I have wrote the code like this
<?php
echo "Add1:".$_POST['Address1'];
echo "<br>";
echo "Add2:".$_POST['Address2'];
?>
<FORM name="myForm" method='post'>
<table>
<tr>
<td>Address 1</td>
<td><input type="text" name="Address1"></td>
</tr>
<tr>
<td>Address2</td>
<td><input type="text" name="Address2"></td>
</tr>
</table>
<input type="submit">
</FORM>
When enter a value as some thing like this
Address1 = test <test1 and Address2=address2:
But in post i got only the Address1 value in both post variable like.
Add1:test
Add2:test1
Any one can help me.
Debug what you're getting via
<?php print_r($_POST);
and you'll see if it is PHP releated problem or the problem lays somewhere else
Based on your example, it doesn't look like it's not working (Add1 and Add2 are different). What happens if you do a var_dump($_POST)?
If $_POST['Address2'] wasn't filled through the form, it is empty, no output should be displayed.
Try var_dump($_POST); to check the whole POST-array and/or use tools like Firebug, to see which data was sent to the actually.
you can have exact idea by writing this code
<?php
echo "<pre>";print_r($_POST);echo "</pre>";
echo "Add1:".$_POST['Address1'];
echo "<br>";
echo "Add2:".$_POST['Address2'];
?>
that will emable you to check what exactly is coming in $_POST
Related
I have created a simple HTML form containing just one field. When I press submit some PHP code that I have written gets called and outputs text that would include submitted data if everything was working. But no submitted text gets printed by the PHP. The form has been created on a Godaddy HTML page and the form is as follows:
<FORM BORDER="1" action="http://www.bestpro.com.au/wordpress/PHB_action.php"
method="post" enctype="application/x-www-form-urlencoded"
eenctype="multipart/form-data" name="PHBForm" accept-charset="ISO-8859-1"
ienctype="text/plain">
<TABLE>
<TR>
<TD>First name:</TD><TD><INPUT type="text" name="firstname" id="firstname"></TD>
<TD></TD><TD></TD>
<TD> </TD><TD> </TD>
</TR>
<TR>
<TD> </TD><TD> </TD>
<TD> </TD><TD></TD>
<TD> </TD><TD><input type="submit" value="Submit"></TD>
</TABLE>
</FORM>
The PHP code output starts as follows:
This is where we end up.
Using `$_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>");
I have the following link on index page
<td align="center">update</td>
and this is the page it goes to ... Update.php
and here is the code
$id=$_GET['keyword'];
<tr><td> Keyword: </td><td> <input type="text" name="keyword" id="keyword" value="<? echo $id['keyword']; ?>">
I want the keyword to be automatically inserted into the form so that the rest of info can be updated. How can i accomplish this? For some reason this is not working. It is putting the keywork in the URL ?id=KEYWORD on the update page but not displaying in the form. Prob something stupid im sure
You need to either use $_GET['id'] or change the link href to ../athena/admin/update.php?keyword=. Currently, you are not looking for the id parameter, which contains the keyword.
You need to change:
<? echo $id['keyword']; ?>
to:
<?php echo htmlspecialchars($id); ?>
Note:
I used normal php tags instead of short tags, just in case;
I used htmlspecialchars to avoid errors and javascript nastiness; always prepare your data for the medium you are outputting to.
Please check your variable name again.
Here, you pass .php?id= as your GET parameter:
<td align="center">update</td>
In the Update.php, I think you were just misnamed it, it was supposed to be:
$id = $_GET["id"]; //not $_GET["keyword"]
The code below will output all the GET parameter, usually useful to debug in a simple PHP app:
<?php
echo '<pre>';
print_r($_GET);
echo '</pre>';
?>
Hope it helped.
Edit: there is no println in php. Use print_r() or var_dump()
$id=$_GET['keyword'];
<tr><td> Keyword: </td><td> <input type="text" name="keyword" id="keyword" value="<? echo $id; ?>">
May be like this?
I have a table with php values (table.php )retrieve from mysql database.
I want to insert this particular value into another database with the insertion code inside another php file (admin.php)
How should I go about doing it?
Code from table.php
<tr>
<td class="brack_under cell_1">
<?php
$row = mysql_fetch_row($data);
echo "<a>".$row[2]."</a>";
?>
</td>
<td class="cell_2"> </td>
<td class="cell_3"> </td>
<td class="cell_4"> </td>
<td class="cell_5"> </td>
<td class="cell_6"> </td>
</tr>
Code from admin.php
<?php
it is something like : $a = $_POST[ $row[2] ]; ???
$sql="INSERT INTO matchTable (schInitial, schName,position)VALUES
('$_POST[ $row[1] ]','$_POST[$row[2] ]','top8')";
mysql_close($con)
?>
According to the comment discussion and your comments, I'm gonna try to help you a bit here (hopefully I got the point)
You need to POST a <form ...> to your script. Unless you do that, there is no $_POST available. E.g.:
<form method="post" action="myscript.php">
<input type="hidden" name="some_var" value="player1" />
<input type="submit" />
</form>
This gives the following in myscript.php where you can do your filtering and insertion and stuff.
print_r($_POST);
// output: array( 'some_var' => 'player1' );
Which means, according to your tries (and to keep it understandable)
$a = $_POST['some_var']
// $a is now 'player1'
So, the only thing you need to do: create a form beneath your table, put some hidden fields in it with data you need to do your math things, and put a submit button in it.
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">