CODEWITHSUNDEEP
javascriptphphtml
Neof
Neof

Reputation: 13

Set option value from php variable

<form method="post" id="employee" action="change.php">
<select id="employeename" name="employeename" onchange="this.form.submit()">
 <?php 
while ($row = mysql_fetch_array($result))
{
    echo "<option value='".$row['name']."'>".$row['name']."</option>";
}
?>    
</select>
</form>

The form will be submitted when select value is changed,

Once the form is submitted, it will redirect to the same page. How do I set the selected option value with the one I just selected?

Upvotes: 0

Views: 2273

Answers (2)

Avinash Raut
Avinash Raut

Reputation: 2113

    <form method="post" id="employee" action="change.php">
    <select id="employeename" name="employeename" onchange="this.form.submit()">
     <?php 
    while ($row = mysql_fetch_array($result))
    {
   if($_POST['employeename']==$row['name'])
        echo "<option value='".$row['name']."' selected>".$row['name']."</option>";
else 
   echo "<option value='".$row['name']."'>".$row['name']."</option>";
    }
    ?>    
    </select>
    </form>

just marked as selected if you find selected key in available list.

Upvotes: 0

Rajdeep Paul
Rajdeep Paul

Reputation: 16963

Change your form, especially the <select> dropdown list in the following way,

<form method="post" id="employee" action="change.php">
    <select id="employeename" name="employeename" onchange="this.form.submit()">
        <?php 
            while ($row = mysql_fetch_array($result)){
                $output = "<option value='".$row['name']."'";
                if($_POST['employeename'] == $row['name']){
                    $output .= " selected='selected'";
                }
                $output .= ">".$row['name']."</option>";
                echo $output;
            }
        ?>    
    </select>
</form>

Upvotes: 2

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