Re: Prepared Statements Insert Problem

On 9/2/2012 12:45 AM, Ethan Rosenberg, PhD wrote:
> Dear List -
>
> I wish to accomplish the following with prepared statements:
>
> FYI -
>
> The Database:
>
> mysql> describe Intake3;
> +--------+-------------+------+-----+---------+-------+
> | Field  | Type        | Null | Key | Default | Extra |
> +--------+-------------+------+-----+---------+-------+
> | Site   | varchar(6)  | NO   | PRI |         |       |
> | MedRec | int(6)      | NO   | PRI | NULL    |       |
> | Fname  | varchar(15) | YES  |     | NULL    |       |
> | Lname  | varchar(30) | YES  |     | NULL    |       |
> | Phone  | varchar(30) | YES  |     | NULL    |       |
> | Height | int(4)      | YES  |     | NULL    |       |
> | Sex    | char(7)     | YES  |     | NULL    |       |
> | Hx     | text        | YES  |     | NULL    |       |
> | Bday   | date        | YES  |     | NULL    |       |
> | Age    | int(3)      | YES  |     | NULL    |       |
> +--------+-------------+------+-----+---------+-------+
>
> The connection to the database was successful
>
> The variables:
>
> $fptr1 = fopen("/home/ethan/PHP/HRecnumSite", "r+");
> fscanf($fptr1,"%d %s",$MedRec, $Site);
>
> $_POST['MedRec'] = $MedRec;
> $_POST['Site'] = $Site;
> $fname = $_POST['Fname'];
> $lname = $_POST['Lname'];
> $phone = $_POST['Phone'];
> $hgt   = $_POST['Height'];
> $sx    = $_POST['Sex'];
> $hx    = $_POST['Hx'];
> $bday  = $_POST['Bday'];
> $age   = $_POST['Age'];
> if($sx==1)$_POST['Sex'] = 'F';
> if($sx==0)$_POST['Sex'] = 'M';
>
> Statement to convert:
>
> $sql1 =  "INSERT INTO Intake3(Site, MedRec, Fname, Lname, Phone, Height,
> Sex, Hx, Bday, Age)
> VALUES('$Site', $MedRec, '$fname', '$lname', '$phone', $hgt, '$sx',
> '$hx', '$bday', '$age')";
>
> My attempt [which failed]
>
>    $stmt = mysqli_stmt_init($cxn);
>       if($stmt = mysqli_stmt_prepare($stmt, "INSERT INTO Intake3 (Site,
> MedRec, Fname, Lname, Phone, Height, Sex, Hx, Bday, Age)
> VALUES(?,?,?,?,?,?,?,?,?,?")!=0)
>      {
>      print_r($stmt);
>      mysqli_stmt_bind_param($stmt, 'sisssisssi', $Site, $MedRec, $fname,
> $lname, $phone, $hgt, $sx, $hx, $bday, $age);
>      mysqli_execute($stmt);
>      mysqli_stmt_bind_result($stmt, $Site, $MedRec, $fname, $lname,
> $phone, $hgt, $sx, $hx, $bday, $age);
>      echo $stmt;
>      mysqli_stmt_fetch($stmt);
>      mysqli_stmt_close($stmt);
>      }
>       else
>           echo "Ouch";
>
> Regrettably, all I see on the monitor is "ouch"!!
>
> Help and advice, please.
>
> Ethan Rosenberg
>
>
>
>
So - no error.  Does that mean it now works?

-- 
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php