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
