Pull DB data into drop down in MYSQL, PHP

Question

I want to make select option by getting DB values into a drop down, following is my code that I have used,

$query = "SELECT apply_year FROM emp_leaves ORDER BY apply_year DESC";
$result = mysql_query($query);
echo "<select name='apply_year' class=\"form-control\"><option>Select Year</option>";
while ($r = mysql_fetch_array($result)) {
    echo "<option value=" . $r['leave_id'] . ">" . $r['apply_year'] . "</option>";
}
echo "</select>"; ?>

It work fine, But I want to show only one value of each years, I want to avoid repeat same year in drop down. Please help

how can you get the leave_id value when you only have "apply_year" in your query? - I also agree with the use of "distinct" as @Dhara Parmar said but you might also try "GROUP BY apply_year" as well. — gavgrif
– gavgrif, Commented Jun 13, 2016 at 7:36

Dhara Parmar · Accepted Answer · 2016-06-13 07:30:24Z

1

Use DISTINCT for apply_year and Create mysql query like, so that it will return apply_year only once in result:

$query = "SELECT DISTINCT apply_year FROM emp_leaves ORDER BY apply_year DESC";

answered Jun 13, 2016 at 7:30

Dhara Parmar

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Keyur Chavda-kc1994 · Accepted Answer · 2016-06-13 07:35:08Z

1

Try this,

$query = "SELECT apply_year FROM emp_leaves GROUP BY apply_year  ORDER BY apply_year DESC";
$result = mysql_query($query);
echo "<select name='apply_year' class=\"form-control\"><option>Select Year</option>";
while ($r = mysql_fetch_array($result)) {
     echo "<option value=" . $r['leave_id'] . ">" . $r['apply_year'] . "</option>";
}
echo "</select>"; ?>

answered Jun 13, 2016 at 7:35

Keyur Chavda-kc1994

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