Re: [PHP-DB] Prepared Statements - Search
Ethan,
9 times out of 10 your answer is in the error statement.
Warning: mysqli_stmt_bind_result(): Number of bind variables doesn't
match number of fields in prepared statement.
GL,
Best,
Karl
On Sep 13, 2012, at 7:09 PM, Ethan Rosenberg, PhD wrote:
Dear List -
Here is another problem I am having with prepared statements. The
last one was INSERT, this one is SELECT.
Here is 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| |
++-+--+-+-+---+
10 rows in set (0.00 sec)
Here is my code:
// Prepare statement
$stmt = mysqli_stmt_init($cxn);
$sql11 = "SELECT 'Fname', 'Lname', 'Phone', Height, Hx, Bday,
Age FROM Intake3 where 1 and (MedRec = ?) and (Site = ?) and (Sex
= ?)";
// Allocates and initializes a statement object suitable for
mysqli_stmt_prepare().
// Prepare statement, bind result variables, execute and place
results into bound result variables
mysqli_stmt_prepare($stmt, $sql11);
mysqli_stmt_execute($stmt);
mysqli_stmt_bind_result($stmt, $Site, $MedRec, $Fname, $Lname,
$Phone, $Height, $Sex, $Hx, $Bday, $Age); //The error is in this
statement.
while (mysqli_stmt_fetch($stmt)) {
printf("%s %s %s %s %s %s %s %s %s %s \n", $Site, $MedRec,
$Fname, $Lname, $Phone, $Height, $Sex, $Hx, $Bday, $Age);
}
I get no output from the printf statement.
I receive the following error:
Warning: mysqli_stmt_bind_result(): Number of bind variables doesn't
match number of fields in prepared statement.
The query, with the values inserted, works on the command line
Help and advice, please.
Ethan
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Karl DeSaulniers
Design Drumm
http://designdrumm.com
[PHP-DB] Prepared Statements - Search -- SOLVED!!!
Dear List -
-->> THANKS TO ALL. See below <--
Here is another problem I am having with prepared statements. The last
one was INSERT, this one is SELECT.
Here is 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| |
++-+--+-+-+---+
10 rows in set (0.00 sec)
Here is my code:
// Prepare statement
$stmt = mysqli_stmt_init($cxn);
$sql11 = "SELECT 'Fname', 'Lname', 'Phone', Height, Hx, Bday, Age
FROM Intake3 where 1 and (MedRec = ?) and (Site = ?) and (Sex = ?)";
// Allocates and initializes a statement object suitable for
mysqli_stmt_prepare().
// Prepare statement, bind result variables, execute and place results
into bound result variables
mysqli_stmt_prepare($stmt, $sql11);
mysqli_stmt_execute($stmt);
mysqli_stmt_bind_result($stmt, $Site, $MedRec, $Fname, $Lname,
$Phone, $Height, $Sex, $Hx, $Bday, $Age); //The error is in this statement.
while (mysqli_stmt_fetch($stmt)) {
printf("%s %s %s %s %s %s %s %s %s %s \n", $Site, $MedRec,
$Fname, $Lname, $Phone, $Height, $Sex, $Hx, $Bday, $Age);
}
I get no output from the printf statement.
I receive the following error:
Warning: mysqli_stmt_bind_result(): Number of bind variables doesn't
match number of fields in prepared statement.
The query, with the values inserted, works on the command line
Help and advice, please.
Ethan
***
Here is my revised code:
// Prepare statement
$stmt = mysqli_stmt_init($cxn);
$sql11 = "SELECT MedRec, Site, Sex,Fname, Lname, Phone, Height,
Hx, Bday, Age FROM Intake3 where 1 and (MedRec = ?) and (Site = ?) and
(Sex = ?)";
/* Allocates and initializes a statement object suitable for
mysqli_stmt_prepare(). */
/* Prepare statement, bind result variables, execute and place results
into bound result variables */
mysqli_stmt_prepare($stmt, $sql11);
mysqli_stmt_bind_param($stmt, 'iss', $_POST['MedRec'],
$_POST['Site'], $_POST['Sex']);
mysqli_stmt_execute($stmt);
mysqli_stmt_bind_result($stmt, $MedRec, $Site, $Sex, $Fname,
$Lname, $Phone, $Height, $Hx, $Bday, $Age);
while (mysqli_stmt_fetch($stmt)) {
printf("%s %s %s %s %s %s %s %s %s %s \n", $Site, $MedRec,
$Fname, $Lname, $Phone, $Height, $Sex, $Hx, $Bday, $Age);
}
Added the bind_parameters statement and it worked. Stupid me - you
can’t perform a query unless the parameters have been inserted into the
query.
Live and learn.
Ethan
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Re: [PHP-DB] Re: Problems w/ insert -- SOLVED!!!
On 9/13/2012 9:15 PM, Ethan Rosenberg, PhD wrote: Dear list - Thanks to all. It now works! The problem, as you correctly noted, was the erroneous inclusion of the bind-results statement. Removed that and it worked!! Thanks again! Ethan Methinks Ethan is thanking the group for assistance on his PREVIOUS problem with the prepared INSERT query. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Re: Problems w/ insert -- SOLVED!!!
Dear list - Thanks to all. It now works! The problem, as you correctly noted, was the erroneous inclusion of the bind-results statement. Removed that and it worked!! Thanks again! Ethan -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Prepared Statements - Search
Does the list of the SELECT fields not have to match the variables you are
binding? E.g. if you do not include MedRec in your SELECT then you have no
MedRec data to bind from your $sql11 variable to the $MedRec varable and
then nothing to print there... or what? Am I just fabulating? :).
Cheers,
Fjalar
On Friday, September 14, 2012, Ethan Rosenberg, PhD
wrote:
Dear List -
>
> Here is another problem I am having with prepared statements. The last
> one was INSERT, this one is SELECT.
>
> Here is 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| |
> ++-+--**+-+-+---+
> 10 rows in set (0.00 sec)
>
> Here is my code:
>
> // Prepare statement
> $stmt = mysqli_stmt_init($cxn);
> $sql11 = "SELECT 'Fname', 'Lname', 'Phone', Height, Hx, Bday, Age
> FROM Intake3 where 1 and (MedRec = ?) and (Site = ?) and (Sex = ?)";
> // Allocates and initializes a statement object suitable for
> mysqli_stmt_prepare().
> // Prepare statement, bind result variables, execute and place results
> into bound result variables
>mysqli_stmt_prepare($stmt, $sql11);
>mysqli_stmt_execute($stmt);
>mysqli_stmt_bind_result($stmt, $Site, $MedRec, $Fname, $Lname, $Phone,
> $Height, $Sex, $Hx, $Bday, $Age); //The error is in this statement.
> while (mysqli_stmt_fetch($stmt)) {
> printf("%s %s %s %s %s %s %s %s %s %s \n", $Site, $MedRec, $Fname,
> $Lname, $Phone, $Height, $Sex, $Hx, $Bday, $Age);
> }
>
> I get no output from the printf statement.
>
> I receive the following error:
>
> Warning: mysqli_stmt_bind_result(): Number of bind variables doesn't match
> number of fields in prepared statement.
>
> The query, with the values inserted, works on the command line
>
> Help and advice, please.
>
> Ethan
>
>
> --
> PHP Database Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
On Friday, September 14, 2012, Ethan Rosenberg, PhD wrote:
> Dear List -
>
> Here is another problem I am having with prepared statements. The last
> one was INSERT, this one is SELECT.
>
> Here is 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| |
> ++-+--**+-+-+---+
> 10 rows in set (0.00 sec)
>
> Here is my code:
>
> // Prepare statement
> $stmt = mysqli_stmt_init($cxn);
> $sql11 = "SELECT 'Fname', 'Lname', 'Phone', Height, Hx, Bday, Age
> FROM Intake3 where 1 and (MedRec = ?) and (Site = ?) and (Sex = ?)";
> // Allocates and initializes a statement object suitable for
> mysqli_stmt_prepare().
> // Prepare statement, bind result variables, execute and place results
> into bound result variables
>mysqli_stmt_prepare($stmt, $sql11);
>mysqli_stmt_execute($stmt);
>mysqli_stmt_bind_result($stmt, $Site, $MedRec, $Fname, $Lname, $Phone,
> $Height, $Sex, $Hx, $Bday, $Age); //The error is in this statement.
> while (mysqli_stmt_fetch($stmt)) {
> printf("%s %s %s %s %s %s %s %s %s %s \n", $Site, $MedRec, $Fname,
> $Lname, $Phone, $Height, $Sex, $Hx, $Bday, $Age);
> }
>
> I get no output from the printf statement.
>
> I receive the following error:
>
> Warning: mysqli_stmt_bind_result(): Number of bind variables doesn't match
> number of fields in prepared statement.
>
> The query, with the values inserted, works on the command line
>
> Help and advice, please.
>
> Ethan
>
>
> --
> PHP Database Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
>
>
[PHP-DB] Prepared Statements - Search
Dear List -
Here is another problem I am having with prepared statements. The last
one was INSERT, this one is SELECT.
Here is 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| |
++-+--+-+-+---+
10 rows in set (0.00 sec)
Here is my code:
// Prepare statement
$stmt = mysqli_stmt_init($cxn);
$sql11 = "SELECT 'Fname', 'Lname', 'Phone', Height, Hx, Bday, Age
FROM Intake3 where 1 and (MedRec = ?) and (Site = ?) and (Sex = ?)";
// Allocates and initializes a statement object suitable for
mysqli_stmt_prepare().
// Prepare statement, bind result variables, execute and place results
into bound result variables
mysqli_stmt_prepare($stmt, $sql11);
mysqli_stmt_execute($stmt);
mysqli_stmt_bind_result($stmt, $Site, $MedRec, $Fname, $Lname,
$Phone, $Height, $Sex, $Hx, $Bday, $Age); //The error is in this statement.
while (mysqli_stmt_fetch($stmt)) {
printf("%s %s %s %s %s %s %s %s %s %s \n", $Site, $MedRec,
$Fname, $Lname, $Phone, $Height, $Sex, $Hx, $Bday, $Age);
}
I get no output from the printf statement.
I receive the following error:
Warning: mysqli_stmt_bind_result(): Number of bind variables doesn't match
number of fields in prepared statement.
The query, with the values inserted, works on the command line
Help and advice, please.
Ethan
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
