I have been given some spaghetti code for a login page to fix.
We've got two input fields ID and Password.
Here's what I've been asked.
So in terms of a user inputting their ID, I want to add '#email.com' onto the end.
Im assume placeholder="#email.com" would work if i could align it to the right, but I also need to it to be added into the POST method. So if the user entered 'ID123' it would post 'ID123#email.com'
Here is the form:
<form action="command.php" formmethod="post">
<div class="ID">User ID:<br><input name="ID" type="text"><br></div>
<div class="Pass"> Password:<br><input name="pwd" type="password"></div>
<input id="submit" type="submit" value="Sign In">
Can anyone help? Is it even possible?
Placeholder in the input type in HTML will just give a hint in a textbox and it will not be added into the input of the user. What you need to do is catch the input type in the command.php you created and add it.
for example, in the command.php
$ID = $_POST['ID'].'#gmail.com'
When you submit the form, your POST data will contain:
$_POST['ID'] = 'test_id'; // Sample Data
$_POST['pwd'] = 'test_pwd'; // Sample Data
So, if you want to add #email.com at the end of the ID, you could simply concatenate the field:
$_POST['ID'] .= '#email.com';
Note: This is a simple solution and doesn't specify SQL injection/vulnerability prevention.
I think you want something like if anyone give ID123#email.com as input it will not change but if only give ID123 then it will be ID123#email.com,then :-
$userId=$_POST['ID'];
if(!filter_var($userId, FILTER_VALIDATE_EMAIL)) {
$userId=$userId.'#email.com';
}
echo $userId;
Related
I created a form by RSForm. I have two textbox in it.
First textbox name is km1 (new_km) and second textbox name is km2 (old_km).
In the first time, user will fill km1 field (new_km) by her car kilometer number.
When the user recourse again and fill the form, km2 (old_km) must shown value that user entered it in previous recourse.
Please guide me?
Best regards.
Right, I know, my English is very bad. sorry.
Like this image: http://persianupload.com/kleeja/do.php?imgf=141839395854871.jpg
I personally use session for that...
Your html:
<?php
session_start();
if(!isset($_SESSION['something'])) $_SESSION['something'] = '';
?>
<input type="text" name="something" value="<?php $_SESSION['something']?>"/>
Your receiver php:
<php
session_start();
if($_POST['something']) $_SESSION['something'] = $_POST['something'];
?>
This is more of a technique question rather than maybe code. I am having a php form with many fields (items to select). Naturally some of the items might be selected and some not. How do I know which ones are selected when i post the data from page 1 to page 2? I thought of testing each one if empty or not, but there are just too many fields and it doesn't feel at all efficient to use or code.
Thanks,
UPDATE EDIT:
I've tried the following and maybe it will get me somewhere before I carry on testing the repliers solutions...
<html>
<body>
<form name="test" id="name" action="testprocess.php" method="POST">
<input type="text" name="choices[shirt]">
<input type="text" name="choices[pants]">
<input type="text" name="choices[tie]">
<input type="text" name="choices[socks]">
<input type="submit" value="submit data" />
</form>
</body>
</html>
and then second page:
<?php
$names = $_POST['choices'];
echo "Names are: <br>";
print_r($names);
?>
This gives out the following:
Names are: Array ( [shirt] => sdjalskdjlk [pants] => lkjlkjlk [tie]
=> jlk [socks] => lkjlkjl )
Now what I am going to try to do is iterate over the array, and since the values in my case are numbers, I will just check which of the fields are > 0 given the default is 0. I hope this works...if not then I will let you know :)
I think what you're looking for is this:
<form action="submit.php" method="POST">
<input type="checkbox" name="checkboxes[]" value="this" /> This
<input type="checkbox" name="checkboxes[]" value="might" /> might
<input type="checkbox" name="checkboxes[]" value="work" /> work
<input type="submit" />
</form>
And then in submit.php, you simply write:
<?php
foreach($_POST['checkboxes'] as $value) {
echo "{$value} was checked!";
}
?>
The square brackets in the name of the checkbox elements tell PHP to put all elements with this name into the same array, in this case $_POST['checkboxes'], though you could call the checkboxes anything you like, of course.
You should post your code so we would better understand what you want to do.
But from what I understood you are making a form with check boxes. If you want to see if the check boxes are selected, you can go like this:
if(!$_POST['checkbox1'] && !$_POST['checkbox2'] && !$_POST['checkbox3'])
This looks if all the three check boxes are empty.
Just an idea:
Create a hidden input field within your form with no value. Whenever any of the forms fields is filled/selected, you add the name attribute of that field in this hidden field (Field names are saved with a comma separator).
On doing a POST, you can read this variable and only those fields present in this have been selected/filled in the form.
Hope this helps.
Try this.....
<?php
function checkvalue($val) {
if($val != "") return true;
else return false;
}
if(isset($_POST['submit'])) {
$values = array_filter(($_POST), "checkvalue");
$set_values = array_keys($values);
}
?>
In this manner you can get all the values that has been set in an array..
I'm not exactly sure to understand your intention. I assume that you have multiple form fields you'd like to part into different Web pages (e.g. a typical survey form).
If this is the case use sessions to store the different data of your forms until the "final submit button" (e.g. on the last page) has been pressed.
How do I know which ones are selected when i post the data from page 1 to page 2?
is a different question from how to avoid a large POST to PHP.
Assuming this is a table of data...
Just update everything regardless (if you've got the primary / unique keys set correctly)
Use Ajax to update individual rows as they are changed at the front end
Use Javascript to set a flag within each row when the data in that row is modified
Or store a representation of the existing data for each row as a hidden field for the row, on submission e.g.
print "<form....><table>\n";
foreach ($row as $id=>$r) {
print "<tr><td><input type='hidden' name='prev[$id]' value='"
. md5(serialize($r)) . "'>...
}
...at the receiving end...
foreach ($_POST['prev'] as $id=>$prev) {
$sent_back=array( /* the field values in the row */ );
if (md5(serialize($sent_back)) != $prev) {
// data has changed
update_record($id, $sent_back);
}
}
Users of my website can generate a custom form. All the fields are saved in a database with a unique ID. When someone visits the form, the fields 'name' attribute is field*ID*, for example
<p>Your favorite band? <input type="text" name="field28"></p>
<p>Your Favorite color? <input type="text" name="field30"></p>
After submitting the form, I use php to validate the form, but I don't know retrieve the value of $_POST[field28] (or whatever number the field has).
<?
while($field = $query_formfields->fetch(PDO::FETCH_ASSOC))
{
$id = $field[id];
//this doesn't work!!
$user_input = $_POST[field$id];
//validation comes here
}
?>
If anybody can help me out, it's really appreciated!
Add some quotes:
$user_input = $_POST["field$id"];
I'd suggest taking advantage of PHP's array syntax for forms:
<input type="text' name="field[28]" />
You can access this in php with $_GET['field'][28]
$user_input = $_POST['field'.$id];
Remember that you are using a string for the first part of the input name, so try something like: $user_input=$_POST['field'.$id];.
Also, I would suggest calling them into an array to retrieve all data:
<?php
$user_inputs=array();
while($field=$query_formfields->fetch(PDO::FETCH_ASSOC)) {
$id=$field['id'];
$user_inputs[]=$_POST['field'.$id];
}
?>
I am trying to create a multi steps form where user will fill the form on page1.php and by submitting can go to page2.php to the next 'form'. What would be the easiest way?
Here is my code:
<?php
if ($_SERVER['REQUEST_METHOD'] == 'POST') {
?>
<form id="pdf" method="post">
New project name:<input type="text" name="pr_name" placeholder="new project name..."><br/>
New project end date:<input id="datepicker" type="text" name="pr_end" placeholder="yyyy-mm-dd..."><br/>
<textarea class="ckeditor" name="pagecontent" id="pagecontent"></textarea>
<?php
if ($_POST["pr_name"]!="")
{
// data collection
$prname = $_POST["pr_name"];
$prend = $_POST["pr_end"];
$prmenu = "pdf";
$prcontent = $_POST["pagecontent"];
//SQL INSERT with error checking for test
$stmt = $pdo->prepare("INSERT INTO projects (prname, enddate, sel, content) VALUES(?,?,?,?)");
if (!$stmt) echo "\nPDO::errorInfo():\n";
$stmt->execute(array($prname,$prend, $prmenu, $prcontent));
}
// somehow I need to check this
if (data inserted ok) {
header("Location: pr-pdf2.php");
}
}
$sbmt_caption = "continue ->";
?>
<input id="submitButton" name="submit_name" type="submit" value="<?php echo $sbmt_caption?>"/>
</form>
I have changed following Marc advise, but I don't know how to check if the SQL INSERT was OK.
Could give someone give me some hint on this?
thanks in advance
Andras
the solution as I could not answer to my question (timed out:):
Here is my final code, can be a little bit simple but it works and there are possibilities to check and upgrade later. Thanks to everyone especially Marc.
<form id="pdf" method="post" action="pr-pdf1.php">
New project name:<input type="text" name="pr_name" placeholder="new project name..."><br/>
Email subject:<input type="text" name="pr_subject" placeholder="must be filled..."><br/>
New project end date:<input id="datepicker" type="text" name="pr_end" placeholder="yyyy-mm-dd..."><br/>
<textarea class="ckeditor" name="pagecontent" id="pagecontent"></textarea>
<?php
include_once "ckeditor/ckeditor.php";
$CKEditor = new CKEditor();
$CKEditor->basePath = 'ckeditor/';
// Set global configuration (will be used by all instances of CKEditor).
$CKEditor->config['width'] = 600;
// Change default textarea attributes
$CKEditor->textareaAttributes = array(“cols” => 80, “rows” => 10);
$CKEditor->replace("pagecontent");
if ($_SERVER['REQUEST_METHOD'] == 'POST')
{
// data collection
$prname = $_POST["pr_name"];
$prsubject = $_POST["pr_subject"];
$prend = $_POST["pr_end"];
$prmenu = "pdf";
$prcontent = $_POST["pagecontent"];
//SQL INSERT with error checking for test
$stmt = $pdo->prepare("INSERT INTO projects (prname, subject, enddate, sel, content) VALUES(?,?,?,?,?)");
// error checking
if (!$stmt) echo "\nPDO::errorInfo():\n";
// SQL command check...
if ($stmt->execute(array($prname, $prsubject, $prend, $prmenu, $prcontent))){
header("Location: pr-pdf2.php");
}
else{
echo"Try again because of the SQL INSERT failing...";
};
}
$sbmt_caption = "continue ->";
?>
<input id="submitButton" name="submit_name" type="submit" value="<?php echo $sbmt_caption?>"/>
</form>
Add the attribute action with the url you'd like to go to. In this case it'd be
<form id="pdf" method="post" action="page2.php">
EDIT: i missed you saying this method doesn't work. What part of it doesn't work?
You should keep the action to the same script, so the POST action is still performed and then redirect with header("Location: page2.php"); when the processing is done.
A basic structure like this will do it:
form1.php:
<?php
if ($_SERVER['REQUEST_METHOD'] == 'POST') {
... process form data here ...
if (form data ok) {
... insert into database ...
}
if (data inserted ok) {
header("Location: form2.php");
}
}
?>
... display page #1 form here ...
And then the same basic structure for each subsequent page. Always submit the form back to the page it came from, and redirect to the next page if everything's ok.
You're probably better off separating the php code from the form. Put the php code in a file called submit.php, set the form action equal to submit.php, and then add the line header('Location: whateverurl.com'); to your code.
The easiest way is to post it to form2.php by giving the form the attribute action="page2.php". But there's a risk in that. It means that form2 must parse the posted data of form1. Also, if the data is wrong (verification) form1 must be shown instead of form2. This will make your code over complicated and creates dependencies between the two forms.
So the better solution (and quite easy as well) is to implement the post-redirect-get pattern.
You post to form1, verify all data and store it. If the data is ok, you redirect to form2. If the data is wrong, you just show form1 again.
Redirecting is done by a header:
// Officially you'll need a full url in this header, but relative paths
// are accepted by all browsers.
header('Location: form2.php');
Save already posted fields in hidden input fields, but don't forget to validate them every time user submits another step of the form as the user may change hidden inputs in source code.
<input type="hidden" name"some_name" value="submitted_value"/>
There are several ways handling the submitted data while jumping between steps.
You will find your reasons for /against writing data to session, database, whatever... after each step or not.
I did following approach:
The form includes always a complete set of input elements, but on page #1 the step-2-elements are hidden ... and other way round.
I built a 6-step-wizard this way. One large template, some JS /Ajax for validating input, additional hidden inputs that hold current step-ID and PHP deciding, which fields to show or hide.
The benfit in my opinion: Data can easily be saved completely, as soon as input is alright and complete. No garbage handling, if users abort after step 1.
I would store it all in a session array (or sub array)
a really rough example where I'm saving all the form names to an array (to be checked later of course):
<?
foreach($_POST as $k => $v){
$session['register'][$k]=$v;}
?>
I've got the following php code printing out the contents of a SQL table.
$query="select * from TABLE";
$rt=mysql_query($query);
echo mysql_error();
mysql_close();
?>
<i>Name, Message, Type, Lat, Lng, File </i><br/><br/>
<?php
while($nt=mysql_fetch_array($rt)){
if($nt[name] != null){
echo "$nt[id] $nt[name] $nt[message] $nt[type] $nt[lat] $nt[lng] $nt[file]";
}
}
?>
How would I implement a button so for each "row" if the button is clicked on that row it'll submit the information of that row to another php file?
I want it looking something like...
details details2 details3 BUTTON
details4 details5 details6 BUTTON
details7 details8 details9 BUTTON
details10 details11 details12 BUTTON
Where if BUTTON was hit on row 1 details1,2,3 would be sent to a php file, on row 2 detals 4,5,6 would be sent etc. How would I do this?
Thanks.
it's going to be something like that, depending on the data you need to send:
while($nt = mysql_fetch_array($rt)) {
if($nt[name] != null){
echo "$nt[id] $nt[name] $nt[message] $nt[type] $nt[lat] $nt[lng] $nt[file] ".'send request<br/>';
}
}
You can either use GET method and send a query string to the second php page and receive the variables there, like
next.php?variable1=value1&variable2=value2&...
or use POST method by making a hidden form for each row and assign a hidden field for each variable you want to send.
<form method="post" action"next.php">
<input type="hidden" name="variable1" value="value1" />
<input type="hidden" name="variable2" value="value2" />
.
.
.
</form>
or instead of sending all the values, just send the row ID (if any) using any of these two methods and run another query in next.php to get the information you need from database.
Instead of submitting the entire data, just send the ID and fetch the results from the database in the other script. If you want to have an input button, you can do
<form action="/other-script.php" method="GET">
<?php printf('<input type="submit" name="id" value="%s" />', $nt["id"]); ?>
</form>
but you could also just add a link, e.g.
printf('Submit ID', $nt["id"]);
If you really want to send the entire row values over again, you have to make them into form inputs. In that case, I'd send them via POST though.