how to display data from db after choosing value from drop down 1

[amir4oracle]

hello,

i need to know how can i display data from database using drop down.

in other words, i have to create a drop down using select-option and depending upon the choosen value i need to display the values in the row using select query.

Your help is highly appreciated.

 

[Vragabond]

What have you done so far? Where are you stuck? We can help with individual problems, but all we can tell you is this:

1. create a dropdown and a button
2. submit
3. on the next page, construct your query in a way to include the sent element.
4. run query.

 

[amir4oracle]

I did exactly as you said Vragabond ... and its working perfectly
gave you the star
thanks a lot

 

[Olavxxx]

Lookie, lookie:

<?php
if (empty($cat)) {
$cat = "fafafafa";
}

///////// Getting the data from Mysql table for first list box//////////
$quer2=mysql_query("SELECT DISTINCT category,cat_id FROM category ORDER BY category ASC");
///////////// End of query for first list box////////////

/////// for second drop down list we will check if category is selected else we will display all the subcategory/////
if(isset($cat) and strlen($cat) > 0){
//her
$query = sprintf("SELECT DISTINCT subcategory FROM subcategory where cat_id=%s order by subcategory",
           quote_smart($cat));
$quer = mysql_query($query);
}else{$quer=mysql_query("SELECT DISTINCT subcategory FROM subcategory ORDER BY subcategory ASC"); }
////////// end of query for second subcategory drop down list box ///////////////////////////
//reload(this.form)
echo "<form method=\"post\" name=\"f1\" action=\"index.php\"  enctype=\"multipart/form-data\">";

/// Add your form processing page address to action in above line. Example  action=dd-check.php////
//////////        Starting of first drop downlist /////////
echo "<select name='cat' onchange=\"document.f1.submit()\"><option value=\"falaffel\">Velg en dyreart</option>";
while($res2 = mysql_fetch_array($quer2)) {

if($res2['cat_id']==@$cat){
echo "<option selected value='$res2[cat_id]'>$res2[category]</option>";}
else{echo  "<option value='$res2[cat_id]'>$res2[category]</option>";}
}
echo "</select>";

echo "<input type=\"text\" name=\"__annen_kategori\" value=\"{$_POST['__annen_kategori']}\" style=\"width:106px; padding-left: 10px;\" /><br />";
//////////////////  This will end the first drop down list ///////////

//////////        Starting of second drop downlist /////////
echo "<select name='subcat' style=\"width:336px;\">";
while($res = mysql_fetch_array($quer)) {
$subcat = str_replace("!", "", $res[subcategory]);
echo  "<option value='$subcat'>{$subcat}</option>";
}
echo "</select>";
?>

I made this based on some js-code I found while googling.
It will submit to self, when choosing from list.

Olav Alexander Mjelde

http://www.volvo-power.net

Admin & Webmaster

 

[Olavxxx]

oh, you need this too:

// Quote variable to make safe
function quote_smart($value)
{
   // Stripslashes
   if (get_magic_quotes_gpc()) {
       $value = stripslashes($value);
   }
   // Quote if not a number or a numeric string
   if (!is_numeric($value)) {
       $value = "'" . mysql_real_escape_string($value) . "'";
   }
   return $value;
}

Olav Alexander Mjelde

http://www.volvo-power.net

Admin & Webmaster