Checkbox with PHP and MYSQL

Question

I have two checkboxes

<form action="" method="post" enctype="multipart/form-data">
<input type="checkbox" name="file_type" value="1"> Filer<br />
<input type="checkbox" name="file_type" value="1"> Statistik
</form>

and i have two rows on my database table das_custermer_files.

row 1 = files row 2 = statistic

how can i check one of them an put the value into my rows.

i have try with this code

if($_POST['file_type'] == 1){
            mysql_query("INSERT INTO das_custermer_files (url, das_file_categories_id, das_custermers_id, name) 
                         VALUE('$fileurl', 1, $user_custermers_id, '$name')"
                    ) or die(mysql_error());
            }

            elseif($_POST['statistik_type'] == 1){
                mysql_query("INSERT INTO das_custermer_files (url, das_custermer_upload_id, das_custermers_id, name) 
                         VALUE('$fileurl', 1, $user_custermers_id, '$name')"
                    ) or die(mysql_error());
            }
            else{
                echo "no choices";
            }

But if i only check one of them it stil put value 1 in both of my rows.

Hope people understand. If not. ask

i have done it.

the finel code:

html

<form action="" method="post" enctype="multipart/form-data">
    <input type="radio" name="file_type" value="1"> Filer<br />
    <input type="radio" name="file_type" value="2"> Statistik
</form>

PHP

$field = false;

            switch($_POST['file_type'])
            {
                case 1:
                    $field = 'das_file_categories_id'; 
                break;
                case 2:
                    $field = 'das_custermer_upload_id'; 
                break;
                default:
                $field = false;
            }
            if($field)
            {
               mysql_query("INSERT INTO das_custermer_files (url, $field, das_custermers_id, name) 
                         VALUE('$fileurl', 1, $user_custermers_id, '$name')"
                    ) or die(mysql_error());
            }

what happens if they're both ticked? maybe use radio buttons instead — DannyTheDev
– DannyTheDev, Commented May 8, 2013 at 11:30

DannyTheDev · Accepted Answer · 2013-05-08 11:58:26Z

1

My approach would be like this:

I would use radio buttons instead, to ensure a type is selected.

Form Bits:

    <input type="radio" name="file_type" value="0" checked="checked"> None<br />
    <input type="radio" name="file_type" value="1"> Filer<br />
    <input type="radio" name="file_type" value="2"> Statistik

PHP Bits:

$field = false;

 // set the field name based on the file type selected in the form
switch($_POST['file_type'])
{
    case 1:
        $field = '`das_file_categories_id`';
    break;
    case 2:
        $field = '`das_custermer_upload_id`'; 
    break;
    default:
    $field = false;
}

// if a field has been set ( i.e file_type != 0 ) then build and run the query
if($field)
{

    $query  =   "INSERT INTO `das_custermer_files` (`url`, " . $field . ", `das_custermers_id`, `name`) ";
    $query .=   "VALUES ('" . $fileurl . "', 1, " . $user_custermers_id . ", '" . $name . "')";
    mysql_query($query);
    echo "Inserting to db: " . $query;
} else {
    echo "No field set: ";
    print_r($_POST);
}

Things worth noting

You're using mysql_query which is outdated, see here: http://php.net/manual/en/function.mysql-query.php
Even though you are using $fileurl, $user_custermers_id,$name i cannot see these being set, so i am assuming they are set somewhere higher in the file

edited May 8, 2013 at 11:58

answered May 8, 2013 at 11:36

DannyTheDev

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2 Comments

Thomas Over a year ago

there are nothing in my database. But i like the radio buttons

DannyTheDev Over a year ago

i've added some echo's to see what its doing.. so run it & see what the output is, it'll probably help diagnose the issue

Amir · Accepted Answer · 2013-05-08 11:44:59Z

1

your check_box names should be different. yours the same.

<input type="checkbox" name="file_type" value="1"> Filer<br />
<input type="checkbox" name="file_type" value="1"> Statistik

answered May 8, 2013 at 11:44

Amir

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Comments

Rajeev Ranjan · Accepted Answer · 2013-05-08 11:54:32Z

Html code :

<form action="test.php" method = 'post'>
<input type="checkbox" name="file_type[]" value="1"> Filer<br />
<input type="checkbox" name="file_type[]" value="2"> Statistik
<input type="submit" value="submit" />

checkbox will contain its value in array $_POST['file_type'] and you may inter its value to db as follows :-

if(isset($_POST['file_type'])){
 foreach($_POST['file_type'] as $keys)
   {     
     mysql_query("INSERT INTO das_custermer_files (url, das_file_categories_id,     das_custermers_id, name) 
                 VALUE('$fileurl', $keys, $user_custermers_id, '$name')"
            ) or die(mysql_error());
    }
 }
    else if(isset($_POST['statistik_type'])){
        mysql_query("INSERT INTO das_custermer_files (url, das_custermer_upload_id, das_custermers_id, name) 
                 VALUE('$fileurl', 1, $user_custermers_id, '$name')"
            ) or die(mysql_error());
    }
    else{
        echo "no choices";
    }

this maybe your requirement...

Atanu · Accepted Answer · 2013-05-08 11:23:14Z

0

Try this .Should work.

 if(isset($_POST['file_type'])){
        mysql_query("INSERT INTO das_custermer_files (url, das_file_categories_id, das_custermers_id, name) 
                     VALUE('$fileurl', 1, $user_custermers_id, '$name')"
                ) or die(mysql_error());
        }

        else if(isset($_POST['statistik_type'])){
            mysql_query("INSERT INTO das_custermer_files (url, das_custermer_upload_id, das_custermers_id, name) 
                     VALUE('$fileurl', 1, $user_custermers_id, '$name')"
                ) or die(mysql_error());
        }
        else{
            echo "no choices";
        }

answered May 8, 2013 at 11:23

Atanu

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1 Comment

Atanu Over a year ago

You are using same name for the two check boxes.That is wrong.Please use unique name

Muhammad Raheel · Accepted Answer · 2013-05-08 11:24:01Z

You approach is totally wrong. You should follow this structure

<form action="test.php" method = 'post'>
    <input type="checkbox" name="file_type" value="1"> Filer<br />
    <input type="checkbox" name="file_type" value="2"> Statistik
    <input type="submit" value="submit" />
</form>

Now with php

$query  =   '';
if($_POST['file_type'] == 1){
    $query  =   "INSERT INTO das_custermer_files 
                (url, das_file_categories_id, das_custermers_id, name) 
                VALUE
                ('$fileurl', 1, $user_custermers_id, '$name')";
}else if($_POST['file_type'] == 2){
    $query  =   "INSERT INTO das_custermer_files 
                (url, das_custermer_upload_id, das_custermers_id, name) 
                VALUE
                ('$fileurl', 1, $user_custermers_id, '$name')";
}

mysqli_query($query);

RePRO · Accepted Answer · 2013-05-08 11:47:01Z

0

You probably forgot to <form action = "..." method = "POST"> </form>.

First step, you should do the test:

print_r($_POST);

What you get by this test? If the first step is correct (not empty array), we can finish:

if (isset($_POST['file_type'])) { 
   ... 
};

// OR use alternative
if ($_POST['file_type'] == 1) { 
   ... 
};

edited May 8, 2013 at 11:47

answered May 8, 2013 at 11:30

RePRO

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Comments

Hexodus · Accepted Answer · 2013-05-08 11:52:54Z

0

Clicking on a checkbox doesen't change the value attribute but adds "checked" in the dome like this:

<input type="checkbox" name="file_type" value="1" checked> Filer<br />

So the submitted value of the checked box will stay forever at "1"

EDIT

When you replace the values with strings you can check against it in php.

HTML

<form action="" method="post" enctype="multipart/form-data">
<input type="checkbox" name="file_type" value="Filer"> Filer<br />
<input type="checkbox" name="file_type" value="Statistik"> Statistik
</form>

PHP

if($_POST['file_type'] == "Filer"){
            mysql_query("INSERT INTO das_custermer_files (url, das_file_categories_id, das_custermers_id, name) 
                         VALUE('$fileurl', 1, $user_custermers_id, '$name')"
                    ) or die(mysql_error());
            }

            elseif($_POST['file_type'] == "Statistik"){
                mysql_query("INSERT INTO das_custermer_files (url, das_custermer_upload_id, das_custermers_id, name) 
                         VALUE('$fileurl', 1, $user_custermers_id, '$name')"
                    ) or die(mysql_error());
            }
            else{
                echo "no choices";
            }

edited May 8, 2013 at 11:52

answered May 8, 2013 at 11:35

Hexodus

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2 Comments

Hexodus Over a year ago

I did as you wished and tried to let your code almost untouched.

Hexodus Over a year ago

Could you please replace echo "no choices"; with echo "no choices - submitted value: ".$_POST['file_type']; and tell me what it says?

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