This is the index.php:
header('Location:media.php?module=home');
I called this media.php:
<html>
<head>
<title>Test media</title>
</head>
<body>
<table width="960" border="0" align="center" cellpadding="0" cellspacing="0">
<tr>
<td colspan="2"><img src="images/cms/header.png" width="780" height="77"></td>
</tr>
<tr>
<td width="200" valign="top" bgcolor="#1e2528">
<?php include "menu.php"; ?>
<p align="center"> </p>
</td>
<td width="760" valign="top" bgcolor="#FFFFFF">
<p>
<?php include "content.php"; ?>
<br>
</p>
</td>
</tr>
</table>
</body>
</html>
And this is the problem, content.php:
<table width="100%" cellspacing=5>
<?php
include_once 'include/config.php';
include_once 'admin/include/date_lib.php';
include_once 'class/class_lib.php';
include_once 'class/paging.php';
$action = new DB();
$action->db_connect();
if ($_GET[module]=='home'){
</td></tr>";
echo "<tr><td align=center>Headline News<br><br></td></tr>";
elseif ($_GET[module]=='request'){
echo "<tr><td class=judul_head>» Hubungi Kami</td></tr>";
echo "<tr><td class=isi>Silahkan hubungi kami secara online:</td></tr>";
echo "<form action='?module=sendemail' method='post'>
<tr><td class=isi>Name : <input type=text name=name size=35></td></tr>
<tr><td class=isi>E-mail : <input type=text name=email size=35></td></tr>
<tr><td class=isi>Subject: <input type=text name=subject size=50></td></tr>
<tr><td class=isi>Message : <br><textarea name=message rows=13 cols=70>
</textarea></td></tr>
<tr><td><input type=submit value=Send></td></tr>
</form>";
echo "<tr><td class=back><br>
[ <a href=javascript:history.go(-1)>Back</a> ]</td></tr>";
}
elseif ($_GET[module]=='sendemail'){
mysql_query("INSERT INTO email(name,
email,
subject,
message,
date)
VALUES('$_GET[name]',
'$_GET[email]',
'$_GET[subject]',
'$_GET[message]',
'$today_date')");
echo "<tr><td class=header_head>» Status Email</td></tr>
<tr><td class=isi>Email has been sent</td></tr>
<tr><td class=back><br>
[ <a href=index.php>Back</a> ]</td></tr>";
}
I have an email form using post method. But when I click the submit button
it will be like this on url address bar
http://staging/media.php?name=Test+name&email=Test+email&subject=test+subject&,message=Test+message
Just like when I use get method. But if change $_POST to $_GET at the query. It doesn't work.
Is there something missing on my script? Or is it because I use the $_GET[module] method to call on same page?
if ($_GET[module]=='home'){
</td></tr>";
It seems to me you're missing echo " statement.
Since you have chosen to have the form method to be set as 'post', the values you're getting from the form in your mysql_query need to be $_POST instead of $_GET. Hoped that helped!
You're also missing a bracket before the elseif ($_GET[module]=='request')
May I recommend a syntax highlighting editor. Becomes very easy to find this stuff.
It's also not good to mix HTML output and script together.
GET and POST are two different things and shouldn't be thought of as interchangeable.
http://php.net/manual/en/reserved.variables.get.php
http://www.php.net/manual/en/reserved.variables.post.php
Using $_GET variables within a PHP script allows you to parse and "get" the URL paramaters (often the query string or folder arguments with mod_rewrite), while $_POST variables allow you to collect the transfered data.
Furthermore, if you are submitting an HTTP request to an API, or your own script, it should be a POST request... because you're submitting data. If you are trying to access an API, or your own script, and only retrieving (not creating/updating/deleting) data you should use a GET request.
when I click the submit button it will be like this on url address bar [...] Just like when I use get method.
Then your form is not correct. Is it the only form on the page (use View Source from your browser) and is it not nested in another form?
Also keep all the hints from the other answers in mind. While they may not immediately solve your problem, it'll become easier to hunt it down when your code is more readable.
Your setting the method of the form to be POST and then in the following code your trying to access the data using $_GET (GET method) which only contains the module=sendmail
To access the data from the submitted form you need to use $_POST (i.e. POST method).
Also the code is not properly written :
Input tag should be like : <input type="value" name="value" />
Form tag should be like : <form method="post" action="http://complete/link/page.php?media=sendmail"></form>
It is a good practice to specify complete URLs for the action attribute.
Related
I have created a simple HTML form containing just one field. When I press submit some PHP code that I have written gets called and outputs text that would include submitted data if everything was working. But no submitted text gets printed by the PHP. The form has been created on a Godaddy HTML page and the form is as follows:
<FORM BORDER="1" action="http://www.bestpro.com.au/wordpress/PHB_action.php"
method="post" enctype="application/x-www-form-urlencoded"
eenctype="multipart/form-data" name="PHBForm" accept-charset="ISO-8859-1"
ienctype="text/plain">
<TABLE>
<TR>
<TD>First name:</TD><TD><INPUT type="text" name="firstname" id="firstname"></TD>
<TD></TD><TD></TD>
<TD> </TD><TD> </TD>
</TR>
<TR>
<TD> </TD><TD> </TD>
<TD> </TD><TD></TD>
<TD> </TD><TD><input type="submit" value="Submit"></TD>
</TABLE>
</FORM>
The PHP code output starts as follows:
This is where we end up.
Using `$_POST["firstname"]` which outputs nothing.
Using `htmlspecialchars($_POST["firstname"])` which also outputs nothing.
Question:
The PHP output doesn't include the value that I entered into the field.
Can anyone see what I am doing incorrectly?
I see nothing wrong here, so I can only assume it is something wrong with how you output it on your PHB_action.php page.
You say that you're placing $_POST['firstname'] on your page, but have you actually made sure to echo or print it to the page?
You can do this like so:
echo $firstname = $_POST['firstname']; // notice the echo placed before
or
$firstname = $_POST['firstname'];
print("$firstname");
EDIT:
I've notice you have put your post data inside of single quotation marks when echoing out to your page.
You must concatenate on your data rather than putting them inside of single quotes when echoing, like so:
echo 'Using' . $_POST['firstname']; // notice the dot in between the string and the post data.
Either that, or you have not installed PHP correctly (or at all) onto your server.
Hope this helps
So, this is pretty straight forward and I have written it up and will explain each bit as i go.
The PHP you need for this is:
<?php
if (isset($_POST['send']))
{
$fname = $_POST['firstName'];
if (!empty($fname))
{
echo "hello $fname";
} else {
echo "Please supply your first name.";
}
}
?>
$_POST['send'] is the name of your submit button, this will be the trigger for your PHP to initiate and run through the rest of the code.
$fname = $_POST['firstName']
This is just where I prefer to store the $_POST as a variable in the event you are going to re use it again it saves time writing the entire thing.
if(!empty)
if the username isn't empty (!empty meaning not empty) then perform the echo of $fname. however if it comes back as empty it will echo the else echo "please supply...;
Now for the form.
<form action="" method="post">
<table>
<tr>
<td>First Name:</td>
<td><input type="text" name="firstName"></td>
</tr>
<tr>
<td><input type="submit" name="send"></td>
</tr>
</table>
</form>
Just a straight forward form with a blank action on mine (I prefer to keep the PHP within the same file however I normally relay it back to a Class within a different file.
Each form input (First Name / Submit) must have a name="" value otherwise the PHP cannot read it and run with it.
I hope this makes sense and isn't too puzzling :)
Your input field should be inside tag and method should be post. Like:
<html>
<body>
<Form method=post>
<input id=mytextfield name=mytextfield type=text />
<input type=submit value=Submit />
</Form>
</body>
</html>
I've been trying to develop a real estate page where people can add listings. I am new to the world of php mysql. I have been over this problem for over a day and can't figure out where the problem is.
I have a form where people can add data. That's good and working. Now I am starting to have a place where people can add / delete / update their info. I am trying to build this step by step.
This is where a user could pull the information. My problem is with the piece of the code:
edit_form.php?idBAR=$row[id].
Full code below.
<table>
<tr>
<td align="center">EDIT DATA</td>
</tr>
<tr>
<td>
<table border="1">
<?php
include"configS_OH.php";//database connection
$order = "SELECT * FROM braaasil_brokerstour.property";
$result = mysql_query($order);
while ($row=mysql_fetch_array($result)){
echo ("<tr><td>$row[id]</td>");
echo ("<td>$row[address]</td>");
echo ("<td>$row[day]</td>");
echo ("<td>$row[hours]</td>");
echo ("<td>Edit</td></tr>");
}
?>
</table>
</td>
</tr>
</table>
Then this tutorial try to pass id through the address bar (I don't know much about php to actually say much)
It tries to upload the data into a new form where a person could edit info.
But I can't load the data into the new form. If I use where id=7, I get the info into the form. But this method of passing the info in the address bar like ?idBAR=8... and then try to catch it in the other code (where id=$idBAR), is not working.
Here is the code:
<table border=1>
<tr>
<td align=center>Form Edit Employees Data</td>
</tr>
<tr>
<td>
<table>
<?php
include "configS_OH.php";//database connection
print $database;
$order = "SELECT * FROM braaasil_brokerstour.property
WHERE id='$idBAR'";
print $idBAR;
$result = mysql_query($order) or die( mysql_error() );
$row = mysql_fetch_array($result);
?>
<form method="post" action="edit_data.php">
<input type="hidden" name="idBAR" value="<?php echo "$row[id]"?>">
<tr>
<td>Address</td>
<td>
<input type="text" name="address"
size="20" value="<?php echo "$row[address]"?>">
</td>
</tr>
<tr>
<td>Date</td>
<td>
<input type="text" name="day" size="40"
value="<?php echo "$row[day]"?>">
</td>
</tr>
<tr>
<td>Time</td>
<td>
<input type="text" name="time" size="40"
value="<?php echo "$row[hours]"?>">
</td>
</tr>
<tr>
<td align="right">
<input type="submit"
name="submit value" value="Edit">
</td>
</tr>
</form>
</table>
</td>
</tr>
</table>
I tried an tried and tried..
Thank you for your time in advance.
WHERE id='$idBAR'
You haven't assigned $idBAR any value. You need to read it from the $_GET array first:
$idBAR = $_GET['idBAR'];
You should, of course, check that this value exists first, and is acceptable.
I don't see anywhere you have actually used the GET data, just the reference name which is used in GET.
If you first query is working and is getting the $row['id'] value ok - you can verify this when you go to edit_form.php, in your browser URL bar at the top, does it say this:
edit_form.php?idBAR=7
(or whatever number should be there)
If so, then you just need to use the PHP GET. Your data is stored in $_GET[], and in this case, the reference name is idBAR. So your id from your previous page query is sent through the link into your URL, and on your edit_form.php page, you'd use that data as:
$_GET['idBAR']
You can use that, but personally I assign the data to a variable, such as:
$strGetId = $_GET['idBAR'];
Then you can use $strGetId throughout your code.
Also, check things like isset(), empty() etc, just so you know you are working with A) something is actually there, and B) it's not empty etc
if you are putting a variable directly in a string without concatenating, it can't be an array variable; you must concatenate those you also need to surr. so this
echo ("<td>Edit</td></tr>");
should be this
echo ("<td>Edit</td></tr>");
also, it looks like your form is sending data with POST. When you pass form data in the url string after the question mark, that is passing with get.
so...in your form where you want to use that variable, you set it up like this
$idBAR=$_GET['idBAR']; //to get the variable if it was part of the URL
$idBAR=$_POST['idBAR']; //if it was sent with post, as is the case with your form
also, request contains both get and post, so
$idBAR=$_REQUEST['idBAR'];
will work in either case.
The problem is the $row[id] is seen as text just like everything else. You want the value of $row[id]. Instead of
echo ("<td>Edit</td></tr>");
try
echo ("<td>Edit</td></tr>");
In the same server, I have a page called mkwars and another called generatedtab. Inside mkwars I have a lot of input fields that contains numeric numbers.I need to transfer the datas from those inputs, to another new inputs located in the page generatedtab.
This is the HTML code:
<table border="0" id="table1" align="center" cellspacing="0" cellpadding="3" width="100%">
<tr>
<td width="50%" valign="top"><b>Home Clan:</b> <input type="text" id="clan1" name="clan1" onchange="nomewar();"/></td>
<td><b>Opponent Clan: </b> <input type="text" id="clan2" name="clan2" onchange="nomewar();"/></td>
</tr>
</table>
//other code
<form method="post" action="savewar.php">
<input type="submit" Value="Generate Table" style="height:70px;width:800px" />
</form>
And here you can see the PHP file:
<?
$percorso = file("war/filedb.txt");
while(list(,$value) = each($percorso)){
list($clan1, $clan2) = split("[:]", $value);
$params["clan1"] = trim($clan1);
$params["clan2"] = trim($clan2);
#print results
echo $params["clan1"]." - ".$params["clan2"]."<br />";
}
?>
war is a folder inside my server. When I click the button Generate Table I can't see the file (war/filedb.txt). Could you help me? I thought that the PHP way was the better, but if you think that I should do something else, tell me.
I'm not exactly clear on what you're trying to do here. I think you want to fill out the html form and have the php script save the new input into a file on the server, and then print out the contents of the file. If that's correct, here are a few things you need to fix.
1) On your html page, the <form> tag must enclose all of the input fields you want to post back to the server. so:
<form method="post" action="savewar.php">
<table border="0" id="table1" align="center" cellspacing="0" cellpadding="3" width="100%">
<tr>
<td width="50%" valign="top"><b>Home Clan:</b> <input type="text" id="clan1" name="clan1" onchange="nomewar();"/></td>
<td><b>Opponent Clan: </b> <input type="text" id="clan2" name="clan2" onchange="nomewar();"/></td>
</tr>
</table>
<input type="submit" Value="Generate Table" style="height:70px;width:800px" />
</form>
2) In your php script, you need to use the superglobal $_POST or $_REQUEST variable to catch the data from the posted form. For example:
$clan1 = $_POST['clan1'];
$clan2 = $_POST['clan2'];
3) In your php script, you need to open the file for writing and append the new data to the end of the file:
$fileappendpointer = fopen("war/filedb.txt",'a');
$newline = $clan1 . " - " . $clan2 . "<br>";
fwrite($fileappendpointer, $newline);
4) Then you can easily read out the contents of the file:
fclose($fileappendpointer);
$filereadpointer = fopen("war/filedb.txt",'r');
$contents = fread($filereadpointer,filesize("war/filedb.txt"));
fclose($filereadpointer);
print $contents;
I'm pretty new to the whole PHP/HTML deal, and I've run into a problem that I don't know how to fix. It's a pretty simple form that lets you enter data into database. The PHP code is as following:
<?
include("../sqlcontext.php");
$foo = mysql_query("SELECT*FROM users WHERE checksum='".$_COOKIE['userCookie']."'");
if($_COOKIE['userCookie'] == '' || !mysql_num_rows($foo)){
echo 'Du er ikke logget ind :( log ind her';
}
else{
if($_POST['genreKnap']){
$nameGenre = $_POST['nameGenre'];
$textGenre = $_POST['textGenre'];
$succes = mysql_query("INSERT INTO genre VALUES('null','$nameGenre','$textGenre')");
if($succes){
echo 'Yay!';
}else {
echo 'Oh no';
}
}
?>
The form is as following:
<form name="form1" method="post" enctype="text/plain" action="">
<table>
<tr>
<td>Genre navn:</td>
<td><input type="text" name="nameGenre" id="nameGenre" style="width:100%; padding-right: 1px" /></td>
</tr>
<tr>
<td>Genre beskrivelse:</td>
<td><textarea name="textGenre" id="textGenre" style="width:100%; padding-right: 1px"></textarea></td>
</tr>
<tr>
<td></td>
<td><input type="submit" name="genreKnap" id="genreKnap" value="Go!"/></td>
</tr>
</table>
</form>
Whenever I press the submit button, it seems as though it acts as if it was a get method and not a post.
Aha!!!
You are not posting the form correctly.
Set the
action=""
to
action="code.php"
Assuming your php page is called code.php. Just change it to the name/path of the php page and the form will send the data to your php code to process.
When you leave action="" to blank, it posts the data to itself (the same page). It is not acting as GET, it is still acting as POST, but posting to the wrong place. I think you worded the title of the question wrong.
What do you mean it is acting like get instead of post.
Can you not read $_POST variables in your PHP?
remove the 'enctype="text/plain"' in your form code.
enctype="text/plain"
Take that out. It is provided for debugging purposes only and doesn't generate anything that is sane to parse with a machine.
Valid form enctypes:
application/x-www-form-urlencoded: This is the default content type
multipart/form-data
The content type "application/x-www-form-urlencoded" is inefficient
for sending large quantities of binary data or text containing
non-ASCII characters. The content type "multipart/form-data" should be
used for submitting forms that contain files, non-ASCII data, and
binary data.
Source: http://www.w3.org/TR/html401/interact/forms.html#h-17.13.3.4
You're all ignoring the primary question and focusing on irrelevent items.
First of all more than anything he's using a short php opener <? not <?php now not every web server accepts short openers first up check that.
Echo out your $_POST vars and see if they're returning the correct items
echo "POSTS BELOW<br />";
echo $_POST['nameGenre']."<br />";
echo $_POST['textGenre']."<br />";
echo "<br />GETS BELOW<br />";
echo $_GET['nameGenre']."<br />";
echo $_GET['textGenre']."<br />";
Put this block of code directly below your php opener see what data it returns.
Also this if($_POST['genreKnap']){ is generally a bad way of doing it as its user input the safest way is a hidden field <input type="hidden" name="action" id="action" value="dopost" /> and change your if clause to if($_POST['action']=="dopost && isset($_POST['action'])){
Also set your form action="" to the actual page name not blank
Give all that a try and if its still not working we'll try something different
If you are send normal data without any files by the form .
Then enctype is not always needed .
But even if you want to include it
The correct way is :
enctype="multipart/form-data"
Also give a url in the action method of the form : <form action='example.php'>
I hope it solves the problem .
I think that there is either an error in my code, or my PHP or Apache is set up incorrectly.
When I submit a form with a hidden field in it, I do not get any data in my $_POST array...
When I comment out the hidden field in my code, the POST data is returned correctly...
HTML FORM
<form action='/utils/login.php ' method='POST'>
<table>
<tr>
<td colspan='2'>
Login
</td>
</tr>
<tr>
<td>
Username
</td>
<td>
<input type='text' name='userid' value='' size='12' />
</td>
</tr>
<tr>
<td>
Password
</td>
<td>
<input type='password' name='password' size='12' />
</td>
</tr>
<tr>
<td>
<input type='hidden' name='formtype' value='login' />
</td>
</tr>
<tr>
<td>
<input type='submit' value='Submit' />
</td>
</tr>
</table></form>
Here is the code that is processing it in PHP...
foreach ($_POST as $var => $value) {
echo "$var = $value<br>";
}
I am using PHP 5 and Apache 2.2 on my server.
Any ideas?
EDIT...
I have narrowed it down to this...
$command = $_POST['formtype'];
When I removed the # sign from my $_POST, I am getting the following error...
Notice: Undefined variable: formtype in C:\webroot\utils\login.php on line 17
If I comment out that line, the POST data is passed into the program without a problem.
I would suggest changing the code you are using to display the contents of $_POST to a single call:
print_r($_POST);
Anytime you are displaying the entire contents of an array, this is better than a loop w/ echo, as it will show every value at every level of the array.
Also, as was mentioned in a comment, make sure you close the form in the html.
You never closed your <form> tag.
And I see now that someone beat me to it by a mile in the comments. Still, this is the right answer.
Have you tried taking the hidden input out of the table and placing it right after the opening form tag?
You can also use:
var_dump($_POST);
...to view the post variables.
Also, if any inputs are being dynamically created or might be missing from the POST variables... you can use:
variable = 'default';
if(isset($_Post['variable'])) $variable = $_POST['variable'];
...to dynamically set variables that could be there or not.
I changed my form to work with Twig. The changed form was not sending the hidden input value with post.
In case someone has the same problem, I solved it by doing the following.
The original line was:
<input hidden name='foo[{{ loop.index }}][id]' value='{{id}}' />
I sold it by making type='hidden':
<input type='hidden' name='foo[{{ loop.index }}][id]' value='{{id}}' />
Please try with:
<form action="..." method="post" enctype="application/x-www-form-urlencoded">