    rc = sqlite3_bind_int(SQL, 1, i);

    rc = sqlite3_step(stmt);

    if (rc == SQLITE_ROW)
    {
       // QUERY retrieve a record from DB
    }

Last thing It would be better to change name of array. Not has the same name of defined struct type. Something like Register registers[8];

Upvotes: 1

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