从mysql数据的下拉列表中预选择值

Question

I have the following code, which creates a dropdown list based on the contents of a mysql database. I also have a variable $ir1text which is the users current selection from that database. How can make the drop down list pre-select the value the user already has. The idea being I want to show them what they have and have the dropdown list right there for them to select a new value.

<?php
include("config.php");
$sql1 = "SELECT * FROM iRadios";
$result1 = mysql_query($sql1);

echo "<select name='Station1'>";
while ($row = mysql_fetch_array($result1)) {
    echo "<option value='" . $row['StationName'] ."'>" . $row['StationName'] ."</option>";
}
echo "</select>";
?>

Answer 1

Here is the solution

<?php
include ("config.php");

$sql1 = "SELECT * FROM iRadios";
$result1 = mysql_query($sql1);
echo "<select name='Station1'>";
while ($row = mysql_fetch_array($result1)) {
    if ($row['StationName'] == trim($ir1text)) {
        echo "<option value='" . $row['StationName'] . "' selected>" . $row['StationName'] . "</option>";
    }
    else {
        echo "<option value='" . $row['StationName'] . "'>" . $row['StationName'] . "</option>";
    }
}
echo "</select>";
?>

Answer 2

Well you can put that user selected value in the first option. if you dont hyave that value in the same table then you can make another query and store that value and put it in the first option of your tag.

Answer 3

If $row[] == $ir1tex show this option selected

while ($row = mysql_fetch_array($result1)) {
    if($row['StationName']==  $ir1text){
        echo "<option value='" . $row['StationName'] .selected"'>" . $row['StationName'] ."</option>";
    }else{
    echo "<option value='" . $row['StationName'] ."'>" . $row['StationName'] ."</option>";
    }

}