declaring a checkbox in php and then updating a mysql db

[bhavsta]

Hi Guys,

I have a html for that consists of drop down menus, text boxes and two checkboxs. I can get all the data to be sent to a mysql db via php. But when i try to insert the checkboxes into the db i get an error stating the following:

Could not insert data: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ') VALUES ('6','a1','b1','3','c1','1234','1234','1234@1234.com

the code for my html form (checkboxes) is below:

<input type="checkbox" border="0" name="dprotection_one" value="1" />

<input type="checkbox" border="0" name="dprotection_two" value="2" />

i have declared the boxes in the php script as follows:

$dprotection_one=$_POST['dprotection_one'];
$dprotection_two=$_POST['dprotection_two'];


Please can someone help me.

Regards,

Bhav

 

[vacunita]

Can we see your complete query, and how you are putting the checkbox values in it?

Also you should know that if a checkbox is not checked, then php will not create the POST variable for it.

so if dprotection_one is not checked but dprotection_two is, you'll only get $_POST['dprotection_two'] as a POST variable.

If you try to use $_POST['dprotection_one'] you'll get an error.

----------------------------------
Ignorance is not necessarily Bliss, case in point:
Unknown has caused an Unknown Error on Unknown and must be shutdown to prevent damage to Unknown.

 

[bam720]

The trick about check boxes in PHP is they return a value of "on" if checked, and nothing if unchecked. You need to convert this to a boolean value... For MySQL I use...

if($Chkbx== "on")
    $Chkbx = 1;
else
    $Chkbx = 0;

 

[bhavsta]

DECLARATIONS

$salutation=$_POST['salutation'];
$first_name=$_POST['first_name'];
$surname=$_POST['surname'];
$status=$_POST['status'];
$job_title=$_POST['job_title'];
$DoB=$_POST['DoB'];
$telephone=$_POST['telephone'];
$email_address=$_POST['email_address'];
$university_college_school=$_POST['university_college_school'];
$start=$_POST['start'];
$end=$_POST['end'];
$dprotection_one=$_POST['dprotection_one'];
$dprotection_two=$_POST['dprotection_two'];


CREATING TABLE

$create ="CREATE table Registration_Data (salutation varchar(50),
first_name varchar(50),
surname varchar(50),
job_title varchar(50),
status varchar(30),
DoB varchar(30),
telephone varchar(20),
email varchar(50),
university_college_school varchar(30)
start varchar(15),
end varchar(15),
dprotection_one CHAR(1),
dprotection_two CHAR(1))";
mysql_query($create);

QUERY INSERT

$query = "INSERT INTO Registration_Data (salutation,
first_name,
surname,
job_title,
status,
DoB,
telephone,
email,
university_college_school,
start,
end,
dprotection_one,
dprotection_two,) VALUES
('".$salutation."','".$first_name."','".$surname."','".$status."','".
$job_title."','".$DoB."','".$telephone."','".$email_address."','".
$university_college_school."','".$start."','".$end."','".$dprotection_one.
"','".$dprotection_two."')";
if (mysql_query($query)) {
header("Location:

http://www.thelawyer.com/l2b/reg_response.html");

exit();
} else {
die ("Could not insert data: " . mysql_error());
}

 

[bhavsta]

Hi Bam720,

How would i declare this as a variable, the if statement makes sense, however, how can i declare this and then get the results to insert into a mysql db after is has been picked up by the php?

Regards,

Bhavesh

Thank you for your help so far.

 

[bam720]

$dprotection_one=$_POST['dprotection_one'];
$dprotection_two=$_POST['dprotection_two']; 

if($dprotection_one== "on")
    $dprotection_one = 1;
else
    $dprotection_one = 0;

if($dprotection_two== "on")
    $dprotection_two = 1;
else
    $dprotection_two = 0;

You insert statement will work fine now

 

[jpadie]

surely the first two lines of bam720's response will throw warnings in the event a checkbox is not checked. These may or may not be visible depending on your setting for display_errors

i would recommend preprofiling your forms (have the controls in an array that you use to create the output, validate and process the input.

absent of a profiled form i'd recommend rather using this construct

$dprotection_one= isset($_POST['dprotection_one']) ? 1 : 0;

 

[bhavsta]

thanx alot people i have cracked it!!!! in the end i ended up using the variable type as a tinyint in the declaration for dprotection. this seemed to do the trick. it even allowed blank ticks to be parsed and to be written into the db!!!! wonderfull

thanx again to everyone who helped me.

Regards,

Bhav