*Display values of the form inside the table on the new page

sharapov

I am generating data from database and displaying it inside input text boxes (for users to make some changes). What I need to do is when user presses "submit" button display all of the results inside the table on the new opened page. How can I do it?

Here is the example of what I am trying to do without actual data and php: http://home.pacbell.net/sharapov/form

ir4z0r

hmm, did you get it to work? it looks okay for me now

sharapov

ir4z0r,

Thank you for your reply. The example on my siite is just pure HTML. There is no PHP there at all. I just did to show what I want my PHP script to do. Can you help?

P.S. May be I shouldn't mention that example. I keep confusing people with it! 😃

ir4z0r

well, if I got you right, you have sth like that on the 1st page:

<?PHP
  ...
  // DB query which gets all data needed
  ...
  $row = mysql_fetch_array($result);
  ...
  echo '<input type=text name=field1 value="' . $row['column1'] . '">';
  ...
?>

on the 2nd page, you would then just fill the table with $POST-values

<?PHP
  ...
  echo '<td>' . $_POST['field1'] . '</td>';
  ...
?>

sharapov

ir4z0r,

How can I fill table with $_POST-values ? I can't reference the field, because the fields on the first page are generated dynamically.

On my second page I am getting all the values from $POST. Now I need to take all those values and distibute them among the cells in the table, and that is where I get stuck. If I simly "echo" values from the $POST, I get something like: Some Data 1, Some Data 2, Some Data 3, etc. Now I need to take all that loop through it somehow and display the data inside cells of the table.

ir4z0r

okay here's my suggestions how you get all passed variables, it's
just outputting the names and values so far:

<html><body>
<?PHP
  $keys = array_keys($_POST);

  for ($i=0; $i < sizeof($_POST); $i++)
  {
    echo $keys[$i] . "=" . $_POST[$keys[$i]] . "<br>";
  }

?>
</body>
</html>

first line gets all names (keys) which are contained in $POST,
in the loop, these names are used as keys for the $POST-array

sharapov

OK. I got that far, except I don't really need the keys just the values.

Let's just say I have a variable $myvalues = "Some Data 1, Some Data 2, Some Data 3, etc."

How can I take that, loop through the values insert it inside the table similar to the one inn my example that I mentioned in my first post.

ir4z0r

you have all values in one varibale? 😕

<?PHP

  ...
  $values = explode(",", $myvalues);
  for ($i=0; $i < sizeof($values); $i++) $values[$i] = trim($values[$i]); // remove spaces before and after value

  $entries_per_row = 6;

  for ($i=0; $i < sizeof($values); $i++)
  {
    if (($i % $entries_per_row) == 0) echo "<tr>"; // new row
    echo "<td>" . $values[$i] . "</td>";
    if ((($i+1) % $entries_per_row) == 0) echo "</tr>"; // close row
  }
  ...
?>

this way?

sharapov

Originally posted by ir4z0r
you have all values in one varibale? 😕

How else can get all of the values if I get them from $_POST.

In any case below is an example of that what I did, and it looks like it works fine. Except one thing (well actually 2 but the first one is much more important then the second 🙂 ). As you can see from my example the second column of my first page has a field with a value "noimage.jpg". When I pass this to my script I dont't want to show the value, instead I want to insert "<img src=" then the value of that field. This will allow me too display the image itself. I guess I will need another loop here. Can you help.

The second question is not that crusial. I woould want to open my second page with the resulting table in the new window with just a menu bar. I know I can do it with java script, but if I do that will I be able to grab all the values from the $_post there.

In any case if you could help me with the first one I'll be more than happy! 😃

<?
echo"<table>";

if(count($_POST) > 0) {
foreach($_POST as $key=>$value)
{ $value_array[] = $value; }

$values = implode(",",$value_array);
}

$values = explode(",",$values);

$total = count($value_array);

  for ($i=0; $i < $total; $i++);

  $entries_per_row = 7;

  for ($i=0; $i < $total; $i++)
  {
    if (($i % $entries_per_row) == 0) echo "<tr>"; // new row
    echo "<td>" . $values[$i] . "</td>";
    if ((($i+1) % $entries_per_row) == 0) echo "</tr>"; // close row
  }


echo"</table>";

ir4z0r

do you really know what you're doing here😃

$values = implode(",",$value_array);
$values = explode(",",$values);

at first you put all values together in one string, after that
you divide it back to an array, this is crap
you're already builing the array in the foreach loop

anyway, try this for the image issue:

   $entries_per_row = 7;
  $images = array(1);  // contains all columns which are a picture, 
                                   // starting with 0
                                   // multiple values separated by ,

  for ($i=0; $i < $total; $i++)
  {
    if (($i % $entries_per_row) == 0) echo "<tr>"; // new row
    if (in_array($i, $images)) echo "<td><img src=\"" . $values[$i] . "\"></td>";
    else echo "<td>" . $values[$i] . "</td>";
    if ((($i+1) % $entries_per_row) == 0) echo "</tr>"; // close row
  }

untested, but should work
sorry for your 2nd quest., I have only basic javascript knowledge

sharapov

do you really know what you're doing here😃

$values = implode(",",$value_array);
$values = explode(",",$values);

at first you put all values together in one string, after that
you divide it back to an array, this is crap
you're already builing the array in the foreach loop

No, not really! 😃 How can I fix it? What I thought I was doing is: 1. Taking all elements in array and insering "," between them. 2. Taking comma delimited string and divide by segment.

I guess I was wrong!

In any case. I tried to use your exapmle where you inserting the image, and the result turned out to be not what I expected it would be. The image only shows on the first row. Third and fourth rows display "noimages.jpg" instead of the actual image. In other words if you look at the source code of the loaded page it shows "<img src="noimages.jpg">" only on the first row, the rest of the rows don't have "<img src="">"

I propose other solution to this problem. Below is my version. What do you think?

<?
echo"<table>";

if(count($_POST) > 0) {
foreach($_POST as $key=>$value)
{ $value_array[] = $value; }

$values = implode(",",$value_array);
}

$values = explode(",",$values);

$total = count($value_array);

  for ($i=0; $i < $total; $i++);

  $entries_per_row = 7;

  for ($i=0; $i < $total; $i++)
  {
    if (($i % $entries_per_row) == 0) echo "<tr>"; // new row
    echo "<td>";

if (file_exists($values[$i])) {
	echo"<img src=\"";
	echo $values[$i];
	echo"\"";
} else {
    echo $values[$i];
}



echo"</td>";

if ((($i+1) % $entries_per_row) == 0) echo "</tr>"; // close row
  }


echo"</table>";

?>

ir4z0r

uhh right, I didn't thought till the end :rolleyes:
so either use your version, which checks for file existance, too
or my "extended version" 😉

for ($i=0; $i < $total; $i++)
  {
    if (($i % $entries_per_row) == 0) echo "<tr>"; // new row
    if (in_array($i%$entries_per_row, $images))
    {
       if (file_exists($values[$i]) echo "<td><img src=\"" . $values[$i] . "\"></td>";
       else echo "<td>image (" . $values[$i] . ") not found.</td>";
    }
    else echo "<td>" . $values[$i] . "</td>";
    if ((($i+1) % $entries_per_row) == 0) echo "</tr>"; // close row
  }

sharapov

ir4z0r,

Thank you very much. Your method works now. Last question. should I do anything with the following? How can I change implode/explode to use it correctly?

if(count($_POST) > 0) {
foreach($_POST as $key=>$value)
{ $value_array[] = $value; }

$values = implode(",",$value_array);
}

$values = explode(",",$values);

ir4z0r

hehe, let's take a closer look:

you have a number, say 17, if you negate it, it's -17, if you
negate it again, it's 17 like in the beginning

long words, short sense: just leave 'em out, you need an array, you
get an array, just rename $value_array[] into $values[]
this should do the trick 🙂