ModX custom package xPDO statement not evaluating correctly - php

I'm developing a custom package that accepts a form input, compares it against a table and display the relevant information on the screen, I have the transport and schema setup correctly (I hope! I followed the guide at BobsGuides.com) and I can read back from the table without issue. My problems began when I started implementing the form input.
Basically the session variables I am posting never seem to get recieved so the isset never evaluates to true and I just see the form over and over again.
I've only just begun with PHP and am a complete newb when it comes to xPDO so I accept theres probably more than a few things I havn't noticed but if someone could point me in the right direction I'd be extremely grateful.
<?php
$path = MODX_CORE_PATH . 'components/dataease/';
var_dump($_POST['submit']);
var_dump($_POST['accNo']);
// get POST variable this is captured
$accNo = $_POST['accNo'];
$output = '';
// Check if form has been submitted
if (isset($_POST['submit'])) {
// Get info from the database
$query = $modx->newQuery('accno');
$query->select($modx->getSelectColumns('Dataease','Dataease','',array('*')));
$query->where(array('accNumber:LIKE' => '$accNo'));
var_dump($query);
// Place it into a variable for output
if (!$query) {
return "Query failed";
} else {
$dataease = $modx->getCollection('Dataease',$query);
// Count the returned rows, should only ever be 1
$output .= '<p>Total: '. count($dataease) . '</p>';
// Show the found data
foreach($dataease as $sql) {
$fields = $sql->toArray();
$output .= $modx->getChunk('showData', $fields);
}
}
return $output;
} else {
// Get the form chunk
$form = $modx->getChunk('dataEntryForm');
return $form;
}
This is my form chunk
<h2>Enter Account Number:</h2>
<form method="POST" action="">
<input name="accNo" type="text"/>
<input name="submit" type="submit" value="submit" />
</form>

Give this a try to see if there has been a post submission
if($_SERVER['REQUEST_METHOD'] == "POST")
and/or give your form a hidden submit field, some browsers [at least some used to] not post the submit field if you just hit return when the submit button is not in focus. Test for that as well.

Do you trying http://rtfm.modx.com/display/ADDON/Rowboat for work with your tables?

Related

Form validation with a php array

I'm hoping someone could help me finish off some php code (the avon guy already kindly helped me with this but I'm still struggling with the last bit).
All it is, is I have a form where I have 10 particular sequences of digits, which if entered, allows the form to redirect to the following page. If anything else is entered I want the page to deny access with some kind of error prompt.
At top of the php, in the part before any php is printed, avon guy suggested an array to check the 10 correct sequences against.
$possibles = array('rva858', 'anothersequence', 'andanother');
$match = $_POST['nextpage'];
if (array_search($match, $possibles) != false) {
//match code in here
} else {
// fail code in here
}
I'm not sure what to put in the //match code in here AND the //fail code in here, bits. Can someone help me with this last bit please?
Many thanks
Jon
If you are just trying to redirect to another page using php, you can use header('Location: mypage.php');. More information on header here.
So for your code example (edited based on comment):
invitation.php
<?php
//invitation.php
$possibles = array('rva858', 'anothersequence', 'andanother');
$match = $_POST['nextpage'];
if (array_search($match, $possibles) === false)
{
//If fail
header('Location: formpage.php?errorMessage=Incorrect code!');
exit();
}
//If success:
//All of the invitation.php html and success code below
formpage.php
<?php
//formpage.php
if(!empty($_GET['errorMessage'])){
echo '<span>' . $_GET['errorMessage'] . '</span>';
}
?>
<form action="invitation.php" method="post">
<input name="rsvp" type="text" />
<input type="submit" value="Submit" name="submit" />
</form>

Running a snippet from Form input Modx Revolution

I try to run a form in Modx Revo that the form action="" then runs a PHP snippet which inserts the input name="" into a url, that page is then parsed and inserted into the db and results are to be displayed on the same page as the form. My problem is the form is either not reading the input or the script isn't running.
My form is as follows:
<form action="[[!charAll]]" method="post">Character: <input type="text" name="char" /><br /> Realm:      <input type="text" name="realm" /><br /> <input type="submit" /></form>
My script is too long to post here but will give pieces of it for example.
The 'charAll' snippet:
$realm = $_POST["realm"];
$char = $_POST["char"];
$realm = rawurlencode($realm);
#$mstats = file_get_contents("http://us.battle.net/api/wow/character/$realm/$char? fields=statistics,achievements", true);
$pvp = json_decode($mstats, true);
$link = mysqli_connect("127.0.0.1","dbuser","dbpw","dbname") or die("Error " . mysqli_error($link));
$link->set_charset("utf8");
$mname = $link->real_escape_string($pvp['name']);
$mmrealm = $link->real_escape_string($pvp['realm']);
$mclass = $pvp['class'];
$mrace = $pvp['race'];
$mgender = $pvp['gender'];
$mlvl = $pvp['level'];
$mid = $pvp[""];
Has several more queries but for sake of space I will summarize
It then echo's (or returns) the data into a table structure that I would like to have on the same page but wouldn't mind it echo'ing on another. I know Modx Revo will be great once I get the hang of it but atm it is giving me a headache lol, any help would be greatly appreciated and I thank you in advance.
The form action has to be a document, not a snippet.
One method to do this is to use your form document, put action="[[~[[*id]]]]" on your form.
Check for a posted variable, then call your charAll snippet. Then display whatever you need.

Form to form with PHP

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;}
?>

Retaining values in forms fields when validation of data fails

I am having problems figuring out how to retain users data when the validation fails. I am somewhat new to PHP so I might be making some huge mistakes in my logic.
Currently if the validation fails all the fields are wiped clean and $_Post data is also gone.
Here is some code assuming the user enters an invalid email I want the Name field to be retained. This code is not working.
<?php
if($_POST['doSubmit'] == 'Submit') {
$usr_name = $data['Name'];
$usr_email = $data['Email'];
if (isEmail($usr_email)==FALSE){
$err = "Email is invalid.");
header("Location: index.php?msg=$err");
exit();
}
//do whatever with data
}
if (isset($_GET['msg'])) {
$msg = mysql_real_escape_string($_GET['msg']);
echo "<div class=\"msg\">$msg</div><hr />";
}
if (isset ($_POST['Name'])){
$reusername = $_POST['Name'];}
else{$reusername = "NOTHING";}//to test
?>
<form action="index.php" method="post" >
<input name="UserName" type="text" size="30" value="<?echo $reusername;?>">
<input name="Email" type="text" size="30">
<input name="doSubmit" type="submit" value="submit">
</form>
}
You can use AJAX to submit your form data to your PHP script and have it return JSON data that specifies whether the validation was successful or not. That way, your fields won't be wiped clean.
Another way is to send back the recorded parameters to the posting page, and in the posting page, populate the fields using PHP.
However, I think the first solution is better.
UPDATE
The edit makes your code clearer and so I noticed something. Your input field is called UserName in the HTML, but you are referring to Name in PHP. That's probably why it's not working. Is your field always being filled with the value NOTHING? Make sure the name of the input field and the subscript you are using in $_POST are the same.
Also, there's no need to redirect to another page (using header) if you have an error. Maintain an $errors array or variable to print error messages in the same page. But like I mentioned before, it's probably better to use the JSON approach since then you can separate your view layer (the html) from the PHP (controller layer). So you'd put your HTML in one file, and your PHP in another file.
EDIT:
Vivin had commented that my assumption regarding the header was incorrect and he was right in that. Further more it looks like what the OP is doing is essentially what i layed out below albeit in a less structured fashion. Further Vivin - caught what is likely the actual problem here - the html name and the array key $_POST do not match.
Its wiped clean because you are using header to redirect to another page. Typicaly you would have a single page that validates the data and if ok does something with it and returns a success view of some sort, or that returns an error view directly showing the form again. By using header youre actually redirecting the browser to another page (ie. starting up an entirely new request).
For example:
// myform.php
if(strtolower($_SERVER['REQUEST_METHOD']) == 'get')
{
ob_start();
include('form.inc.php'); // we load the actual view - the html/php file
$content = ob_get_clean();
print $content; // we print the contents of the view to the browser
exit;
}
elseif(strtolower($_SERVER['REQUEST_METHOD']) == 'post')
{
$form = santize($_POST); // clean up the input... htmlentities, date format filters, etc..
if($data = is_valid($form))
{
process_data($data); // this would insert it in the db, or email it, etc..
}
else
{
$errors = get_errors(); // this would get our error messages associated with each form field indexed by the same key as $form
ob_start();
include('form.inc.php'); // we load the actual view - the html/php file
$content = ob_get_clean();
print $content; // we print the contents of the view to the browser
exit;
}
}
so this assumes that your form.inc.php always has the output of error messages coded into it - it just doesnt display them. So in this file you might see something like:
<fieldset>
<label for="item_1">
<?php echo isset($error['item_1']) ? $error['item_1'] : null; ?>
Item 1: <input id="item_1" value="<?php echo $form['item_1'] ?>" />
</label>
</fieldset>
Could do something similar to if failed then value=$_POST['value']
But vivin's answer is best. I don't know much about AJAX and wouldn't be able to manage that.
Ok, firstly header("Location: index.php?msg=$err"); is not really required. It's best practice not to redirect like this on error, but display errors on the same page. Also, redirecting like this means you lose all of the post data in the form so you can never print it back into the inputs.
What you need to do is this:
<input name="Email" type="text" size="30" value="<?php print (!$err && $usr_email ? htmlentities($usr_email, ENT_QUOTES) : '') ?>">
Here I'm checking whether any errors exist, then whether the $usr_email variable is set. If both these conditions are matched the post data is printed in the value attribute of the field.
The reason I'm using the function htmlentities() is because otherwise a user can inject malicious code into the page.
You appear to be processing the post on the same page as your form. This is an OK way to do things and it means you're nearly there. All you have to do is redirect if your validation is successful but not if it fails. Like this
<?php
if( isset( $_POST['number'] ) ) {
$number = $_POST['number'];
// validate
if( $number < 10 ) {
// process it and then;
header('Location: success_page.php');
} else {
$err = 'Your number is too big';
}
} else {
$number = '';
$err = '';
}
?>
<form method="POST">
Enter a number less than 10<br/>
<?php echo $err ?><br/>
<input name="number" value="<?php echo $number ?>"><br/>
<input type="submit">
</form>

Post Back response from PHP to javascript

I'm new to forms and post data ... so I don't know how solve this problem!
I've a php page (page1) with a simple form:
<form method="post" action="/page2.php">
<input type="search" value="E-Mail Address" size="30" name="email" />
<input type="submit" value="Find E-Mail" />
</form>
How you can notice ... this form post the 'email' value to the page2. In the page2 there is a small script that lookup in a database to check if the email address exist.
$email = $_POST['email'];
$resut = mysql_query("SELECT * FROM table WHERE email = $email");
.
.
.
/* do something */
.
.
.
if($result){
//post back yes
}
else{
//post back no
}
I don't know how make the post back in php! And how can I do to the post back data are read from a javascript method that shows an alert reporting the result of the search?
This is only an example of what I'm trying to do, because my page2 make some other actions before the post back.
When I click on the submit button, I'm trying to animate a spinning indicator ... this is the reason that I need to post back to a javascript method! Because the javascript function should stop the animation and pop up the alert with the result of the search!
Very thanks in advance!
I suggest you read up on AJAX.
Here's a PHP example on W3Schools that details an AJAX hit.
Hi i think you can handle it in two ways.
First one is to submit the form, save the data in your session, check the email, redirect
back to your form and display the results and data from session.
Like
session_start();
// store email in session to show it on form after validation
$_SESSION['email'] = $_POST['email'];
// put your result in your session
if ($results) {
$_SESSION['result'] = 'fine';
header(Location: 'yourform.php'); // redirect to your form
}
Now put some php code in your form:
<?php
session_start();
// check if result is fine, if yes do something..
if ($_SESSION['result'] == 'fine) {
echo 'Email is fine..';
} else {
echo 'Wrong Email..';
}
?>
More infos : Sessions & Forms
And in put the email value back in the form field
<input type="search"
value="<?php echo $_SESSION['email']; ?>"
size="30"
name="email" />
Please excuse my english, it is horrible i know ;)
And the other one the ajax thing some answers before mine !
As a sidenote, you definitly should escape your data before using it in an SQL request, to avoid SQL injection
As you are using mysql_* functions, this would be done with one of those :
mysql_escape_string
or mysql_real_escape_string
You would not be able to post in this situation as it is from the server to the client. For more information about POST have a look at this article.
To answer your question you would want to do something like this when you have done your query:
if(mysql_num_rows($result)){ //implies not 0
$data = mysql_fetch_array($result);
print_r($data);
}
else{
//no results found
echo "no results were found";
}
The print_r function is simply printing all the results that the query would have returned, you will probably want to format this using some html. $data is just an array which you can print a single element from like this:
echo $data['email'];
I hope this helps!
<?php
echo " alert('Record Inserted ');"
OR
echo " document.getElementByID('tagname').innerHtml=$result;"
?>
OR
include 'Your Html file name'

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