Checkbox doesn't work

Question

HasNor 0 Light Poster

12 Years Ago

Hi, actually i've tried the simple checkbox code but it doesn't work. My point is want to insert the selected checkbox value into the same database row.

Example : one,three --> if there are selected.

This is my code. Anybody can help me?

<?php
include 'connection/db_connect.php';

if(isset($one))
 $val1=$one.";";
 else
 $val1=NULL;

if(isset($two))
 $val2=$two.";";
 else
 $val2=NULL;
 
if(isset($three))
  $val3=$three.";";
  else
 $val3=NULL;
 
$printVal=$val1."".$val2."".$val3;

$val=array($printVal);


if(isset($_POST['Submit'])) {

    if($error ==0){
	
	
	foreach($val as $values)
	{
		
	
	if(!empty($values)){	
	$sql="INSERT INTO `request` (detail) values('$values')";
        
    mysql_query($sql);}
	
if (mysql_error()) trigger_error(mysql_error());
	}
} }?>

<html>
<body>
<form name="frm" method="POST" action="<?php $_SERVER['PHP_SELF']; ?> "  enctype="multipart/form-data">
<input name="one" type="checkbox" value="$one" /> One
<input name="two" type="checkbox" value="$two" /> Two
<input name="three" type="checkbox" value="$three" /> Three
<input type="submit" name="submit" value="submit">
</form>

</body>
</html>

php

3 Contributors
3 Replies
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4 Days Discussion Span
Latest Post 12 Years Ago Latest Post by Atli

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pzuurveen 90 Posting Whiz in Training

12 Years Ago

the result of your form will be in $_POST[ ]

if(isset($_POST['one']))
      $val='one;';
else
      $val=NULL;

if(isset($_POST['two']))
      $val.='two;';

if(isset($_POST['tree']))
      $val.='tree;';

you now have a string $val : one;three;
just put it into your database

if(!empty($val)){
   $sql="INSERT INTO `request` (detail) values('$val')";
   mysql_query($sql)or die(mysql_error());
   }

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HasNor 0 Light Poster · Answer 1 · 2012-01-31T06:43:39+00:00

yaaa!! thanks to pzuurveen.. i got it! Now my problem is solved. Thank you for the help.

Atli 182 Posting Pro · Answer 2 · 2012-01-31T10:01:43+00:00

My point is want to insert the selected checkbox value into the same database row.
Example : one,three --> if there are selected.

I think it's worth pointing out that this is a very bad idea. The first rule of relational database design is to: Never store repeating groups of data.

What that means is:

Don't put more than one value into a single field.
Don't put more than one identical column into a table. (Such as "phone1", "phone2", etc...)

Ideally you would want to create one "request" table, and one "request_details" table. The first table is to store each request, and the second is to store each checkbox selected for that request. Something like:

+---------+     +---------------------+
| request |     | request_details     |
+---------+     +---------------------+
| id (PK) |1---*| request_id (FK, PK) |
| etc...  |     | detail (PK)         |
+---------+     | etc...              |
                +---------------------+

The PHP code to work with that could look something like:

<?php
// Stores the " checked" attribute for the boxes that are
// selected. Will be filled in in the for loop.
$checkedBoxes = array("", "", "");

// What data do you *need* to be set for this to work? 
// The submit button? NO! The checkboxes.
// My point: Don't check if the submit button is
// set, make sure the *data* is set.
if (isset($_POST["checks"])) 
{
    // Stores the values for the selected checkboxes, the once
    // we want inserted into the database.
    $selectedvalues = array();
    
    // We have three boxes, so we loop three times.
    for ($i = 0; $i < 3; ++$i) 
    {
        // See if the user selected this checkbox.
        if (isset($_POST["checks"][$i])) 
        {
            // Fetch the value for the database.
            $selectedvalues[] = mysql_real_escape_string($_POST["checks"][$i]);
            
            // Set the "checked" attribute on the checkbox
            // in the form.
            $checkedBoxes[$i] = " checked";
        }
    }
    
    // Only proceed if there is anything to add to the database.
    if ($count($selectedvalues) > 0) 
    {
        // Insert the request. I assume there is more data
        // that you just left out?
        $sql = "INSERT INTO `request`() VALUES()";
        $result = mysql_query($sql);
        if (!$result || mysql_affected_rows($result) != 1)
            trigger_error("Failed to insert request: " . mysql_error(), E_USER_ERROR);
        
        $requestID = mysql_insert_id();
        
        // And then add each checkbox value for the request to the
        // request_details table.
        $sql = "INSERT INTO `request_details`(`request_id`, `value`)
                VALUES (%d, '%s')";
        
        // We want ONE ROW FOR EACH OPTION, so we loop through
        // each of them. To link the options to the request, we
        // set the "requestID" as well. (A Foreign Key.)
        foreach ($selectedValues as $value) {
            $result = mysql_query(sprintf($sql, $value)) or die(mysql_error());
            if (!$result || mysql_affected_rows($result) != 1)
                trigger_error("Failed to insert request option: " . mysql_error(), E_USER_ERROR);
        }
    }
}
?>
<!DOCTYPE html>
<html>
<head>
    <title>Example</title>
    <meta charset="UTF-8"/>
</head>
<body>
    <form name="frm" method="POST" action="<?php $_SERVER['PHP_SELF']; ?> "  enctype="multipart/form-data">
        <input name="checks[]" type="checkbox" value="one"<?php echo $checkedBoxes[0]; ?>> One
        <input name="checks[]" type="checkbox" value="two"<?php echo $checkedBoxes[1]; ?>> Two
        <input name="checks[]" type="checkbox" value="three"<?php echo $checkedBoxes[2]; ?>> Three
        <input type="submit" name="submit" value="submit">
    </form>
</body>
</html

Note that the value attribute of a <input type="checkbox"> does not determine it's state. It's essentially a static value passed only if the box is checked, and ignored if it's not. - To have a checkbox initially checked, add the checked attribute to it.

Checkbox doesn't work

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