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drop down with db

digitalgod

By digitalgod
in PHP Coding Help

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digitalgod Regular Member

digitalgod

Posted
Posted
hey guys,

I've been pulling my hair out trying to figure this out...

I have 2 drop downs, first 1 is populated from a db, second one displays the sub categories of what has been selected in the first drop down

here's what I got so far

[code]
<?php
$sql_clubs='SELECT * FROM clubs ORDER BY name';
$req_clubs=mysql_query($sql_clubs) or die(query_error());
$numRows = mysql_num_rows($req_clubs);
$clubList = array();
?>
<script type="text/javascript">
function fillclub(){
// this function is used to fill the club list on load
<?php
while($data = mysql_fetch_assoc($req)) {
array_push($clubList,$data['name']);
echo 'addOption(document.form.club, "'.$data['name'].'", "'.$data['name'].'", "");';
}
?>
}

function Selectnight(){
// ON selection of club this function will work

removeAllOptions(document.form.night);
addOption(document.form.night, "", "Choose one", "");

<?php
for ($i=0;$i<=$numRows;$i++) {
$sql_nights='SELECT * FROM clubnights WHERE club = '.$clubList[$i].'ORDER BY night';
$req_nights=mysql_query($sql_nights) or die(query_error());
$night_data = mysql_fetch_array($req_nights);
echo 'if(document.form.club.value == '.$clubList[$i].'){
addOption(document.form.night,"'.$night_data['night'].'", "'.$night_data['night'].'");
}';
}
?>

function removeAllOptions(selectbox)
{
var i;
for(i=selectbox.options.length-1;i>=0;i--)
{
//selectbox.options.remove(i);
selectbox.remove(i);
}
}


function addOption(selectbox, value, text )
{
var optn = document.createElement("OPTION");
optn.text = text;
optn.value = value;

selectbox.options.add(optn);
}
[/code]

but it's not working... any ideas what's wrong?
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https://forums.phpfreaks.com/topic/15626-drop-down-with-db/
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thepip3r Member

thepip3r

Posted
Posted
this would be more of a javascript question than a PHP one; but, considering you don't have any <html> or <head> tags anywhere (where your javascript functions MAY need to go) that would be the first place i'd look.

edit:

AND this is not correct:

[code]or die(query_error());[/code]

it should be:

[code]or die(mysql_error());[/code]



edit 2:

unless you have $req called somewhere else in the page that you didn't copy and paste, it looks like your loop should be mysql_fetch_assoc'ing ($req_clubs) instead of just $req...

[code]<?php
$sql_clubs='SELECT * FROM clubs ORDER BY name';
$req_clubs=mysql_query($sql_clubs) or die(query_error());
$numRows = mysql_num_rows($req_clubs);
$clubList = array();
?>
<script type="text/javascript">
function fillclub(){
// this function is used to fill the club list on load
<?php
while($data = mysql_fetch_assoc($req)) {
array_push($clubList,$data['name']);
echo 'addOption(document.form.club, "'.$data['name'].'", "'.$data['name'].'", "");';
}
?>[/code]

should look like:

[code]<?php
$sql_clubs='SELECT * FROM clubs ORDER BY name';
$req_clubs=mysql_query($sql_clubs) or die(query_error());
$numRows = mysql_num_rows($req_clubs);
$clubList = array();
?>
<script type="text/javascript">
function fillclub(){
// this function is used to fill the club list on load
<?php
while($data = mysql_fetch_assoc($req_clubs)) {
array_push($clubList,$data['name']);
echo 'addOption(document.form.club, "'.$data['name'].'", "'.$data['name'].'", "");';
}
?>[/code]
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https://forums.phpfreaks.com/topic/15626-drop-down-with-db/#findComment-63662
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digitalgod Regular Member

digitalgod

Posted
  • Author
Posted
well I only placed the part that I know has errors in, the form itself is just 2 drop downs and a submit button... first drop down has a onChange="fillnight();"

I think it's php related because I'm positive I'm doing something wrong in php, it works fine if I use predefined values and only javascript
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https://forums.phpfreaks.com/topic/15626-drop-down-with-db/#findComment-63664
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digitalgod Regular Member

digitalgod

Posted
  • Author
Posted
thanks thepip3r but it's still not working...

*edit*

I know it's this part that doesn't work because if I take it off the page is displayed...

[code]
<?php
for ($i=0;$i<=$numRows;$i++) {
$sql_nights='SELECT * FROM clubnights WHERE club = '.$clubList[$i].'ORDER BY night';
$req_nights=mysql_query($sql_nights) or die(query_error());
$night_data = mysql_fetch_assoc($req_nights);
echo 'if(document.form.club.value == '.$clubList[$i].'){
addOption(document.form.night,"'.$night_data['night'].'", "'.$night_data['night'].'");
}';
}?>
[/code]

still can't find why though
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ryanlwh Newbie

ryanlwh

Posted
Posted
need a space before ORDER, and add quotes too:
[code]$sql_nights='SELECT * FROM clubnights WHERE club = "'.$clubList[$i].'" ORDER BY night';[/code]

Now if $clubList contains strings:
[code]echo 'if (document.form.club.options[document.form.club.selectedIndex].value == "'.$clubList[$].'"){ [/code]
Link to comment
https://forums.phpfreaks.com/topic/15626-drop-down-with-db/#findComment-63701
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digitalgod Regular Member

digitalgod

Posted
  • Author
Posted
I think we're getting somewhere, the page is displayed but there's nothing in the drop downs, it's giving me a javascript error saying that there's something wrong here

selectbox.options.add(optn);

what's wrong now? :(
by the way is that part any good

[code]
<?php
for ($i=0;$i<=$numRows;$i++) {
$sql_nights='SELECT * FROM mtlguest_clubnights WHERE club = "'.$clubList[$i].'" ORDER BY night';
$req_nights=mysql_query($sql_nights) or die(query_error());
$night_data = mysql_fetch_assoc($req_nights);
echo 'if (document.form.club.options[document.form.club.selectedIndex].value == "'.$clubList[$i].'"){ 
addOption(document.form.night,"'.$night_data['night'].'", "'.$night_data['night'].'");
}';
}
?>[/code]

I have a feeling that it will only display 1 sub category...


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digitalgod Regular Member

digitalgod

Posted
  • Author
Posted
yes I do have document.form.night

[code]
<div id="article">
<form name="form" id="form" action="admin.php?a=delclubnight" method="post">
<input type="hidden" name="process" value="yes" />
    <table border="0">
      <tr>
        <td>Club Name:</td>
        <td><select name="club" id="club" onChange="Selectnight();">
          <option>Choose one</option>       
        </select></td>
      </tr>
      <tr>
        <td>Night::</td>
        <td><select name="night" id="night">
          <option>Choose one</option>
        </select></td>
      </tr>
    </table>
<br />
<input type="submit" value="Delete Night" />
</form>
</div>
[/code]

but it's the first drop down that isn't displaying anything anymore, If I remove the part of code that populates the subcat the first drop down shows me the list of clubs available in the database
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