Checkboxes in Web forms

emme85

I had to make a web form for user to fill in. my form has many textfields and a lot of checkboxes too. anyone can help me on how to store checkboxes data into database??
i have 2 type of checkboxes group named 'Type of Accident' and 'Location of the Accident'. currently i'm using wamp5 server to create table.
i have only one table named Report(1 form). is it possible to store all data in one table??

timus

Hey

I've had a very quick look at your code, if you're trying to store all the different values in type_of_accident check boxes i think you'll only receive the last one that has been checked from the form. Following example will show you.

<?php 
if(isset($_POST['submit']))
{
	var_dump($_POST);
}
else
{
	print'
	<form action="" method="post" target="_self">
	test1<input name="test" type="checkbox" value="test1" />
	test2<input name="test" type="checkbox" value="test2" />
	<input name="submit" type="submit" value="submit" />
	</form>	';
}
?>

If your users are only supposed to select one option, you might be better using radio buttons with the same name and different values, if they can select more than one option you'll need to change the names of the check boxes.

Hope this helps

Tim

emme85

Thanks Tim.. but my users supposed to select multiple option for Type of Accident n Location of the Accident.. so, am i supposed to create an array for the table?? or, split the Report table into 3 tables(Report, TypeOfAccident, LocationOfAccident)??
How is it??

timus

When you submit the form, have you tried displaying the values you have for $_POST['TypeOfAccident'] ? I'm guessing even if you tick multiple boxes, you only get one value on screen. Is that correct?

Once you have all the values being returned correctly you could use serialze and unserialize to store the values (possible you've put them in an array) in the database.

Hope this helps
Tim

emme85

Timus, i'd already tried to key in my data, but the datas weren't stored into the database.. why is it?? this is my coding after i alter it.. (sory 4 my english.. im not fluent)..

timus

Can you add a line before

[/$Result1 = mysql_query($insertSQL, $Care_Pair) or die(mysql_error());PHP]

to say print $insertSQL; ?

This will show you what you're trying to add to the database before it goes in. Do you get an error?

Tim

emme85

Yupp.. i got an error.. when the first time i keyed in the data, there's nothing in the database.. when i tried it for the second time, it had an error saying 'duplicate data entry'.. is it because of this coding??

 function GetSQLValueString($theValue, $theType, $theDefinedValue = "", $theNotDefinedValue = "") 
{
  $theValue = (!get_magic_quotes_gpc()) ? addslashes($theValue) : $theValue;

  switch ($theType) {
    case "text":
      $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL";
      break;    

    case "long":
    case "int":
      $theValue = ($theValue != "") ? intval($theValue) : "NULL";
      break;
    case "double":
      $theValue = ($theValue != "") ? "'" . doubleval($theValue) . "'" : "NULL";
      break;
    case "date":
      $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL";
      break;
    case "defined":
      $theValue = ($theValue != "") ? $theDefinedValue : $theNotDefinedValue;
      break;
  }
  return $theValue;
}

i dont know whether it can be used..

timus

Did you do the print $insertsql? what was the sql statement before you tried to run it? something must be going into the database for you to be getting a duplicate key error. I'm guessing you have a problem with your sql and that function you mentioned isn't being used properly, did you write it?

Tim

emme85

Hey Tim.. i'd already did the print $insertsql before, but it didn't work too. then, i change the coding a little bit:

but there problems such as:
1) no data were stored in Report table
2) for the 1st time, data could be stored in Position, Type n Location table(those with multiple data, but only one choice was stored).
3)for the 2nd time, it has error called 'Duplicate entry '' for key 1'.
What would be the problem??

big_nerd

From what I can tell (and seems to be common) there is no ID column in your SQL syntax.

I did this (MySQL):

CREATE TABLE `care_pair_new` (
  `report_id` INT NOT NULL AUTO_INCREMENT PRIMARY KEY, 
  `Report_No` VARCHAR(50) NOT NULL, 
)
ENGINE = myisam;

Etc Etc..

This should eliminate the duplicate entry (making each incremented part even with IDENTICAL information different)

Do you already have data stored in this table?

emme85

I'd already add ID column just like you do, but it still had errors..
may i ask you?? could my const_location, const_type n position table (those with checkboxes) to have the same primary key, ID? n maybe that the cause of the duplicate entry?? because the 1st time i enter data, the datas were stored, but not for the 2nd time..

timus

Its difficult to say where exactly its failing, You may want to try and print each of your sql statements rather than executing them and see if any of it looks wrong. You need to break the script down and try and find where you think the problem is

Tim