PHP postrgres SQL executing a query

Question

i'm having problems in executing a query inside php. I have the following query in my php code:

    $nome = pg_escape_string($_POST['cnome']);

$obtem_idb = "SELECT idb FROM banda WHERE nome = $nome";

echo("$nome");

$idb = pg_query($connection, $obtem_idb);
if(!$idb){
        die("Error in SQL query: " . pg_last_error());
    } else { 
echo("o idb que vem da query é $idb");
}

The name i print is correct, but when trying to execute the query i get the following error:

Error in SQL query: ERROR: column "anthrax" does not exist LINE 1: SELECT idb FROM banda WHERE nome = Anthrax ^

Can anyone help, i can't seem to find the error.

When i put the variable $nome in single quotes the value of it changes to 'Resource id #2' and i get the following error:

Error in SQL query: ERROR: insert or update on table "edicao" violates foreign key constraint "edicao_idb_fkey" DETAIL: Key (idb)=(Resource id #2) is not present in table "banda".

Marc B · Accepted Answer · 2011-11-25 19:33:13Z

1

You've forgotten quotes around the value:

$obtem_idb = "SELECT idb FROM banda WHERE nome = '$nome'";
                                                 ^-----^--- must be quoted

answered Nov 25, 2011 at 19:33

Marc B

362k44 gold badges433 silver badges508 bronze badges

Sign up to request clarification or add additional context in comments.

4 Comments

Cláudio Ribeiro Over a year ago

When i put nome in single quotes i get the following error: Resource id #2Error in SQL query: ERROR: insert or update on table "edicao" violates foreign key constraint "edicao_idb_fkey" DETAIL: Key (idb)=(Resource id #2) is not present in table "banda".

Marc B Over a year ago

that's a whole other query causing that error. a select query can NOT produce a foreign key violation.

Cláudio Ribeiro Over a year ago

When i execute the query, the result that comes from it is "Resource id #2" and "B-7" as it should...

Marc B Over a year ago

That's because a query() call returns a result HANDLE, not the data you've requested from the DB. You still have to actually FETCH a row of data and extract the value you want before you can insert that value into another query.

Julien · Accepted Answer · 2011-11-25 19:34:06Z

1

you need to quote $nome in your sql query

$obtem_idb = "SELECT idb FROM banda WHERE nome = '$nome'";

answered Nov 25, 2011 at 19:34

Julien

3103 gold badges20 silver badges56 bronze badges

1 Comment

Cláudio Ribeiro Over a year ago

When i put nome in single quotes i get the following error: Resource id #2Error in SQL query: ERROR: insert or update on table "edicao" violates foreign key constraint "edicao_idb_fkey" DETAIL: Key (idb)=(Resource id #2) is not present in table "banda".

Jake Feasel · Accepted Answer · 2011-11-25 19:34:13Z

1

You need to put your $nome reference in single quotes:

$obtem_idb = "SELECT idb FROM banda WHERE nome = '$nome'";

answered Nov 25, 2011 at 19:34

Jake Feasel

17k5 gold badges55 silver badges68 bronze badges

1 Comment

Cláudio Ribeiro Over a year ago

When i put nome in single quotes i get the following error: Resource id #2Error in SQL query: ERROR: insert or update on table "edicao" violates foreign key constraint "edicao_idb_fkey" DETAIL: Key (idb)=(Resource id #2) is not present in table "banda".

Collectives™ on Stack Overflow

PHP postrgres SQL executing a query

3 Answers 3

4 Comments

1 Comment

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

4 Comments

1 Comment

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related