[Last Call] Learn how to a build a cloud-first strategyRegister Now

x
  • Status: Solved
  • Priority: Medium
  • Security: Public
  • Views: 652
  • Last Modified:

dynamic list with checkbox mysql php

Hi - I'm very new to both php and mysql so will need detailed explanation - sorry!
I have created an update page which lists projects from a table in my database. The table has a field called 'live' which i'm trying to use to determine whether the project is to be included on the site or not. I'm using a checkbox nect to each project to allow the user to toggle live between the values 'y' and'n'. I have the list working just fine and i've managed to get the checkbox to show the current value in the database but i'm stumped as to how I then update the database with the chosen value. the field 'live' needs to be sent both 'y' and 'n' values for each project_id. I've tried various ways and think maybe foreach is the way to go but at present have deleted all attemps and am close to giving up. I'd appreciate any help I can get. thanks I've included the code for my form....
0
heather-allen
Asked:
heather-allen
  • 6
  • 4
1 Solution
 
heather-allenAuthor Commented:
Hi - i've upped the points - can anybody help me? Not sure if the code got added so here it is again just in case...

<form method="POST" action="">


<ul id="project_list">

 
   
      <?php // query projects table and print out each item

$result = mysql_query("SELECT project_id, project_name, client, thumbnail, live, project_order FROM projects ORDER BY project_order");



// print the list items

while ($row = mysql_fetch_assoc($result)) {

if($row['live']=="y") {$live="checked=\"checked\"";} else {$live="";}

echo '<li>

<input type="hidden" name="form_order[]" value="'. $row['project_id'] .'" />
<table width="900" border="0" id="projectform">
  <tr>
    <td width="200">'. $row['project_name'] .'</td>
    <td width="200">'. $row['client'] .'</td>
    <td width="200">'. $row['thumbnail'] .'</td>
    <td width="54" align="center"><input type="checkbox" name="live_check[]"'. $live.'value="y"/></td>
    <td width="82"><img src="../images/thumb/'. $row['thumbnail'] .'" alt="'. $row['project_name'] .'" width="82" height="50" /></td>
    <td width="82" align="center"><a href="project_update.php?project_id='. $row['project_id'] .'">EDIT</a></td>
    <td width="82" align="center"><a href="project_delete.php?project_id='. $row['project_id'] .'">DELETE</a></td>
</tr>
</table>
</li>';

}

?>
     
</ul>


<div class="holder">
  <input type="submit" name="reorder" value="update projects" />
</div>

</form>



I just need to know how to post the result to my database for each item whether checked or not.

Thank you
0
 
Richard QuadlingSenior Software DeverloperCommented:
I would recommend using a radio group per project.

An unchecked checkbox sends no data back to the server, so you you will have to do a lot more work to determine what you need to set.

With a radio group per project, you get the appropriate value.

Something like this ...


// Get SQL to provide is_live and is_not_live ...

$result = mysql_query
      (
      "SELECT
            project_id,
            project_name,
            client,
            thumbnail,
            live,
            project_order,
            CASE WHEN live = 'y' THEN ' checked' ELSE '' END AS is_live,
            CASE WHEN live <> 'y' THEN ' checked' ELSE '' END AS is_not_live
      FROM
            projects
      ORDER BY
            project_order"
      );

// Use the radio groups like this.
<input type="radio" name="live_check[{$row['project_id']}]" value="1"{$row['is_live']} />
<input type="radio" name="live_check[{$row['project_id']}]" value="0"{$row['is_not_live']} />



At this stage, I would prove to myself that I understand what I'm getting out of the form.

Using ...

<?php
echo '<pre>';
print_r($_POST);
echo '</pre>';



as the form handler, will allow you to see the values returned from the form.

Then it should be just a matter of iterating $_POST['live_check'] ...



foreach($_POST['live_check'] as $i_ProjectID => $i_ProjectState)
 {
 $s_SQL = 'UPDATE table SET live=' . intval($i_ProjectState) . ' WHERE project_ID = ' . intval($i_ProjectID);

 // Run the query to update the live state.
 }



sort of thing (untested code snippets).
0
 
heather-allenAuthor Commented:
Thanks RQuadling but I'd really like to use a checkbox - I don't mind changing the values to '1' and '0' if that helps - has anyone got any ideas?

Have upped the point value again as this seems to be harder than I thought

thanks
0
Technology Partners: 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!

 
heather-allenAuthor Commented:
Don't worry I have worked out a way to do it myself.

Basically when 'update' is clicked I update all checkboxes to value 'n' and then run this code

if(isset($_POST['update'])) {

$live = $_POST['live_check'];

foreach ($live as $live_id) {

$result = mysql_query("UPDATE projects SET live='y' WHERE project_id=". mysql_real_escape_string($live_id));

}
}


seems to do the trick

:)
0
 
Richard QuadlingSenior Software DeverloperCommented:
Like I said ...

" you will have to do a lot more work to determine what you need to set."

How about this instead ...

$s_Lives = implode(', ', $_POST['live_check']);
$result = mysql_query("UPDATE projects SET live='y' WHERE project_id in (" . mysql_real_escape_string($s_Lives) . ")");

1 SQL statement rather than x number.

Or ...

$s_Lives = implode(', ', $_POST['live_check']);
$result = mysql_query("UPDATE projects SET live= CASE WHEN project_id IN (" . mysql_real_escape_string($s_Lives) . ") THEN 'y' ELSE 'n' END");


No need to turn them all to no first. 1 SQL command sets ALL the projects in 1 go, either to 'y' or 'n'.


0
 
heather-allenAuthor Commented:
That makes more sense but I tried it (version 2 as need to update to 'n' as well) and if there are no projects checked it throws up the error

Notice: Undefined index: live_check in /blah/blah
Warning: implode() [function.implode]: Invalid arguments passed in /blah/blah

Any ideas?

:)
0
 
Richard QuadlingSenior Software DeverloperCommented:
Yes. Oops.

$_POST['live_check'][] = -1;
$s_Lives = implode(', ', $_POST['live_check']);
$result = mysql_query("UPDATE projects SET live= CASE WHEN project_id IN (" . mysql_real_escape_string($s_Lives) . ") THEN 'y' ELSE 'n' END");

I assume that your id are automatically created and positive.
0
 
heather-allenAuthor Commented:
yes

fantastic - works like a dream!

thanks for all your help...

:)
0
 
heather-allenAuthor Commented:
although I got it working your way was far better! thanks so much :)
0
 
Richard QuadlingSenior Software DeverloperCommented:
Refactoring and improving the code is often much more fun than the original development.

Just make sure you add adequate comments so that this level of shortcutting doesn't catch you out in years to come.

And thank you for the points.
0

Featured Post

Restore individual SQL databases with ease

Veeam Explorer for Microsoft SQL Server delivers an easy-to-use, wizard-driven interface for restoring your databases from a backup. No expert SQL background required. Web interface provides a complete view of all available SQL databases to simplify the recovery of lost database

  • 6
  • 4
Tackle projects and never again get stuck behind a technical roadblock.
Join Now