hi all,
i am fetching recods from mysql in a combobox.but only one record display .... i want all record display in combobox ......................i write query below and need help how to sote fetch row in array i declered "$va[]"

    // echo "Connected MySql";
  $res1 = mysql_query("SELECT gp FROM gp order by gp") or die("Invalid query: " . mysql_query());
  //echo ".mysql_query()";
   $va[] =$row['gp'];/?????????????????????????????????????????????
 echo '
	<select name="gp" id="gp">';
for($x = 0; $x < count($va); $x++)
	  // write "selected" if the value matches the one posted
	  if($va[$x] == $_POST['gp'])
		  $selected = 'selected';
		  $selected = '';
	   // print the option
	   echo '
		<option value="'.$va[$x].'"'.$selected.'>'.$va[$x].'</option>';
}echo '</select>';
8 Years
Discussion Span
Last Post by Rhyan

First of all you have to read all records returned from the DB. You cannot fetch the whole result table from the database using a single php function. PHP can retrieve only one record at a time, so if you want to retrieve all rows from the SQL result, you have to type this many times mysql_fetch_array(), as many rows have been returned from the sql query.
Please check my post here, explaining result retrieval.

Then, once you have all your data, you can proceed with selected / non-selected verification of values in the <select> element.
Instead of loading a variable with name $selected, in my opinion it is better that you do the following

if ($result[key] == $_POST[whatever]
 echo '<option value="'.$result[key].'" selected="selected">Whatever text here</option>';
echo '<option value="'.$result[key].'">Whatever text here</option>';
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