$sql = "SELECT DISTINCT Branch FROM student_main";
$result = mysql_query($sql);
$row_num = mysql_num_rows($result);
$rows = mysql_fetch_array($result);
echo "<select name='Branch'>";
for($i=0;$i<=$row_num-1;$i++){
echo "<option value='".$rows[$i]."'>".$rows[$i]."</option>";
}
echo "</select>";
echo "<input type='submit' Value='submit' />";
echo "</form>";
I am trying to create a dropdown using the above code for my form. But its not working. There are 3 distinct values in the Branch column but in the dropdown, it shows only one value(the first one) and the next two as blank values.
However when in echo $row_num, its shows 3.
Thats means its fetching the three rows, but then why its not showing in the dropdown list.
If I run the same query in phpmyadmin it shows the correct answer i.r it returns 3 distinct Branch values.
You should do something like this:
$sql = "SELECT DISTINCT Branch FROM student_main";
$result = mysql_query($sql);
echo "<select name='Branch'>";
while ($row = mysql_fetch_array($result)) {
echo "<option value='".$row[0]."'>".$row[0]."</option>";
}
echo "</select>";
echo "<input type='submit' Value='submit' />";
echo "</form>";