i have a file 'colecting-data.html' with this code :
<html>
<body>
<form action="collect.php" method="get">
name : <input type="text" name="name">
email : <input type="text" name="email">
telephone: <input type="text" name="telephone">
<input type="submit" value="save">
</form>
</body>
</html>
i when to creat "collect.php" file to collect all the data (name email telephone)
and save then in another file "save.html" inside a table .
it will help me a lot if any one know how to do that .
Very quick & dirty answer, remember to sanitize your inputs.
<?php
//collect.php
$html = "<html><head><title>Test</title></head><body><table>";
foreach($_REQUEST as $k=>$v) {
$html .= "<tr><td>$k</td><td>$v</td></tr>";
}
$html .= "</table></body></html>";
file_put_contents('save.html',$html);
?>
As you can use this to learn, start with the basic way to do this before going for more advanced options...
First off, your form needs to be a POST, not a GET.
<form action="collect.php" method="POST">
This will POST the values to collect.php.
On collect.php, you'll need to collect the values;
<?php
if(isset($_POST['name']) {
// Do Something
;}
if(isset($_POST['email']) {
// Do Something
;}
if(isset($_POST['telephone']) {
// Do Something
;}
?>
From here, you can set the posted values as variables and then do whatever you want with them.
Related
This is my first post and I'm complete beginner so please be gentle :)
I'm trying to create a form that after submitting an account name would check and return a CNAME of the host (account+domain.com)
The problem is that I want to do it all on the same website so it will either display the form if nothing has been posted or display the result otherwise.
This is what I've created, it seems that I'm not calling the POST correctly, but I can't really get what am I doing wrong.
Please help
<?php
if(isset($_POST[DomainSubmit])){
$AccountName = $_POST[ClientDomain];
$CName = dns_get_record($AccountName."domain.com", DNS_CNAME);
echo '<h1>'.$CName.'<h1>';
}
echo'<form action="index.php" method="POST" ">
<input type="text" name="ClientDomain">
<input type="submit" name="DomainSubmit">
</form>'
?>
Try to add else so it will be displayed one or another stuff
<?php
if(isset($_POST['DomainSubmit']) && isset($_POST['ClientDomain'])){
$AccountName = $_POST['ClientDomain'];
$CName = dns_get_record($AccountName."domain.com", DNS_CNAME);
echo '<h1>'.$CName.'<h1>';
} else {
echo'<form action="index.php" method="POST" ">
<input type="text" name="ClientDomain">
<input type="submit" name="DomainSubmit">
</form>'
}
?>
Edit:
You forgot to properly write array POST (missing quotes)
$_POST[DomainSubmit]
And it should be
$_POST['DomainSubmit']
I want form to post automatically if zip variable is passed from URL.
URL looks like: www.sitename.com/maps/zipsearch.php?zip=90210
Form looks like:
<form method="post">
Zipcode:
<input name="zip" value="<?php echo (isset($_GET["zip"]))? $_GET["zip"]:"";?>" />
<input type="submit" name="subbut" value="Find instructors" />
</form>
So it fills the input box with zip code but I would like it to post automatically to see results again if zip is passed.
Maybe an IF / THEN?
Any help would be appreciated.
You mean to echo the value passed in GET parameter?
<input type="submit" name="subbut" value="<?php echo isset($_GET['zip'])?$_GET['zip']:'Find'; ?>" />
EDIT
Or, if you are asking about submitting the form, then something like this might work I believe:
<input type="submit" name="subbut" value="<?php echo isset($_GET['zip'])?$_GET['zip']:'Find'; ?>" />
<?php if( isset( $_GET['zip'] ) ) { ?>
<script>
document.forms["name_of_the_form_here"].submit();
</script>
<?php } ?>
like this:
<form id="form" action="form.php" method="post">
Zipcode:
<input name="zip" value="<?php echo (isset($_GET["zip"]))? $_GET["zip"]:"";?>" />
<input type="submit" name="subbut" value="Find instructors" />
</form>
<?php if (isset($_GET["zip"])): ?>
<script>document.getElementById('form').submit()</script>
<?php endif; ?>
since passing data via URL means GET method, so i think you have a little misconception with your question.
if you would like to post automatically you dont need to show form.
just put this code in your zipsearch.php
if ($_GET['zip'] != ""){
// do what you want if zip parameter is not null
}else{
// do what you want if zip parameter is null
}
It looks like your form is submitting to itself. (Eg. zipsearch.php displays HTML form. When user submits form, it is posted back to zipsearch.php which displays the search results).
If this is the case, you don't have to post anything, because you are already inside the file that handles the form submission. You could do something like this:
<?php
if (isset ($_POST['zip'])) {
$zip = $_POST['zip']; /* Form was submitted */
} else if (isset ($_GET['zip'])) {
$zip = $_GET['zip']; /* "?zip=" parameter exists */
}
if (isset ($zip)) {
/* Display search results */
} else {
/* Display form */
}
Using the following code I am attempting to:
Test to see if one of the dynamically assigned field names has been submitted;
Use the "Actionable Code" to process the submitted information.
My problem lies in I am incapable of retrieving the appropriate dynamic variable name. $this->get_field_name('email_to') will output a name variable such as widget-mywidget[3][email_to]; but to access this value via PHP I need it in the form of $_POST['widget-mywidget'][3]['email_to'].
How can I go about solving this dilemma?
OUTPUTTED HTML:
<form id="widget-mywidget-3-osiris_contact" method="post" action="">
<fieldset>
<input type="text" name="widget-mywidget[3][user_name]">
<input type="text" name="widget-mywidget[3][user_email]">
<textarea name="widget-mywidget[3][user_message]"></textarea>
</fieldset>
<fieldset>
<input type="hidden" name="widget-mywidget[3][email_to]" value="">
<input type="hidden" name="widget-mywidget[3][email_subject]" value="">
<button type="submit" name="widget-mywidget[3][email_send]">Send</button>
</fieldset>
</form>
PROCESSING PHP:
if(in_array($this->get_field_name('email_to'), $_POST)){ // <--- Where I need help.
// Actionable Code
}
This is what $this->get_field_name does:
function get_field_name($field_name) {
return 'widget-' . $this->id_base . '[' . $this->number . '][' . $field_name . ']';
}
I suggest that you print_r($_POST) and compare it visually for better debugging...
(Or use a debugger...)
$thing = "widget-mywidget[3][email_to]";
$exp = explode("[", $thing);
$get_it = $_POST['".$exp[0]."[".$exp[1]."[".$exp[2]."'];
Try, if it works.
i'm new to php , i have been searching for a tutorial regarding inserting form's input(text) , radio and selection data to MySQL database's table using php. i found some tutorials but most are confusing. So i decided to ask.
Okay here's what i want to do. I have a form which have two types of input and a selection
1. input type text
2. input type radio
3. selection
Here's the HTML code :
<form action="" method="post" enctype="multipart/form-data">
<strong>Your Name: </strong><br>
<input type="text" name="myname" value="" />
<br /><br/>
<strong>Which class type you want:</strong><br>
<select name="selection">
<option value="A">A</option>
<option value="B">B</option>
<option value="C">C</option>
</select>
<strong>Do you agree?</strong><br>
<input type="radio" name="agree" value="Yes"> or
<input type="radio" name="agree" value="No">
<input type="submit" name="submit" value="Submit">
</form>
I have set the form action to blank because the php code will be in the same file as the HTML (it's a php file btw)
MySQL table : info
structure :
1. name
2. class
3. agree
I want the php code to insert myname into name , selection's selected data into class , radio selected data into agree
P/S Yes i have added a connect to database php script , i just want to know how to get the form data into mysql.
Can someone write a php code example on how can i do this?
Thanks and have a nice day . I hope i have provided enough information. Thanks again if you help.
1. There is a problem with your radio element. The name should be the same for both options.
It should be like this:
<input type="radio" name="agree" value="Yes"> or
<input type="radio" name="agree" value="No">
2. You can access everything in the $_POST array, since you are using the method post for the form.
$name = $_POST['myname'];
$selection = $_POST['selection'];
$agree = $_POST['agree'];
3. If you are not using parametrized SQL with a library such as PDO, MySQLi, etc... you must always escape the data, which will be used in query using mysql_real_escape_string(), in order to protect against SQL injection.
This would be a sample code, to do the escaping and the query.
// write a function somewhere, to use as a shortcut
// for escaping data which will be used in a query
function sql_escape($str){
return "'".mysql_real_escape_string($str)."'";
}
// build the query
$query = sprintf('INSERT INTO table_name(name, class, agree) VALUES(%s, %s, %s)',
sql_escape($_POST['myname']),
sql_escape($_POST['selection']),
sql_escape($_POST['agree']));
// finally run it
$result = mysql_query($query);
I've taken it a little further here, there is still plenty more that can be done and many way's to do it, for instance you could extend the $errors array to include a field id and then highlight the HTML form field so the user can see exactly where they went wrong.
Considering your form is fairly simple you would not need this.
#Shef's code would certainly do the job but I thought you might be interested in some more.
<?php
// check the form has been submitted
if (isset($_POST['submit'])) {
// escape the form fields and assign them to variables
// validate myname to ensure the user entered data
if (isset($_POST['myname']) && $_POST['myname']!='') {
$myname = mysql_real_escape_string($_POST['myname']);
} else {
// create an error variable array to store errors to display
$errors[] = 'Please enter your name';
}
// no need to validate selection here as it alway's has a value
$classtype = mysql_real_escape_string($_POST['selection']);
// validate agree unless you want to add 'checked' to one of the values
if (isset($_POST['agree']) && $_POST['agree']!='') {
$agree = mysql_real_escape_string($_POST['agree']);
} else {
$errors[] = 'Please tell us if you agree?';
}
//if errors found tell the user else write and execute the query
if ($errors) {
$message = '<p class="error">We found a problem:</p><ul>';
foreach($error as $msg){
$message .= '<li>'.$msg.'</li>';
}
$message .= '</ul><p>Please fix the error/s to continue.</p>';
} else {
// write the query
$query = "INSERT INTO table (myname, classtype, agree) VALUES ";
$query .= "('$myname','$classtype','$agree')"
// run the query
mysql_query($query);
$message = '<p class="sucessful">Thanks '.htmlspecialchars($myname).'. Your selection has been saved.</p>';
}
}
// print the message
// show the variables in the form field so they don't need re-input
if ($message!='') { echo $message; }
?>
<form action="" method="post" enctype="multipart/form-data">
<strong>Your Name: </strong><br>
<input type="text" name="myname" value="<?php echo htmlspecialchars($myname) ?>" />
<br /><br/>
<strong>Which class type you want:</strong><br>
<select name="selection">
<option value="A"<?php if ($classtype=='A') { echo ' selected'; } ?>>A</option>
<option value="B"<?php if ($classtype=='B') { echo ' selected'; } ?>>B</option>
<option value="C"<?php if ($classtype=='C') { echo ' selected'; } ?>>C</option>
</select>
<strong>Do you agree?</strong><br>
<input type="radio" name="agree" value="Yes"<?php if ($agree=='Yes') { echo ' checked'; } ?>> or
<input type="radio" name="agree" value="No"<?php if ($agree=='No') { echo ' checked'; } ?>>
<input type="submit" name="submit" value="Submit">
</form>
Also: #sqwk, Don't point people towards w3schools, see this: http://w3fools.com/
Check whether there is any data in the $_POST array and get the values from it.
Have a look here—the second example down is what you need: http://www.w3schools.com/php/php_mysql_insert.asp
(You do have to make the changes that Shef suggested, though.)
Also remember to check your data-integrity, otherwise people could use your insert to run malicious code.
check this simple example:
<form action="welcome.php" method="post">
Name: <input type="text" name="name" />
Sname: <input type="text" name="sname" />
<input type="submit" />
</form>
after you submit form, you can take name and sname.
welcome.php::
<?php
$name= $_POST["name"];
$sname= $_POST["sname"]; ?>
now you can use this variables as if you want.
How would I send information from a form to a block of PHP code and then back to a text area? I can not find this answer.
To print out the text you entered in the textfield on the next request would look like this assuming you render the same page (i.e. myform.php):
<?php
$fieldValue = htmlentities($_POST['myfield']);
?>
<form action="myform.php" method="post">
<label for="myfield">Your textfield:</label>
<input type="text" name="myfield" id="myfield" value="<?php echo $fieldValue; ?>" />
</form>
A very basic example.
Assuming you have a PHP file named index.php
<?php
$val = 'Nothing in POST';
if (!empty($_POST)) { //$_POST is where stuff posted from the FORM is saved
$val = isset($_POST['text']) ? $_POST['text'] : '';//you're looking for the data with key text which is the name of your textarea element
$val = 'Got something from POST : ' . $val;
}
?>
<form action='index.php' method='post'>
<textarea name='text'><?php echo $val ?></textarea>
</form>
Have a look at this tutorial for the basics : http://net.tutsplus.com/articles/news/diving-into-php/
use ajax!(and use jquery for that ajax!)
suppose you have this html :
<input type="text" id="input">
<textarea id="result"></textarea>
than the script should be:
$('#input').keypress(function(e){
if(e.wich != 13)//not enter
return;
$.get(your_php_file.php,{param1:val1,param2:val2},function(result){
$('#result').val(result);
});
});
when you hit enter on the input field,the ajax function is being called that requests the file your_php_file.php?param1=val1¶m2=val2. With the php's result, the callback function is being called which updates your textarea