如何通过从另一个下拉列表中选择值来填充下拉列表？

Question

I have been trying to populate the contents of one drop down list with the selection of an item in another, using php,html and mysql database and Its proved to be a lot harder than i thought for me. I'm trying to get one dropdown list show names of all mobile phones based on company selections in the first drop down list.

The code i've provided below has no errors, but the 2nd dropdown list is not getting populated. the variable '$submittedValue' is not getting the right value and hence '$name' also has no content in it. I can't seem to be able to figure out a reason for this.

what i've done is, to try to use the selected option in the first drop down list, put that into a variable and pass that as a parameter to the 2nd query.

I solution to this would be greatly appreciated.

<?php

 $con=mysqli_connect("localhost","root","","dbmobiles");

 // Check connection
 if (mysqli_connect_errno($con)) {
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
 }

 $options = '';
 $submittedValue = '';
 $name=" ";

 $q1=mysqli_query($con,"select distinct compName from umobile order by compName ASC");
 while($row = mysqli_fetch_array($q1)) {
    $options .="<option value=". $row['compName'] ." >" . $row['compName'] .    "</option>";
 }

 $menu1 = "<form id='filter' name='filter' method='post' action=' '>    
           <p><label>Select Company</label></p>
           <select name='filter' id='filter'>
           " . $options . "
           </select>
          </form>";

 if (isset($_POST["filter"])) {
    $submittedValue = $_POST["filter"];
 }
 echo $menu1;

 if ($q2=mysqli_prepare($con,"SELECT DISTINCT mobname FROM umobile WHERE compName=? "))
 {
    mysqli_stmt_bind_param($q2,"s",$submittedValue);
    $s=$submittedValue;
    mysqli_stmt_execute($q2);
    mysqli_stmt_bind_result($q2,$r);
 }  

 while(mysqli_stmt_fetch($q2)) {
    $name .="<option>".$r['mobname']."</option>";
 }

 echo ".$name.";
 $menu2 = "<form id='moblist' name='moblist' method='post' action=' '>
            <p><label>Select Mobile</label></p>
            <select name='moblist' id='moblist'>
            " . $name . "
            </select>
           </form>";

 echo $menu2;
?>

Answer 1

Don't forget to quote .$r['mobname'].

$name .="<option value=\"".$r['mobname']."\" >".$r['mobname']."</option>";

Your $_POST tests only .

if (isset($_POST["filter"])) {
    $submittedValue = $_POST["filter"];
}

What about if $_POST["filter"] is NOT set ?
Your code did not take care of this fact. He always runs through the section $menu2.

Your $_POST is as the name implies, of a post. To evaluate a $_POST must be done a post.

 <input type="submit" name="Submit" value="Submit" />

for example

 $menu1 = "<form id='filter' name='filter' method='post' action=' '>    
             <p><label>Select Company</label></p>
            <select name='filter' id='filter'>
             " . $options . "
            </select>
            <input type=\"submit\" name=\"Submit\" value=\"Send\" />
          </form>";

Put all the logic for the second menu in your code if (isset($_POST["filter"])) {

 echo $menu1;

 if (isset($_POST["filter"])) {
    $submittedValue = $_POST["filter"];
      if ($q2= ....)
       {
         mysqli_stmt_bind_param($q2,"s",$submittedValue);
         $s=$submittedValue;
         mysqli_stmt_execute($q2);
         mysqli_stmt_bind_result($q2,$r);

         while(mysqli_stmt_fetch($q2)) {
         $name .="<option value=\"".$r['mobname']."\" >".$r['mobname']."</option>";
         }

         $menu2 = "<form id='moblist' name='moblist' method='post' action=' '>
            <p><label>Select Mobile</label></p>
            <select name='moblist' id='moblist'>
            " . $name . "
            </select>
           </form>";

        echo $menu2;
       }  
   }

Update:

You can not use $r['mobname'] with a bind.

mysqli_stmt_bind_result($q2,$r);
....
while(mysqli_stmt_fetch($q2)) {
$name .="<option value=\"".$r['mobname']."\" >".$r['mobname']."</option>";

}

Use it like .

mysqli_stmt_bind_result($q2,$r);
....
while (mysqli_stmt_fetch($q2)) {
$name .="<option value=\"".$r."\" >".$r."</option>";
}


mysqli_stmt_close($q2);

Update 2

you forget to quote that to

 $options .="<option value=". $row['compName'] ." >" . $row['compName'] .    "</option>";

right

 $options .="<option value=\"". $row['compName'] ."\" >" . $row['compName'] .    "</option>";

Answer 2

$menu1 = "<form id='filter' name='filter_form' method='post' action=' '>    
           <p><label>Select Company</label></p>
           <select onchange='filter_form.submit()' name='filter' id='filter'>
           " . $options . "
           </select>
          </form>";

there is no submit button in your form , either place a submit button or try to submit on selection on dropdown values. Only then if($_POST['filter']) will be true and your variable will successfully get assignment