如何使客户端上的按钮重定向到带有数据库信息的php页面

Question

I am brand new to programming and I had a unique (I think) question.

I made a MySql database for storing employee records. I am able to submit new employee data to the db and am able to retrieve employee data from the db and display it in an HTML table.

I want to create buttons on the table for each ID#. Each button would redirect you to a different php page and display the employee information associated with the ID# whose button was clicked.

Does anyone know how would I go about doing this?

echo '<table border="0" cellspacing="25" cellpadding="2">
      <tr>
          <td>ID</td>
          <td>First</td>
          <td>Last</td>
          <td>SIN</td>
          <td>Password</td>
      </tr>';

function loadList() {
include_once 'includes/dbh.inc.php';

try {
  $sqlarray = "SELECT * FROM employee;";
  $result = mysqli_query($conn, $sqlarray);
  $resultCheck = mysqli_num_rows($result);

  if ($resultCheck > 0) {
    while ($row = mysqli_fetch_assoc($result)) {
      $employeeID = $row['id'] . "<br>";
      $employeeFN = $row['firstname'] . "<br>";
      $employeeLN = $row['lastname'] . "<br>";
      $employeeSIN = $row['sin_'] . "<br>";
      $employeePASS = $row['pass_'] . "<br>";

      echo '<tr>
                <td>'.$employeeID.'</td>
                <td>'.$employeeFN.'</td>
                <td>'.$employeeLN.'</td>
                <td>'.$employeeSIN.'</td>
                <td>'.$employeePASS.'</td>
            </tr>';
    }
  }
  else {
    echo "error no data";
  }

} catch (\Exception $e) {
  echo ("error");
}
}

Answer 1

Mate, welcome in the programming world. You can create a anchor tag which will redirect user to a new php page with employee id in URL. check below code:

echo '<tr>
          <td>'.$employeeID.'</td>
          <td>'.$employeeFN.'</td>
          <td>'.$employeeLN.'</td>
          <td>'.$employeeSIN.'</td>
          <td>'.$employeePASS.'</td>
          <td><a href="details.php?id='$employeeID.'>View Details</a></td>              
       </tr>'  

Here you need to create a new php page with details.php name. There you need to get employee id by GET global variable and need to fetch details from database to show. Check below code:

<?php
include_once 'includes/dbh.inc.php';
$employeeId = $_GET['id'];

$sqlarray = "SELECT * FROM employee where id = '".$employeeId."';";
$result = mysqli_query($conn, $sqlarray);
if (!$result) {
    echo 'Could not run query: ' . mysql_error();
    exit;
}
$employeeDetails = mysql_fetch_row($result);

Here $employeeDetails has all data which you can use to render in HTML. But please read little bit about SQL injection before using the same code to avoid crashes. Hope it helps you.

Answer 2

You could do like this by adding <a></a> tag

echo '<tr>
          <td><a href="show.php?id='$employeeID.'>Show</a></td>
          <td>'.$employeeID.'</td>
          <td>'.$employeeFN.'</td>
          <td>'.$employeeLN.'</td>
          <td>'.$employeeSIN.'</td>
          <td>'.$employeePASS.'</td>
       </tr>';

And in show.php page get this id by $_GET['id']