I am learning PHP from a book called PHP and MySQL web development. I am a newbie to PHP , I am pretty well versed with C and HTML, I find syntax of PHP to be pretty same of C.
Look at the code:
<html>
<form action = "processorder.php" method = "post">
<p>tires:<input type=”text” name=”tireqty” size=”3” maxlength=”3” /></p>
<input type=”submit” value=”Submit Order” /></td>
</form>
</html>
This is the HTML code now I will type in the php code and save it as processorder.php
<?php
$tireqty = $_POST[‘tireqty’];
echo "<p>Your order is as follows: </p>";
$tireqty." tires<br />";
?>
After doing this , i go to the html page where I have to enter the value for tires and click submit after doing this it redirects to me mu processorder php page where my raw php code gets displayed and not the proper output which is:
Your order is as follows:
2 tires
But i get a different o/p which is as shown below:
$tireqty = $_POST[‘tireqty’];
echo "<p>Your order is as follows: </p>";
$tireqty." tires<br />";
what is going on here what is wrong??? and i also get an error message is processorder script saying that:
parse error: undefined index 'tireqty' in line something and undefined var tireqty
Hey guys i dont know whats wrong but cant post comments that button is not working or my browser is not enabling me to write comments, I am answering to all the calls here:
Yeah I am using XAMPP what is wrong in that, what should i do?
I copied your codes and yes there are no errors in it but when i go to html page and click submit the browser opens a dialog box saying that "I have chosen to open process.php" and what should it do with it so i select open with gedit because that is the only default option available there, and it redirects to my php script page :(:(:(
WHat is this it is so tough, Is PHP really that difficult or am i working on a poor platform and learning from a bad book?
Is there any good book on php
<?php
$tireqty = $_POST[‘tireqty’];
echo "<p>Your order is as follows: </p>";
$tireqty." tires<br />";
?>
You closed your echo tags in the first 2nd row already, therefor you need to add another echo tag to the 3rde line.
<?php
$tireqty = $_POST[‘tireqty’];
echo "<p>Your order is as follows: </p>";
echo $tireqty." tires<br />";
?>
Always make sure that your variables have been set before you use them:
<?php
if(isset($_POST['tireqty'])){
$tireqty = $_POST['tireqty'];
echo "<p>Your order is as follows: <br />";
echo $tireqty." tires.</p>";
}
?>
Also, your HTML had a random closing table column tag in it:
<html>
<form action = "processorder.php" method = "post">
<p>tires:<input type="text" name="tireqty" size="3" maxlength="3" /></p>
<input type="submit" value="Submit Order" />
</form>
</html>
Sounds like your PHP environment is not working as it should. Do you have a web server with PHP running or XAMP (http://www.apachefriends.org/en/xampp.html)?
Always use
if ( isset($_POST['var']) ) { ... }
Same with $_GET, $_SESSION etc. You variable tireqty isn't initialized until form is submitted, that's why throws errors
Related
I'm learning MySQL and PHP and got a problem with the input of the form. So I wrote a small test code, but it still cannot work. My PHP version is 5.6.
The code:
<html>
<body>
<form action ="2.php" method ="post">
Name: <input type="text" name="username" />
<input type ="submit" value="ok" />
</form>
</body>
</html>
and
<html>
<?php
if(isset($_POST['username'])){
$user=$_POST['username'];
echo $user;
echo " is your name";
}
else{
$user=null;
echo "error";
}
?>
</html>
The output of the project is always error, can't output the input before.
I tried single quote and double quote for username, both can't work.
I also tried to set always_populate_raw_post_data in php.ini to 0, -1, 1, all can't work.
I don't know where the problem is, though it might be very silly.
As what it look it is correct and should run without any problem. Make sure the above code is what you actually have. From my experience most of the form submission can be
you don't have correct name (username)
you might send incorrect http verb (post)
you submit to wrong endpoint (2.php)
From you code above everything look fine. if you still don't have the right result, you better debug it with var_dump, or print_r function with these built-in
$_POST, $_GET, $_REQUEST and check whether they contains you form variable name username
You are using isset as a variable, but it is a function that returns a boolean.
Change $user=isset($_POST['username']); to $user=$_POST['username'];
Another thing is that in both case you will end up in the IF condition even if there is no value added to the field so you can do something like this too:
<html>
<?php
if(isset($_POST['username']) && !empty($_POST['username'])){
$user=$_POST['username'];
echo $user;
echo " is your name";
}
else{
$user=null;
echo "error";
}
?>
</html>
I'm redesigning a site that's mainly static content, so a CMS is really not necessary. The only thing that changes is the 'events' page, which my client frequently updates. He does this now by going into the HTML, copying the tags of a previous event, changing whatever needs to be changed and uploading it back to the server. I want to make this easier for him (it's a pain in the a**, as he puts it), but without using a CMS I'm kind of lost.
Is there a way to have a form he can fill in (with place, date, etc.) and then display it on the proper page on submit? I'm thinking it should be possible with PHP but I don't know how to do it.
DEMO
The cleanest way to do this if you have PHP available is to set up a form that uses PHP to write to a text file, and then subsequent form posts would overwrite the same file. This is a very basic form created with the idea that the user would be the administrator. if this form was user facing to the public you would want toimplement a little more security.
You can format the output in the php script to match the site as needed.
Use this to read in the txt file on the page php file to display the text file:
<?php
readfile("Post.txt");
?>
HTML Form
<h1>Event Post </h1>
<form name="blogs" action="eventpost.php" method="post" enctype="multipart/form-data">
<label for="titlePost">Post Title </label>
<input type="text" name="titlePost">
<label for="commentPost">Comment: </label>
<textarea type="text" name="commentPost" rows="5" cols="35"></textarea>
<input type="submit" name="submitPost" width="200px" value="Submit"/>
</form>
PHP Script
<?php
global $output;
$title = $_POST['titlePost'];
$comment = $_POST['commentPost'];
$tagDate = date('l, M d, Y');
$content = "<div><h2>$title</h2><span class=\"dateStamp\"> $tagDate</span><br><br><span>$comment</span>\n</div><hr>\n\r\r";
$file = "Post.txt";
if($_POST['titlePost'] = !"" && $_POST['commentPost'] != ""){
if (isset($_POST['submitPost'])){
if (file_put_contents($file, $content) > 0){
$output = "The post titled <b>$title</b> was accepted. Here is what was posted:<br><br>$comment<hr><br>";
} else{
$output = "<em>Unfortunately ".$title."</em> did not post appropriately.";
}
} else {
$output = "Your form is not filled out <u>completely.</u>";
}
echo "<span>".$output."</span>";
}
?>
Set it all up like this:
<?
$EventIs = "Event name"
$EventDate = "date"
//etc...
?>
in your HTML:
<p><span class="eventName">Event: <? echo $EventIs ?></span><br />
<span class="eventDate">Date: <? echo $EventDate ?></span></p>
That's really oversimplifying it. But it gives you the basic idea. You guy can just edit the variables at the top of the file, and they will appear wherever you want in the code.
If you want these variables set from a post from a form, you will have to save those variables in your database. But that's the basic idea.
From a high level perspective, you could create a form for him to fill out which would, on the back end, store the form contents into a file or database. Then, on the front end, you would read in that file, or database, parse the content, and display it however you like. That's a very common thing to do using PHP.
It seems my code is correct, however the posted variables in the form will not echo in the update user settings page in the form. I have echoed the posted ids from the input in the database but I cannot get the variables to show.
I have done this in Codeigniter fine but am trying to do it in pure php with my own framework.
$users = new Users($db); comes from my init.php file that is called at the beginning of the file below.
when I
<?php var_dump($user['first_name'])?>
I get Null
<input type="text" name="first_name" value="<?php if (isset($_POST['first_name']) )
{echo htmlentities(strip_tags($_POST['first_name']));} else { echo
$user['first_name']; }?>">
Hoi Stephen,
Try print_r($_POST["first_name"]); instead of var_dump();
or just for all:
print_r($_POST);
best regards ....
add this at the top of your html page
#extract($_REQUEST);
and put is just to check and after checking remove the below line
print_r($_REQUEST);
hope this help .
I'm trying to make a simple captcha in PHP, but it does not work. The query is not currently executing. This is my current code:
<?php
$Random = rand(1, 100);
$Random2 = rand(1,100);
echo "Result: ".$Random." + ".$Random2." ?";
?>
<input type="text" name="r_input"/><br />
$Cap = mysql_real_escape_string($_POST['r_input']);
$Result = $Random+$Random2;
if(isset($_POST['myButton']) and trim($Var) and trim($Var2) and trim($Var3) and $Cap==$Result){
//My Query
}
When you use rand() to generate 2 values, and show those 2 values, and give the form for the user to enter the answer, ...
... the user enters the answer and submits back to the server ...
... the server gets the answer, and then GENERATES 2 NEW VALUES, that don't correspond to the answer given by the user.
Try using session variables to store the generated values in, and match against when the user submits the form!
<?php
session_start();
$captcha_id = 'captcha_' . rand();
$_SESSION['$captcha_id']['val1'] = rand(1,1000);
$_SESSION['$captcha_id']['val2'] = rand(1,1000);
echo "
<form action='' method='post'>
<p>Result: {$_SESSION['$captcha_id']['val1']} + {$_SESSION['$captcha_id']['val2']} = </p>
<input type='hidden' name='captcha_id' value='{$captcha_id}' />
<input type='text' name='captcha_answer' />
<p>?</p>
</form>
";
if (
isset($_POST['captcha_id'])
&& isset($_SESSION[$_POST['captcha_id']])
&& isset($_POST['captcha_answer'])
&& $_SESSION[$_POST['captcha_id']]['val1'] + $_SESSION[$_POST['captcha_id']]['val2'] == intval($_POST['captcha_answer'])
) {
unset($_SESSION[$_POST['captcha_id']]); // don't let this answer be reused anymore.
// do allowed stuff
}
?>
Because $Random and $Random2 have a different value each time.
When you show the form for the first time, they may have the values $Random = 12 and $Random2 = 26. The User sees those, adds them up correctly and types in 38 (which is the correct answer for those two values). The answer is sent to the script again, the values of $Random and $Random2 are generated again (this time as $Random = 23 and $Random2 = 30 which equals 53) and the answer the user has sent is not correct any more.
So you would need to store those values in hidden fields and add these up, instead of the generated ones, like so:
<input type="hidden" name="rand_1" value="<?php echo $Random; ?>">
<input type="hidden" name="rand_2" value="<?php echo $Random2; ?>">
<?php
if ($_POST['rand_1'] + $_POST['rand_2'] == $_POST['r_input']) {
// Query etc.
EDIT: As suggested by #nl-x you should use the Session variables instead of hidden fields to prevent abuse of the captcha:
<?php
$Random = $_SESSION['rand_1'] = rand(1, 100);
$Random2 = $_SESSION['rand_2'] = rand(1,100);
echo "Result: ".$Random." + ".$Random2." ?";
?>
And check those values against the given result afterwards:
<?php
$Cap = mysql_real_escape_string($_POST['r_input']);
$Result = $_SESSION['rand_1'] + $_SESSION['rand_2'];
if ($Result == $Cap) {
// ...
You never re-enter PHP mode after you output your form field:
<input type="text" name="r_input"/><br />
<?php // <----this is missing
$Cap = mysql_real_escape_string($_POST['r_input']);
Pardon me, but you are not making a real captcha. The purpose of the captcha is to distinguish the human from the bots. I would highly suggest you to pick a image database, and randomize a function to call a image. Internally, i would check if the text/description of the image matches with what the user typed.
The only thing you will rand() is what image to load from your image database.
That's a not-healthy way to do it, and there are plenty of better ways to do this. But it's more closer to a captcha than just your current code.
There is also a lot of libraries and engines that can do the job for you.
I'm not a pro at PHP, or even programming at all, but i think you're going to the wrong side - your code won't block any... malicious actions at all, or whatever kind of action that you will try to prevent with the captcha.
Search google for the libraries. PhpCaptcha is one of them. And here is a very simple quickstart guide for phpcaptcha.
Here's a code example, extracted from PHPCaptch that I linked above.
At the desired position in your form, add the following code to display the CAPTCHA image:
<img id="captcha" src="/securimage/securimage_show.php" alt="CAPTCHA Image" />
Next, add the following HTML code to create a text input box:
<input type="text" name="captcha_code" size="10" maxlength="6" />
[ Different Image ]
On the very first line of the form processor, add the following code:
<?php session_start(); ?>
The following php code should be integrated into the script that processes your form and should be placed where error checking is done. It is recommended to place it after any error checking and only attempt to validate the captha code if no other form errors occured. It should also be within tags.
include_once $_SERVER['DOCUMENT_ROOT'] . '/securimage/securimage.php';
$securimage = new Securimage();
This includes the file that contains the Securimage source code and creates a new Securimage object that is responsible for creating, managing and validating captcha codes.
Next we will check to see if the code typed by the user was entered correctly.
if ($securimage->check($_POST['captcha_code']) == false) {
// the code was incorrect
// you should handle the error so that the form processor doesn't continue
// or you can use the following code if there is no validation or you do not know how
echo "The security code entered was incorrect.<br /><br />";
echo "Please go <a href='javascript:history.go(-1)'>back</a> and try again.";
exit;
}
Following the directions above should get Securimage working with minimal effort.
This code is included here as well.
Good luck!
I have researched many places to find an answer to this question, but they never quite answer my real question: What is the best/approved way to move to a new page within the same website? I have read that it is bad to use window.location because search engines will think you are hiding something. But, when I don't want to open a new window (window.open), then I don't know how else to do it. I use href anchors in links and form actions, where appropriate. But when I have menus or buttons with onclick, then I need something else.
Here's an snippet of my code:
my javascript: (with one option commented)
function gotoCat() {
var catcdF = document.catSelect.catcd.value;
<?php
echo "window.location.href='http://www.mysite.org".$pgmdir."services/busMenu.php?catF='+catcdF; ";
/*
echo "window.open('http://www.mysite.org".$pgmdir."services/busMenu.php?catF='+catcdF,'','resizable=1,scrollbars=1,toolbar=1,top=50,left=300,width=950,height=800,location=0'); ";
*/
?>
}
My dynamic SELECT list in a form (within PHP):
echo " <select name='catcd' id='catcd' size='8' onclick=gotoCat() > \n";
// display list of categories
if ($numcats == 0) { // print message text only
echo "<option value='0' >".$catMsg."</option> \n";
}
else {
for ($i=1; $i<=$numcats; $i++) {
$catcd_db = $catAry[$i][1];
$catName_db = $catAry[$i][2];
echo "<option value='".$catcd_db."'> ".$catName_db." </option> \n";
}
}
echo "</select>";
So, as you can see, I just want a method to allow the user a choice and then automatically go to the correct web page once selected. This is not always in a select list. Often it's when they want to exit or get an error:
if (mysqli_connect_errno()) {
echo "<br/> <p style='text-align:center;'> <button type='button'
class='buttonStyle' style='padding: 4px 20px;' value='Exit' ";
echo "onClick=\"window.location.href='http://www.mysite.org/services/catSelbus.php?rc=1&func=Rev'\" > ";
echo "Exit </button></p> ";
}
I cannot use "go back" because they need to go to a prior page, not the form they came from.
So, unless my navigation methods are really off-the-mark, I guess I need to know the acceptable method for using javascript onClick to move to the next page in the same website. Is window.location okay, or should I use something else?
Any opinions or suggestions are welcome!
To navigate to another page using Javascript, use:
window.location.href = "url";
That's how it's done and there's nothing wrong about it.
For the sake of argument, you could create a hidden link and simulate a click on it, but as I said, there's really no need.
You can use php header('location') instead:
<form action="submit.php">
<input type="hidden" value="test" name="hidden1" />
<input type="submit" Value="Exit" ... />
submit.php
<?php
if (isset($_POST['hidden1'])
{
header('Location: http://www.mysite.org/services/catSelbus.php?rc=1&func=Rev');
exit;
}
?>
More info about header('Location ...');:
http://php.net/manual/en/function.header.php
Instead of a hidden, you use your select's value and get it via the $_POST variable.