2

I want to submit drop-down with dynamic value. I concatenate all variables inside the string but data not submit any one help me how is it done

if(isset($_POST['submit'])){
     $name = $_POST['name'];
     $value1 = $_POST['value1'];
     $value2 = $_POST['value2'];


        $data = "<select name=''$name''>";
        $data .="<option>Selet data</option>";
        $data .="<option value='**$value1**'> **$value1**</option>";
        $data .="<option value='**$value2**'> **$value2**</option>";
        $data .="</select>";

        mysql_query("insert into options (value) values ('".$data."')");
    }

With variable string not submit in to the database.

HTML is below

<form action="" method="POST">

        <input type="text" name="name">
        <input type="text" name="value1">
        <input type="text" name="value2">
        <input type="submit" name="submit">

</form>
shan
  • 21
  • 5

1 Answers1

0

Check with below:

    if(isset($_POST['submit'])){
     $name = $_POST['name'];
     $value1 = $_POST['value1'];
     $value2 = $_POST['value2'];

        $data = "<select name='$name'>";
        $data .="<option>Selet data</option>";
        $data .="<option value='**$value1**'> **$value1**</option>";
        $data .="<option value='**$value2**'> **$value2**</option>";
        $data .="</select>";

        if(!mysql_query("insert into options (`value`) values (".json_encode($data).")")) {
          echo("Error description: " . mysql_error());
        }
}
Rupal Javiya
  • 591
  • 5
  • 14