显示所有匹配的值而不是第一个

Question

I have created a query which I am using to display reports. The query gets data from a few different tables which is fine, although I cannot figure out how to echo more than one matching value within a table. For example the table vtiger_addisa has more than one newcurrentamount which I need displayed.

$sql = $adb->query("SELECT *
                FROM vtiger_isa, vtiger_addisa, vtiger_contactdetails
                WHERE vtiger_isa.relatedclient = vtiger_addisa.addrelatedclient
                AND vtiger_addisa.addrelatedclient = vtiger_contactdetails.contactid
                AND vtiger_isa.relatedclient = $relatedclient
                AND vtiger_isa.policynumber = $policynumber");


            //Uncomment lines below for testing sql             
             //echo $sql;
            //exit; 
            while ($sql->fetchInto($row)) {
    // Assuming DB's default fetchmode is DB_FETCHMODE_ORDERED
    echo $row['firstname'] . "
"; 
    echo $row['lastname'] . "
"; 
    echo $row['policynumber'] . "
"; 
    echo $row['newcurrentamount'] . "
"; 
    echo $row['newcurrentamount'] . "
"; 
    echo $row['currentamount'] . "
"; 



   exit;

} 

Answer 1

Use 'as' operator

select newcurrentamount as a, newcurrentamount as b from vtiger_addisa

P.S- Try avoiding ' * ' and use only the necessary columns.