Question

I’m trying to get my postgresql to work. I’ve worked with sql before, but never postgres.

I feel like the following should display a nice table of outputs.

I’ve tested the php in several places, and it seems that the result variable is not getting anything returned, and so is a boolean of false value, so there’s nothing retreived. If anyone could help me find the error I’d be grateful.

if($queryType != NULL)

    echo'<table>;
    if($limit == NULL && $offset == NULL)
    {
    $query = 'SELECT * FROM $1';
    $stmt = pg_prepare($connection, "limitoffset", $query);
    $result = pg_execute($connection, "limitoffset", array($queryType));

    if($queryType == 'city'){
    while($row = pg_fetch_assoc($result)){
    echo '<tr>'; 
    echo '<td>' . $row['id] . '</td>';
    echo '</tr>';
    }
 echo '</table>';

again, it compiles and runs, but it seems $row is coming back as false after the query, and so the loop is not running to print out the results, because it has none.

thanks!

Answer 1

You can’t use a placeholder for an identifier (i.e. table or column name) so this:

select * from $1

won’t work. If you want the table name to be a variable, I think you’re stuck with string interpolation:

$table = pg_escape_identifier($queryType);
$query = "SELECT * FROM $table";

You’ll want to use pg_escape_identifier to avoid problems with strange values in $queryType but keep in mind that pg_escape_identifier will double quote the identifier and that could give you case sensitivity problems.

You also have two missing single quotes in your question: one in echo'<table>; and another in . $row['id] . but those are presumably just in your question text since you say that the code compiles and runs.