Form, array and mysql help

clickernono

I am a newbie to php and mysql and am trying to set up forms for user input that would link responses. I thought it might be a pretty standard code, but I have not been able to find it anywhere. I apologize if it’s out there somewhere and I just don’t know what I’m looking for, but here is my problem.

I have 2 tables in a mysql database, one with registered user information that has an autonumber field called “ID” and another table that has visitor comments and an “ID” field (the two tables are, of course, connected by that ID field).

Visitors search for a certain first name from the registered user mysql table and the results are returned for them in an html table. This part works fine:
Form page:

<form action="searchresults.php" method="post" id="formstyle">
<input type="text" name="fname" size="25"
<br />
<input type="submit" value="Search" />
</form>

Searchresults.php

<?php  

$con = mysql_connect("localhost","user","password");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("advisor1_advisor", $con);

$term = $_POST['fname'];  

$result = mysql_query("select * from advisor where firstname like '%$term%'");

echo "<h3>Your search &quot;" . $term . "&quot; returned the following results:</h3>";

echo "<table border='1'>
<tr>
<th>Firstname</th>
<th>City</th>
<th>State</th>
</tr>";

while($row = mysql_fetch_array($result))
  {
  echo "<tr>";
  echo "<td name='Firstname' align='center'>" . $row['firstname'] . "</td>";
  echo "<td name='City' align='center'>" . $row['city'] . "</td>";
  echo "<td name='State' align='center'>" . $row['state'] . "</td>";
  echo "<td name='ID' type='hidden'>". $row['ID'] . "<td>";
  echo "</tr>";
  }
echo "</table>";

mysql_close($con);
?>

I then have a form field later in this searchresults.php page for visitors to provide feedback.

Share your feedback:<br />
<form action="comments.php" method="post" id="formstyle">
<textarea cols="20" name="comments" rows="2"></textarea><br />

I would like “comments.php” to pull the ID from the html table, attach the “comments” from the form field and put it into the visitor comments table described above. Also, I would like the “ID” in the html table to be hidden, and the “hidden” type does not do that.

Is this possible?

Thank you in advance for your assistance. I do not do this full time, so I may not be able to respond to questions immediately.

Pikachu2000

You are echoing the value outside of the tag.
echo "<td name='ID' type='hidden'>". $row['ID'] . "<td>";
Should be

echo "<td><input name=\"ID\" type=\"hidden\" value=\"" . $row['ID'] . "\"></td>";

Actually, what is the reason all the <td> tags have a name= ? Name isn't even a valid attribute for a <td> tag.

clickernono

Thank you. That worked perfect!

The reason for the names is my biggest problem is to pull the results from this hidden ID field and insert it into a mysql table with the value from the form field submitted by the user. I thought maybe by giving the ID field a name I could identify it for the insert with the form field. Any idea how to do that?

I appreciate the help.

Pikachu2000

Whatever value is in the <input type="hidden" field will be available (after the form has been submitted) as an element of the $POST array, with its name="" value as the name of the index. So, in this case, the value would be available in $POST['ID'].

clickernono

Thank you for the response, but I can’t get that to work. I’ve tried the following script, but only the “comments” field passes to the mysql table. Here is the form code
<form action="comments.php" method="post" id="formstyle">
<textarea cols="20" name="comments" rows="2"></textarea><br />
<br />
<input name="Submit1" type="submit" value="Submit" /></form>

And here is the comments.php script:
$sql="INSERT INTO advisor (VID, Comments)
VALUES
('$POST[ID]','$POST[comments]')";

Again, the form field (comments) passes every time but the ID field does not.

Thank you.

Pikachu2000

You don't have a field named ID in that form. It can't pass what isn't there . . .

clickernono

How do I get the ID from the previous table into the form. As indicated above, this is the table where the ID is located (with the fix that you indicated):
while($row = mysql_fetch_array($result))
{
echo "<td><input name=\"ID\" type=\"hidden\" value=\"" . $row['ID'] . "\"></td>";
}

Each value in the ID result from this table (which is an array) needs to have the "comment" value submitted by the form assigned to it in the mysql table. I tried entering the code you supplied, above, to the form, but it does not pass the ID value to the mysql table.

Thank you so much for hanging in there with me. I truly appreciate your help.

Pikachu2000

In order to give you an accurate answer, I'm going to need more info. How does this db insert form sequence flow? How many separate forms are you using, in what order? This is getting too broken up to decipher, so if you can post the code for each one in its own [noparse]

[/noparse] BBCode block (without passwords, email address, etc.) it would make this a lot easier to work with.

clickernono

No problem.

This is a picture of what I am trying to do:

Your search “Florida” returned the following results:

Date From Date To ID (hidden column)
4/12/2010 5/6/2010 12
6/3/2010 6/20/2010 22
1/10/2011 3/6/2011 87

Share your feedback:
            Congratulations and good luck!	

Submit

I want the information submitted from the “Share your feedback” form field (comments) to be posted to the feedback mysql table with each hidden ID value. So in this case the feedback mysql table would receive the following values:

VID Comments
12 Congratulations and good luck!
22 Congratulations and good luck!
87 Congratulations and good luck!

There are 2 mysql tables with the following fields:
member – ID, Location, DateFrom, DateTo
feedback – ID, VID, Comments
The ID field in Member and the VID field in Feedback are connected so that the comments in the feedback table are related to a specific member in the member table. I can get the comments into the feedback table, but I cannot get the VID into it (i.e., the ID from the member table).

A user searches the member mysql table for members located in a certain city (Location). After submission, the user is taken to the searchresults.php page which returns the results and provides a form field (comments) for the user to provide feedback to the members in that location:

searchresults.php

<?php  

$con = mysql_connect("localhost","","");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("", $con);

$term = $_POST['location'];  

$result = mysql_query("select * from member where Location like '%$term%'");

echo "<h3>Your search &quot;" . $term . "&quot; returned the following results:</h3>";

echo "<table border='1'>
<tr>
<th>From Date</th>
<th>To Date</th>
</tr>";

while($row = mysql_fetch_array($result))
  {
  echo "<tr>";
  echo "<td align='center'>" . $row['DateFrom'] . "</td>";
  echo "<td align='center'>" . $row['DateTo'] . "</td>";
  echo "<td><input name=\"ID\" type=\"hidden\" value=\"" . $row['ID'] . "\"></td>"; 
  echo "</tr>";
  }
echo "</table>";

mysql_close($con);
?>

Right below this code on the same searchresults.php page is the following form for the user to complete:

<form action="comments.php" method="post" id="formstyle">
<textarea cols="20" name="comments" rows="2"></textarea><br />
<br />
<input name="Submit1" type="submit" value="Submit" /></form>

The form is processed by comments.php which submits the results to the Feedback mysql table:
comments.php

<?php

$con = mysql_connect("localhost","","");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("", $con);

$sql="INSERT INTO feedback (VID, Comments)
VALUES
('$_POST[ID]','$_POST[comments]')";

if (!mysql_query($sql,$con))
  {
  die('Error: ' . mysql_error());
  }
header("Location: post_advice.htm");

mysql_close($con)

?>

I hope that helps.