Post variables from a while loop from a second table in form

[jgreen]

I have 2 tables.  I'm trying post 3 variables in order to insert them in table2. The loop works I see the variables but when I use POST outside the loop I'm not getting the variables.  

I'm using a SELECT to get 3 variables from table1 which I'm getting those variables.  I want to insert these variables in table2 but can't see the input value when looking at the source.

The loop works. I get results in the loop but the input post echo's nothing. I've tried many different ways but can't seem to get it to work. Can someone help?

<?php

include('theconnection.php');

$con = mysqli_connect($host,$user,$pass,$dbName);

if (!$con)

{

die('cannot connect: ' . mysqli_error($con));

}

mysqli_select_db($con,"thebooks");

$result = mysqli_query($con,"SELECT * FROM books");

$result = mysqli_query($con,"SELECT b.id, b.name, b.cover, b.pageno FROM books");

while($row = mysqli_fetch_array($result))

{

echo ("<input type='text' value='$row[name]' name='name' id='name'>");

echo ("<input type='text' value='$row[cover]' name='cover' id='cover'>");

echo ("<input type='text' value='$row[pageno]' name='pageno' id='pageno'>");

}

?>


<input type="hidden" name="name" id="name" value="<?php echo "{$_POST['name']}"; ?>">

<input type="hidden" name="cover" id="cover" value="<?php echo "{$_POST['cover']}"; ?>">

<input type="hidden" name="pageno" id="pageno" value="<?php echo "{$_POST['pageno']}"; ?>">

[Barand]

Forget about your "solution" for now - what is it you want to achieve?

[jgreen]

?  I don't have a solution.  What I'm trying to achieve with the code above is to post the variables you see in my first post and add them to the db.  The variables you see are coming from a different table.  All I need to do is grab those variables and post them.  The hidden inputs are not populating with those variables.  The loop is working <  I see the variables but the inputs not populating. 

[Barand]

So you want to transfer three columns of data from table1 to table 2?

 

Why do you have form involved? They are hidden fields so the user cannot see or edit them?

[jgreen]

Yes three columns of one row of data per selection.  The form is involved because that's where they start.  They fill out the form select a book name and everything with that book name needs to be added  cover, page-no to table2.  Yes they are hidden fields so the user can only see the name.  I have text in the input in the loop for testing.  I did this to see if it was working but name will be the only thing the user will see... the other variables will be hidden.  

[Barand]

A database should hold data in a single place and the only things that are replicated in other tables are key values. So the only item you transfer to table2 should be the book_id. You get the book name etc by referencing the book table via the id value.

[jgreen]

Yep so I guess your not going to look at my code and answer my question.   Can someone look at the code and answer my original question?  It would be greatly appreciated.   

[Barand]

You still only need the id, then when the form is posted

INSERT INTO table2 (book_id, name, cover, pageno) 
SELECT book_id, name, cover, pageno 
FROM table1
WHERE book_id = $posted_id

[akphidelt2007]

Where is the form script that you are using to submit these POST variables?

[jgreen]

I found away that works for the code that I have.  There's one issue.  

 

1.  If there's more than one book title it will echo all book titles for name, cover all cover names, and all pageno for all books.

2.  When the db is populated it does the same as the echo, it populates the columns with all - names, covers, and pageno's for all books instead of the specific book selected.  

 

I'm not sure how to loop through the inputs and print only the book selected and populate the db with what was selected only. I hope that makes  sense.  

<select name="name">
<option value="<?php echo "{$_POST['name']}"; ?>"> </option>                      
       
<?php
include('theconnection.php');
$con = mysqli_connect($host,$user,$pass,$dbName);
if (!$con)
{
die('cannot connect: ' . mysqli_error($con));
}
mysqli_select_db($con,"thebooks");
$result = mysqli_query($con,"SELECT * FROM books");
$result = mysqli_query($con,"SELECT b.id, b.name, b.cover, b.pageno FROM books");

$cover = '';
$pageno = '';

while($row = mysqli_fetch_array($result))
{
echo ("<option value='$row[name]'>$row[name] $row[cover], $row[pageno] </option>"); 

$cover .= "$row[cover]";
$pageno .= "$row[pageno]";

}

?>
                      
</select>

<input type="hidden" name="cover" value="<?php echo $cover; ?>">
<input type="hidden" name="pageno" value="<?php echo $pageno; ?>">

Edited August 13, 2015 by jgreen