php - get all mysql row data after selecting name from dropdown

Below is the php code I'm using to populate a select dropdown with the variable "full_name".

$sql = "SELECT * FROM Entries";
$result = mysql_query($sql);

echo "";
while ($row = mysql_fetch_array($result)) {

echo "" . $row['full_name'] ." ";
}
echo "";

I would like to be able to select a name from the dropdown list and display all the relevant data for that specific row so it can be easily seen by the admins.

Basically

select name (details populate below)


Name: $full_name
DOB: $dob
Work Phone: $work_phone
etc...

Maybe something like this?

while($row = mysql_fetch_array($result, MYSQL_ASSOC)) {

  echo "Name :{$row['full_name']}  
 ".
     "DOB : {$row['dob']} 
 ".
     "Work Phone : {$row['work_phone']} 
 ".
     "--------------------------------
";
  }
  echo "Fetched data successfully\n";

I just don't know how to tie the dropdown selection to correctly display the information for that person's name. Any help is appreciated!

Thanks

---

$connection = new mysqli("XXXX", "XXXX", "XXXX", "XXXX");

if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
}
$info = ''; /* WE'LL BE STORING THEIR INFORMATION HERE */


$stmt = $connection->prepare("SELECT full_name, dob, work_phone FROM Entries"); 
if ( false===$stmt ) { die('prepare() failed: ' . htmlspecialchars($mysqli->error)); }
$stmt->execute();
$stmt->bind_result($full_name, $dob, $workphone);
while($stmt->fetch()){

    echo ''.$fullname.'';
    $info .= '
'.$dob.' - '.$work_phone.'
';

}

$stmt->close();

echo '';

echo $info; /* DISPLAY THE INFOs, BUT NOT REALLY BECAUSE THEY ARE HIDDEN */
?>

id
requested_action
full_name
birth_date
sex
work_location
work_phone
hire_date

coverage_choice
network_choice  
plan_choice
dependant_name_1
dependant_relationship_1 
dependant_dob_1
dependant_sex_1 
dependant_name_2 
dependant_relationship_2    
dependant_dob_2 

dependant_sex_2 
dependant_name_3 
dependant_relationship_3
dependant_dob_3 
dependant_sex_3 
dependant_name_4
dependant_relationship_4 
dependant_dob_4 
dependant_sex_4 
spouse_coverage 

employee_enroll 
signature
date_today  
reg_date






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Answer

You can achieve this using Javascript. But we will be using a Javascript library called jQuery.

Oh, and don't use deprecated mysql_* extension. We will be using mysqli_* extension instead.

Let's prepare first your HTML by adding an id tag on your field.

echo '';



Then on your while loop, let's get the other information from each row.

$info = ''; /* WE'LL BE STORING THEIR INFORMATION HERE */

$stmt = $connection->prepare("SELECT id, full_name, birth_date, work_phone FROM Entries"); /* SEE HOW TO ESTABLISH CONNECTION TO YOUR DATABASE USING mysqli AT THE BOTTOM */
$stmt->execute();
$stmt->bind_result($id, $fullname, $dob, $workphone); /* CORRESPONDS TO THE SELECTED COLUMNS FROM YOUR QUERY */
while($stmt->fetch()){


    echo ''.$fullname.'';
    $info .= '

                  Date of Birth: '.$dob.'
                  Tel. Phone (work): '.$workphone.'
              
';

}
$stmt->close();


echo '';

echo $info; /* DISPLAY THE INFOs, BUT NOT REALLY BECAUSE THEY ARE HIDDEN */

Then, let's create the script to display the information of selected full_name:

This is just a simple trick. But I think it is better than doing an Ajax call from every select of full_name.

---

Establish your connection to your database using mysqli_* extension also:

$connection = new mysqli("Host", "User", "Password", "Database");

if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
}