PHP - Moved: Onchange

Good day, I new to PHP I am having problems with a two dropdowns on a form, can someone please tell me where I'm going wrong.


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Hey guys
I wasnt sure if this is a php topic or java script
im trying to make my drop down menu's activate onchange. what confuses me is how do I make php know that the java script is being run. Or have the java script trip my php code.

To know a button has been pressed i use "if (isset['_POST'])"


 echo '<select name="siege_list" id="siege_list">';

foreach($name as $key => $value) {        

echo '<option value="' . $value['name'] . '" ' . ($value['name'] == $current1
['name'] ? 'selected="selected"' : '') . '> ' . $value['name'] . '</options>';
}                           
echo '</select><input type="submit" id="siege_planet" onchange="this.form.submit()">';    
         

Hello:

I have a SELECT dropdown menu that works like this:
Code: [Select]
<?php

$states_arr = array('AL'=>"Alabama",'AK'=>"Alaska",'AZ'=>"Arizona",'AR'=>"Arkansas",'CA'=>"California",'CO'=>"Colorado",'CT'=>"Connecticut",'DE'=>"Delaware",'DC'=>"District Of Columbia",'FL'=>"Florida",'GA'=>"Georgia",'HI'=>"Hawaii",'ID'=>"Idaho",'IL'=>"Illinois", 'IN'=>"Indiana", 'IA'=>"Iowa",  'KS'=>"Kansas",'KY'=>"Kentucky",'LA'=>"Louisiana",'ME'=>"Maine",'MD'=>"Maryland", 'MA'=>"Massachusetts",'MI'=>"Michigan",'MN'=>"Minnesota",'MS'=>"Mississippi",'MO'=>"Missouri",'MT'=>"Montana",'NE'=>"Nebraska",'NV'=>"Nevada",'NH'=>"New Hampshire",'NJ'=>"New Jersey",'NM'=>"New Mexico",'NY'=>"New York",'NC'=>"North Carolina",'ND'=>"North Dakota",'OH'=>"Ohio",'OK'=>"Oklahoma", 'OR'=>"Oregon",'PA'=>"Pennsylvania",'RI'=>"Rhode Island",'SC'=>"South Carolina",'SD'=>"South Dakota",'TN'=>"Tennessee",'TX'=>"Texas",'UT'=>"Utah",'VT'=>"Vermont",'VA'=>"Virginia",'WA'=>"Washington",'WV'=>"West Virginia",'WI'=>"Wisconsin",'WY'=>"Wyoming");
    
    function showOptionsDrop($array){
        $string = '';
        foreach($array as $k => $v){
            $string .= '<option value="'.$k.'"'.$s.'>'.$v.'</option>'."\n";
        }
        return $string;
    }
?>

<form>

<select name="states">
    <option value="0">Choose a state</option>
    <?php echo showOptionsDrop($states_arr); ?>
</select>


</form>

I would like to show all the towns/zipcodes for each individual state with an "onchange" from the SELECT menu.

The table is called "zip_codes" and the City/Zip Codes are called "city" and "zip"

What I would like to do is display them as links, so a user can click into each town and either create a new entry (a restaurant) or update an existing entry.

What would be the best approach for this?

I appreciate the help.

How can I make my on change in the form pass through the year thats been selected, then use that in the second query?

Code: [Select]
<select name="mySelect" onchange="">
<?php $result= mysql_query('SELECT DISTINCT Year FROM MonthlySales'); ?>
        <?php while($row= mysql_fetch_assoc($result)) { ?>
            <option value="<?php echo htmlspecialchars($row['Year']);?>">
                <?php echo htmlspecialchars($row['Year']); ?>
            </option>
        <?php } ?>
  </select>
       
    <?php

$query="SELECT * FROM MonthlySales WHERE Year = PASS YEAR SELECTED VARIABLE TO THIS BIT HERE";
$results = mysql_query ( $query, $conn);

Hey phpFreaks, im having some troubles getting my script to work correctly and im also not sure if this issue is in the right section of the forum. but heres what i have going on.
I have a query result that displays a list of images with a checkbox and a couple buttons for edit and delete. everything works fine other than the checkbox stuff. I had it working when i was using a submit button, but i wanted to get rid of that button cuz it was only dealing with the checkboxes. so whats going on when using the checkboxes is that when checked or uncheck it would submit the form. It works to submit but its not submitting any data to the database.

heres what i got for code for this section of checkbox.
Code: [Select]
<?php
echo '<form method="post">';
if(isset($_POST['submit'])){
foreach($_POST['id'] as $id){
$value = (isset($_POST['location'][$id]) && $_POST['location'][$id]=="0" ? '0' : '1');
$insert = mysql_query("UPDATE items SET location='$value' WHERE id='$id'") or die('Insert Error: '.mysql_error());
}
}
     $result = mysql_query("SELECT * FROM items")
or die("Query Failed: ".mysql_error());
$counter = 0;
echo '<div class="specialcontainer">';
 while($row = mysql_fetch_array($result)){
list($id, $item_info, $item_img, $price, $sale, $location) = $row;
if($location == '0'){
$set_checked = 'checked="checked"';
}else{
$set_checked = '';
}
if($counter % 5==0) {
echo '</div>';
echo '<div class="specialcontainer">';
}
echo '<div class="special"><img src="../images/items/'.$item_img.'" width="130" /><br />'.$item_info.'<br />$'.$price.'<br />$'.$sale.'<br />Slide Show: <input type="checkbox" id='.$id.' value="0" name="location['.$id.']" '.$set_checked.'  onchange="this.form.submit()"/><br /><input type="button" value="Edit" name="edit" id="'.$id.'" onclick="window.location.href=\'specials.php?action=edit&id='.$id.'\'"><input type="button" value="Delete" name="Delete" id="'.$id.'" onclick="window.location.href=\'specials.php?action=delete&id='.$id.'\'"><input type="hidden" name="id[]" value='.$id.' /></div>';
$counter++;
 }
 echo '</div>';
 echo '</form>';
?>

Hi there,

I am having a PHP form with simple combo box. Here is the code:

<form name="form1" method="get" action="">
  <select name="select" onChange="form1.submit()">
    <option value='value1'>My Value1</option>
    <option value='value2'>My Value2</option>
    <option value='value3'>My Value3</option>
  </select>
</form>

Now the form is successfully being submitted upon 'onChange' event. The only thing I am trying to do is to retain the value of the new option. As every time I select the value from the combo, it goes back and displays the 'value 1' even when I select value 3 or 2.

How do I make the combo box retain the value as 3 (when I select option 3) after the form is submitted.

Please reply.

Thank you!

Hi.

Need some help on one simple task yet so hard. I have puzzled on this for days and no luck. I have a great script i got from somewhere.

Code: [Select]
function get_time_difference( $start, $end )
{
    $uts['start']      =    strtotime( $start );
    $uts['end']        =    strtotime( $end );
    if( $uts['start']!==-1 && $uts['end']!==-1 )
    {
        if( $uts['end'] >= $uts['start'] )
        {
            $diff    =    $uts['end'] - $uts['start'];
            if( $days=intval((floor($diff/86400))) )
                $diff = $diff % 86400;
            if( $hours=intval((floor($diff/3600))) )
                $diff = $diff % 3600;
            if( $minutes=intval((floor($diff/60))) )
                $diff = $diff % 60;
            $diff    =    intval( $diff );           
            return( array('days'=>$days, 'hours'=>$hours, 'minutes'=>$minutes, 'seconds'=>$diff) );
        }
        else
        {
            trigger_error( "Ending date/time is earlier than the start date/time", E_USER_WARNING );
        }
    }
    else
    {
        trigger_error( "Invalid date/time data detected", E_USER_WARNING );
    }
    return( false );
}
Now the other part is that i have 3 textboxes and i woul like to add the first one like 15:00:00 the second one 18:30:00 and the third would display the difference in
lets say seconds. At least the function displays the results in seconds. But the problem is how to get the textbox to work whitout posting the info but to use onchange

Any ideas?
Code: [Select]
<input type="text" value="" name="start_time"/>
<input type="text" value="" name="end_time"/>
<input type="text" value="______" name="difference"/>

Hi everyone.

I have a combo box which lists usernames and onchange, the username value is passed to a textbox. however, I have 3 textboxes i need to populate based on the selection of the combobox: username. department and email. i have the username going into a textbox, but i'm not sure how to pass department and email into two other textboxes.

I'd appreciate any help you could provide. Thanks.

Code: [Select]
<script>
function CBtoTB()
{document.getElementById("username").value=document.getElementById("usernameselect").value}
 </script>

<?php  
$result=mysql_query("select Username, EMail, Department from users"); 
$options=""; 
while ($row=mysql_fetch_array($result)) { 
$username=$row["Username"]; 
$options.="<OPTION VALUE=\"$username\">".$username.'</option>'; 
} 
?>
<select name="usernameselect" id="usernameselect" onchange="CBtoTB()">
<option value="">&lt; select user &gt;<?php echo $options ?></option>
</select>
 <input name="username" type="text" id="username" value="<?php echo $username ?>" size="25" readonly="readonly" />

Hello, first time poster..  I've looked the web over for a long time and can't figure this one out.

- Below is basic code that successfully checks MySQL for a match and displays result.  I was debugging and forced the "height" and "width" to be 24 and 36 to make sure that wasn't the problem.  That's good.. 
- I'd like to give the user ability to select width and height from a form.. and have it do an onchange this.form.submit so the form can be changing as fields are altered (thus the onchange interaction)
- In a normal coding environment I've done this numerous times with no "Page cannot be displayed" problems.  It would simply change one select-option value at a time til they get down the form and click submit...  but in WordPress I'm having trouble making even ONE single onchange work!
- I've implemented the plugins they offer which allows you to "copy+paste" your php code directly into their wysiwyg editor.  That works with basic tests like my first bullet point above.
- I've copied and pasted the wordpress url (including the little ?page_id=123) into the form "action" url... that didn't work...  tried forcing it into an <option value=""> tag.. didn't work.  I'm just not sure. 

I've obviously put xx's in place of private info..

Why does this form give me Page Cannot Be Displayed in WordPress every time?  It won't do anything no matter how simple.. using onchange..

Code..

$con = mysql_connect("xxxx.xxxxxxx.com","xxxxxx","xxxxx");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
mysql_select_db("xxxxxx", $con);
$myprodwidth=24;
$myprodheight=36;
$result = mysql_query("SELECT * FROM product_sizes WHERE prodwidth='$myprodwidth' and prodheight='$myprodheight'");
while($row = mysql_fetch_array($result))
  {
  echo $row['prodprice'];
  }
mysql_close($con);


<form method="post" action="">
<select name="myheight" onchange="this.form.submit();">
<option selected="selected" value="">select height</option>
<option value="xxxxxxxxx.com/wordpress/?page_id=199&height=36">36</option>
<option value="xxxxxxxxx.com/wordpress/?page_id=199&height=36">48</option>
</select>

Hi.  I have a page that allows users to upload some files, using <input type="file">...

What I would like is to be able to upload the file "onchange" of this input field using ajax so that (1) users can upload their second file while their first one is uploading, and (2) so that they can see their file on the page as soon as it's done uploading.  But I can't seem to find a way to populate the $_FIlLES array without actually submitting the form.

I've seen some stuff online where people use iframes to do this kind of thing, but I was hoping there was another way to do it.  Any help would be great.

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