Go Premium for a chance to win a PS4. Enter to Win

x
?
Solved

Query is updating my DB using only the last row of data in a dynamically created form?

Posted on 2011-09-19
6
Medium Priority
?
351 Views
Last Modified: 2013-12-13
Experts,

I have a dynamically created form which contains a table.  The rows of this table contain information pulled from the database.  Each row of the table has been given a unique id using the 'ID' filed contained within the DB.  Additionally, each row of the table contains two dropdown menus which I would like my users to be able to update.

The problem is that when processing this form via a foreach loop, the database is only updated using the very last row of data and whatever has been selected using this row's dropdown menus.

Therefore, changes that I may make to row 5 are overwritten by the changes in the last row, row 7 upon form processing.

Here is my code which identifies via a foreach loop, the ID of each row, and attempts to process the dropdown data from each row.

Any suggestions on how I can straighten this out?  Cheers!

This is an abbreviate version of my form:

echo'<td>';
echo'<select name="expected" id="expected" style="font-size:10px">';
echo'<option value="'.$result3['expected'].'">'.$result3['expected'].'</option>';
echo'<option value="Yet to Discuss">Yet to Discuss</option>';
echo'<option value="Undecided">Undecided</option>';
echo'<option value="Plans to Work">Plans to Work</option>';
echo'<option value="Will Not be Working">Will Not be Working</option>';
echo'</select>';
echo'</td>';
echo'<td>';
echo'<select name="coding" id="coding" style="font-size:10px">';
echo'<option value="'.$result3['coding'].'">'.$result3['coding'].'</option>';
echo'<option value="LOA">LOA</option>';
echo'<option value="Sick">Sick</option>';
echo'<option value="Sick - CESLA">Sick - CESLA</option>';
echo'<option value="Sick - FMLA">Sick - FMLA</option>';
echo'<option value="No Call No Show">No Call No Show</option>';
echo'<option value="Worked">Worked</option>';
echo'</select>';
echo'</td>';
echo'<td><textarea name="notes" id="notes" cols="15" rows="1" style="font-size:10px;">'.$result3['notes'].'</textarea></td>';
echo'<input name="row_number[]" type="hidden" value="'.$result3['ID'].'" />';


This is the portion of my form processing script which should save the data to each row of the database based on the unique ID for each row of the table:

//THE NAME OF DROPDOWN 1
$expected = $_POST['expected'];

//THE NAME OF DROPDOWN 2
$coding = $_POST['coding'];

//THE NAME OF TEXTFIELD 1
$notes = $_POST['notes'];

//THE UNIQUE ID OF EACH TABLE ROW
$row_number = $_POST[row_number];

foreach($_POST[row_number] as $row_number){
      $update_row      = "UPDATE schedule SET expected='$expected', coding='$coding', notes='$notes' WHERE ID='$row_number'";
      $update_row_query = mysql_query($update_row) or die ("Could not run query: ".$update_row."<br />\n".mysql_error());
      echo $update_row;
}
0
Comment
Question by:evibesmusic
  • 3
  • 2
6 Comments
 
LVL 8

Accepted Solution

by:
PranjalShah earned 2000 total points
ID: 36563511
First suggestion I would make looking at your code is try to write the html tags only in html rather than using php echo. It will be difficult for you to debug.

I believe your dropdown code is under the while loop of your query. Here, you need to have a unique name for each of the drop downs for each of the row. So what you can do is change your dropdown names to,

<select name="expected<? echo $result3['ID'];?>" id="expected<? echo $result3['ID'];?>" style="font-size:10px">

And now in your foreach loop change your query to,

$update_row = "UPDATE schedule SET expected='".$_post['expected'.$row_number]."' WHERE ID='".$row_number."'"

Do the same thing for other dropdowns
0
 
LVL 1

Expert Comment

by:jastacdoss
ID: 36563520
In short your foreach is only looping once because $_POST[row_number] is not an array. You also need to identify each field using a unique id...
$row_count = 0;

// LOOP THROUGH RESULTS AND PRINT THE FORM
echo '<td><textarea name=" . $row_count . '|notes'" cols="15"></td>';
echo '<input name="' . $row_count  . '|row_number" type="hidden" value="' . $result3['ID'] . '">';

// COUNT NUMBER OF ROWS FOR THE LOOP
$row_count++; 

// OUTSIDE OF YOUR LOOP ADD THIS INPUT
echo '<input name="row_count" type="hidden" value="' . $row_count . '">';

Open in new window


then your for loop
for ($i = 0; $i < $_POST['row_count']; $i++) {

// SET YOUR VARIABLES
$expected = $i . "|expected";
$id = $i . "|row_number";

// ABBREVIATED NEW UPDATE STATEMENT
$update_row  = "UPDATE schedule SET expected='$_POST[$expected]' WHERE ID='$_POST[$id]'";
}

Open in new window

0
 

Author Comment

by:evibesmusic
ID: 36563917
@PranjalShah:

If the following: $update_row = "UPDATE schedule SET expected='".$_post['expected'.$row_number]."' WHERE ID='".$row_number."'"

Supposed to be: $update_row = "UPDATE schedule SET expected='".$_POST['expected'.$row_number]."' WHERE ID='".$row_number."'"
0
Independent Software Vendors: We Want Your Opinion

We value your feedback.

Take our survey and automatically be enter to win anyone of the following:
Yeti Cooler, Amazon eGift Card, and Movie eGift Card!

 
LVL 8

Expert Comment

by:PranjalShah
ID: 36563950
yes, the $_POST should be in uppercase. And try echoing your queries before executing it so that you get the idea of what is being posted.
0
 

Author Comment

by:evibesmusic
ID: 36564275
@PranjalShah:

Thank you very much.  Your solution worked.  Here come the points.
0
 

Author Closing Comment

by:evibesmusic
ID: 36564276
Thanks!
0

Featured Post

New feature and membership benefit!

New feature! Upgrade and increase expert visibility of your issues with Priority Questions.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

These days socially coordinated efforts have turned into a critical requirement for enterprises.
This article discusses how to implement server side field validation and display customized error messages to the client.
The viewer will learn how to count occurrences of each item in an array.
The viewer will learn how to look for a specific file type in a local or remote server directory using PHP.
Suggested Courses

876 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question