displaying in forms data inserted into database - php

Question

I created this table that part of it takes input from the user and save it in the database..
I'm trying to make the data put by the user in the tags remain shown. Basically the input form becoming both input/output source. Any help on how to do so?
switch ($selected){
case 'University':
$stmt = $conn->prepare("SELECT employees.afnumber,employees.name,employees.actualpost,university.brevet FROM employees,university WHERE employees.status='Employed' AND employees.afnumber=university.afnumber ORDER BY employees.afnumber DESC LIMIT :start,:end");
$stmt->bindParam(':start', $pages->limit_start, PDO::PARAM_INT);
$stmt->bindParam(':end', $pages->limit_end, PDO::PARAM_INT);
$stmt->execute();
$result = $stmt->fetchAll();
$selectedtable = "<form method='post' action=''>\n";
$selectedtable .= "<table class='sortable'>\n<tr><th>Description</th><th>A</th><th>B</th><th>C</th><th>D</th></tr>\n";
foreach($result as $row) {
$selectedtable .= "<tr><th>Brevet</th><td><input type='text' name='Brevet1' style=' padding: 10px; margin-top:1px; border: solid 2px #c9c9c9; width:50px; height:2px;'></td><td>".$row[0]."</td><td>".$row[2]."</td><td>".$row[3]."</td></tr>
<tr><th>Baccalaureat/BT</th><td><input type='text' name='Baccalaureatbt' style=' padding: 10px; font-size:16px; margin-top:1px; border: solid 2px #c9c9c9; width:50px; height:2px;'></td><td>".$row[1]."</td><td>$row[2]</td><td>$row[3]</td></tr>
<tr><th>License/TS</th><td><input type='text' name='Licensets' style=' padding: 10px; margin-top:1px; border: solid 2px #c9c9c9; width:50px; height:2px;'></td><td>".$row[1]."</td><td>".$row[2]."</td><td>$row[3]</td></tr>
<tr><th>M1</th><td><input type='text' name='M1' style=' padding: 10px; margin-top:1px; border: solid 2px #c9c9c9; width:50px; height:2px;'></td><td>".$row[1]."</td><td>".$row[2]."</td><td>".$row[3]."</td></tr>
<tr><th>Master's Degree</th><td><input type='text' name='Mastersdegree' style=' padding: 10px; margin-top:1px; border: solid 2px #c9c9c9; width:50px; height:2px;'></td><td>".$row[1]."</td><td>".$row[2]."</td><td>".$row[3]."</td></tr>
<tr><th>PHD</th><td><input type='text' name='Phd' style=' padding: 10px; margin-top:1px; border: solid 2px #c9c9c9; width:50px; height:2px;'></td><td>".$row[1]."</td><td>".$row[2]."</td><td>".$row[3]."</td></tr>";
}
$selectedtable .= "</table>\n";
$selectedtable .= "<input type='submit' name='submit' value='Submit' style='width:80px; height:30px; text-align:center; padding:0px;'>\n";
$selectedtable .= "</form>\n";
if(isset($_POST['submit']))
{ $brevet1 = $_POST['Brevet1'];
$baccalaureatbt = $_POST['Baccalaureatbt'];
$licensets = $_POST['Licensets'];
$m1 = $_POST['M1'];
$mastersdegree = $_POST['Mastersdegree'];
$phd = $_POST['Phd'];
$sql1="SELECT Brevet1,Baccalaureatbt,Licensets,M1,Mastersdegree,Phd FROM university";
if ($result=mysqli_query($con,$sql1))
{
$rowcount=mysqli_num_rows($result);
}
if($rowcount==0)
{
$sql="INSERT INTO university(Brevet1,Baccalaureatbt,Licensets,M1,Mastersdegree,Phd) VALUES('$brevet1','$baccalaureatbt','$licensets','$m1','$mastersdegree','$phd')";
$result = mysql_query($sql);
}
else
{
$sql2 = "UPDATE university SET Brevet1 = '$brevet1' , Baccalaureatbt = '$baccalaureatbt', Licensets = '$licensets', M1 = '$m1', Mastersdegree = '$mastersdegree', Phd = '$phd'";
$result2 = mysql_query($sql2);
}
}
break;

Answer 1

You can set a value to each input like this :
<input type='text' name='Brevet1' value="<?php echo $_POST['brevet_1']; ?>">
With a ternary condition :
echo (!empty($_POST['brevet_1']) ? $_POST['brevet_1'] : '');
EDIT
Fetch the data from database and set the value of input just like above

Answer 2

This is what I've always done in these kind of situations. Suppose your page name is mypage.php. Now, you open mypage.php with an id (like mypage.php?id=1, and you can get the id as $_GET['id']). This id would be the auto_increment value of the table you need to insert / update.
Now, on mypage.php, write this code :-
<form action="someEndFile.php" method="post">
<?php echo $command; ?>
<?php if($command == 1)
{
$userData = fetch via $_GET['id'];
?>
Your input Tags Here
Submit Button Here
<input type='text' name='somename' value='<?php echo $userdata->valueFromTheDatabase; ?>'>
<?php
}
else
{
?>
<input type='text' name='somename'>
<?php
}
?>
</form>
In the if(data_exists) block, check if any record in the table exists with the id id (which you received as $_GET['id']). If the record exists, it will go inside the if block and the command value would be 1, that means update.
If the ID didn't existed, it would go to else and the value would be 2, which means insert new.
Now, in your form, echo $command and put all the fields. Here check the value of $command and echo the value inside the input tags accordingly.
Now on the someEndFile.php file, have a code like this :-
<?php
$command = isset($_POST['command']) ? $_POST['command'] : '';
switch($command)
{
case 1:
//update case
write your update query here;
break;
case 2:
//insert case
write your insert query here;
break;
}
?>
You can now execute your needed query. This is what I always do in these type of situations.

Answer 3

Here's an example where the input form data is displayed when the form is submitted. As a bonus, it also shows some basic PHP form validation as well:
w3schools Example of showing the form entry on the same page as entered
echo out the variables for the values entered on the page. You can place the echo statements as needed, within tags formatting as you like.
Example:
if your form contains:
Name: <input type="text" name="name">
Then display the tag containing 'name' anywhere on the page like this:
<?php
//check if name has changed:
if ($_SERVER["REQUEST_METHOD"] == "POST")
{
$name = test_input($_POST["name"]);
}
?>
<?php
//echo the new field entry
echo $name;
?>