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PHP Form without Java

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Join Date: Oct 2009

Posts: 6

Reputation: catcoffee is an unknown quantity at this point

Solved Threads: 0

catcoffee

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Newbie Poster

PHP Form without Java

Oct 15th, 2009

Hi all!

Сonditions of a problem:

1) in one form post info and upload file
2) validation only php without js
3) associated fields: if the value is not the same, we don't need to validate field
4) associated fields: if the value is the same, we need echo new input

And i don't know how make it all in one form without java.
Google and others resource didn't help me

May be you know how or where i can see tutorial for it problem?

I've come to know all w3school, php.net, tizag, tutorials on pixel2life and tutsearch.net, nettuts and even http://www.benjaminkeen.com/software/php_validation/ don't help me.
despair deeply

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Join Date: Aug 2007

Posts: 793

Reputation: darkagn has a spectacular aura about

Solved Threads: 110

darkagn

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Master Poster

Oct 15th, 2009

Hi catcoffee and welcome to DaniWeb

1) Can be achieved with a <input type='file'> in a PHP/HTML form. PHP then uses an array, $_POST, to retrieve all posted data and the $_FILE array to access uploaded files.
2) By validation do you mean making sure that the inputted data is acceptable? If so, PHP provides regex functions that can be performed on your $_POST array as well as other ways of validating your form data.
3) and 4) I don't follow what you need here?

There are no stupid questions, only those too stupid to ask for help.
echo is a web developer's best friend.

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Join Date: Oct 2009

Posts: 14

Reputation: webexpertzhere is an unknown quantity at this point

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webexpertzhere webexpertzhere is offline

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Newbie Poster

Oct 15th, 2009

Did you have a look at http://www.w3schools.com/PHP/php_file_upload.asp in w3schools site? The code clearly explains all steps required to upload a file using HTML and PHP alone.
What are your doubts in points 3 and 4? Can you elaborate on it? Unable to understand and so unable to help

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Join Date: Oct 2009

Posts: 6

Reputation: catcoffee is an unknown quantity at this point

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catcoffee

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Newbie Poster

Oct 15th, 2009

Hi darkagn! Thank you for your answers

All problems are in associeted fields

example:

 Help with Code Tags 
 PHP Syntax (Toggle Plain Text) 
<form action='<?php echo $_SERVER['SCRIPT_NAME'] ?>' method='post'>
//...//
<input type="checkbox" name="pregnancy" value="1"/>
<input type="checkbox" name="onebaby" value="2"/>
<input type="checkbox" name="twobaby" value="3"/>
<input type="text" name="namebaby" />
<input type="text" name="agebaby" />
<input type="text" name="namebaby1" />
<input type="text" name="agebaby1" />
//...//
<input type="submit" name="btn" value="send"/>
</form>
<form action='<?php echo $_SERVER['SCRIPT_NAME'] ?>' method='post'>
//...//
<input type="checkbox" name="pregnancy" value="1"/>
<input type="checkbox" name="onebaby" value="2"/>
<input type="checkbox" name="twobaby" value="3"/>
<input type="text" name="namebaby" />
<input type="text" name="agebaby" />
<input type="text" name="namebaby1" />
<input type="text" name="agebaby1" />
//...//
<input type="submit" name="btn" value="send"/>
</form>

If user choice pregnancy field - fields "namebaby" and "agebaby" not required for user.
If user choice onebaby field - fields "namebaby" and "agebaby" required for user, but fields "namebaby1" and "agebaby1" not required for user.
All fields required for user with 2 babies.

I tried insert script in form and call each field

 Help with Code Tags 
 PHP Syntax (Toggle Plain Text) 
$value = '1';
echo "<input type='checkbox' name='pregnancy' value='1'/> Pregnancy?";
 
if (isset($_POST['pregnancy'])) {
    // 
    if ($_POST['pregnancy'] == $value) {
        $pregnanc = true;
} else {
    // No value
    $pregnanc = false;
}
 
if ($pregnanc) {
    print 'happy';
} else {
    echo "<input type='text' name='namebaby' />";
}
$value = '1';
echo "<input type='checkbox' name='pregnancy' value='1'/> Pregnancy?";

if (isset($_POST['pregnancy'])) {
    // 
    if ($_POST['pregnancy'] == $value) {
        $pregnanc = true;
} else {
    // No value
    $pregnanc = false;
}

if ($pregnanc) {
    print 'happy';
} else {
    echo "<input type='text' name='namebaby' />";
}

it was bad idea

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Join Date: Oct 2009

Posts: 99

Reputation: jomanlk is an unknown quantity at this point

Solved Threads: 18

jomanlk

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Junior Poster in Training

Form validation

Oct 16th, 2009

First off, your HTML form needs to look like this

 Help with Code Tags 
 PHP Syntax (Toggle Plain Text) 
<form action='<?php echo $_SERVER['SCRIPT_NAME'] ?>' enctype="multipart/form-data" action='' method='post'>
//...//
<input type="radio" name="num_babies" value="0"/>
<input type="radio" name="num_babies" value="1"/>
<input type="radio" name="num_babies" value="2"/>
<input type="text" name="namebaby" />
<input type="text" name="agebaby" />
<input type="text" name="namebaby1" />
<input type="text" name="agebaby1" />
//...//
<input type="submit" name="btn" value="send"/>
</form>
<form action='<?php echo $_SERVER['SCRIPT_NAME'] ?>' enctype="multipart/form-data" action='' method='post'>
//...//
<input type="radio" name="num_babies" value="0"/>
<input type="radio" name="num_babies" value="1"/>
<input type="radio" name="num_babies" value="2"/>
<input type="text" name="namebaby" />
<input type="text" name="agebaby" />
<input type="text" name="namebaby1" />
<input type="text" name="agebaby1" />
//...//
<input type="submit" name="btn" value="send"/>
</form>

Notice the enctype="multipart/form-data" , this tells the browser that a file needs to be encoded with this form.

Next up, you need to change the 'baby' field to a radio button so the user can only select one option. Set the appropriate values (0, 1 and 2) so you know which one is selected.

Your PHP validation should be something like this

 Help with Code Tags 
 PHP Syntax (Toggle Plain Text) 
<?php
<?php	$babyField	= isset($_POST['num_babies']) ? $_POST['num_babies'] : 0; //if num babies is set, it takes that value
																			//otherwise it assigns 0
 
	switch ($babyField) { //This will manage our baby types more easily
		case 0: //Type is preganancy
				//No names are required
			break;
 
		case 1:	//One baby
			if ($_POST['namebaby'] == "") {}//Error do something
			if ($_POST['agebaby'] == "") {}//Error do something
			break;
 
		case 2:	//two babies
			if ($_POST['namebaby'] == "") {}//Error do something
			if ($_POST['agebaby'] == "") {}//Error do something
			if ($_POST['namebaby1'] == "") {}//Error do something
			if ($_POST['agebaby1'] == "") {}//Error do something
			break;
 
		default:	//Just in case someone got smart and put an invalid field. so we
			break;
	}
 
 
?>
?>
<?php
<?php	$babyField	= isset($_POST['num_babies']) ? $_POST['num_babies'] : 0; //if num babies is set, it takes that value
																			//otherwise it assigns 0

	switch ($babyField) { //This will manage our baby types more easily
		case 0: //Type is preganancy
				//No names are required
			break;

		case 1:	//One baby
			if ($_POST['namebaby'] == "") {}//Error do something
			if ($_POST['agebaby'] == "") {}//Error do something
			break;

		case 2:	//two babies
			if ($_POST['namebaby'] == "") {}//Error do something
			if ($_POST['agebaby'] == "") {}//Error do something
			if ($_POST['namebaby1'] == "") {}//Error do something
			if ($_POST['agebaby1'] == "") {}//Error do something
			break;

		default:	//Just in case someone got smart and put an invalid field. so we
			break;
	}


?>
?>

Obviously this is a very simple form handling technique. But it should be good enough for your purposes. Close thread if this is the solution or continue with your query

Last edited by jomanlk; Oct 16th, 2009 at 11:26 am. Reason: accidentally submitted before I was done