我的php表单仅保存表的最后一行。

Question

I have a table within a form (new_tear.php) that takes data from a single user. The first cloumn is populated from a database table (ProfileTable) with names. The user has to select an option from a dropdown for each row (each name). When the form is submitted (POST to tear_done.php), I want to store each row in my database. However, I am only seeing the last row posted each time

new_ter.php

<?php
$sqlT1 = "SELECT * FROM ProfileTable WHERE dept = 'T1'";
$queryT1 = mysqli_query($conn, $sqlT1);
if (!$queryT1) {
  die ('SQL Error: ' . mysqli_error($conn));
}
?>
<form action="tear_done.php" method="post">
<table>
<?php
    $no     = 1;
    while ($row = mysqli_fetch_array($queryT1))
    {
        echo '<tr>
                <td>'.$no.'</td>
                <td>'.$row['name'].'</td>
                <td><select id="Dz_M" name="Dz_M">
                    <option value="4pm">4pm</option>
                    <option value="8am">8am</option>
                    <option value="Rest">Rest</option>
                </select><br>
                </td>
                <td><select id="Dz_T" name="Dz_T">
                    <option value="4pm">4pm</option>
                    <option value="8am">8am</option>
                    <option value="Rest">Rest</option>
                </select><br>
                </td>
                <td><select id="Dz_W" name="Dz_W">
                    <option value="4pm">4pm</option>
                    <option value="8am">8am</option>
                    <option value="Rest">Rest</option>
                </select><br>
                </td>
                <td><select id="Dz_F" name="Dz_F">
                    <option value="4pm">4pm</option>
                    <option value="8am">8am</option>
                    <option value="Rest">Rest</option>
                </select><br>
                </td>
                <td><select id="Dz_S" name="Dz_S">
                    <option value="4pm">4pm</option>
                    <option value="8am">8am</option>
                    <option value="Rest">Rest</option>
                </select><br>
                </td>                   
                </tr>';
        $no++;
        $no_cont = $no;
    }?> 
</table>
<br>
<button value="save" type="submit" name="save">Save Tear</button>       
</form>

tear_done.php

<?php
$sqlT1 = "SELECT * FROM ProfileTable WHERE dept = 'T1'";
$queryT1 = mysqli_query($conn, $sqlT1);
if (!$queryT1) {
    die ('SQL Error: ' . mysqli_error($conn));
}
$no = 1;
while ($row = mysqli_fetch_array($queryT1))
{
$Name = $row ['name'];
$Dept = $row ['dept'];
$DZ_M = $_POST ["Dz_M"];
$DZ_T = $_POST ["Dz_T"];
$DZ_W = $_POST ["Dz_W"];
$DZ_F = $_POST ["Dz_F"];
$DZ_S = $_POST ["Dz_S"];

$sql = "INSERT INTO TearTable (name, Mdy, Tdy, Wdy, Fdy, Sdy, dept)
VALUES ('$Name', '$DZ_M', '$DZ_T', '$DZ_W', '$DZ_F', '$DZ_S', '$Dept')";    

echo $no;
echo '<br>';
echo $sql; //for testing
echo '<br>';
$no++;
}   
?>

What I am getting from the echo is the sql with all the names and dept correct, however the other fields $DZ_M, $DZ_T etc only repeat the last row entered from the table in new_tear.php. What is it am I missing please?

Answer 1

Inside While loop Just concat the insert query strings

while ($row = mysqli_fetch_array($queryT1))
{
......

$sql .= "INSERT INTO TearTable (name, Mdy, Tdy, Wdy, Fdy, Sdy, dept)
VALUES ('$Name', '$DZ_M', '$DZ_T', '$DZ_W', '$DZ_F', '$DZ_S', '$Dept');";    // Note .= and semicolon at the end 

   .....
    } 
// Outside the loop
execute your query here..

You should use prepared statements for security reasons

Answer 2

Best would be to create one single big query statement.

$sql .= "INSERT INTO TearTable (name, Mdy, Tdy, Wdy, Fdy, Sdy, dept) VALUES"
while ($row = mysqli_fetch_array($queryT1)) {  
  $Name = $row ['name'];
  $Dept = $row ['dept'];
  $DZ_M = $_POST ["Dz_M"];
  $DZ_T = $_POST ["Dz_T"];
  $DZ_W = $_POST ["Dz_W"];
  $DZ_F = $_POST ["Dz_F"];
  $DZ_S = $_POST ["Dz_S"]; 
  $sql .= "('$Name', '$DZ_M', '$DZ_T', '$DZ_W', '$DZ_F', '$DZ_S', '$Dept'),"
}
$result = mysqli_query($conn, $sql); //run your INSERT query.

This single query inserts all that data to the DB. The issue with this way of doing it is that it becomes harder to debug if your INSERT fails.

Answer 3

I was able to solve my problem by adding a var ($no) to the new_tear.php select id and name.

new_tear.php

<?php
    $no     = 1;
    while ($row = mysqli_fetch_array($queryT1))
    {
        echo '<tr>
                <td>'.$no.'</td>
                <td>'.$row['name'].'</td>
                <td><select id="Dz_M'  . $no . '" name="Dz_M'  . $no . '">
                    <option value="4pm">4pm</option>
                    <option value="8am">8am</option>
                    <option value="Rest">Rest</option>
                </select><br>
                </td>
                <td><select id="Dz_T'  . $no . '" name="Dz_T'  . $no . '">
                    <option value="4pm">4pm</option>
                    <option value="8am">8am</option>
                    <option value="Rest">Rest</option>
                </select><br>
                </td>
                <td><select id="Dz_W'  . $no . '" name="Dz_W'  . $no . '">
                    <option value="4pm">4pm</option>
                    <option value="8am">8am</option>
                    <option value="Rest">Rest</option>
                </select><br>
                </td>
                <td><select id="Dz_F'  . $no . '" name="Dz_F'  . $no . '">
                    <option value="4pm">4pm</option>
                    <option value="8am">8am</option>
                    <option value="Rest">Rest</option>
                </select><br>

                </td>
                <td><select id="Dz_S'  . $no . '" name="Dz_S'  . $no . '">
                    <option value="4pm">4pm</option>
                    <option value="8am">8am</option>
                    <option value="Rest">Rest</option>
                </select><br>
                </td>                   
                </tr>';
        $no++;
        $no_cont = $no;
    }?> 

And adding the var ($no) to the $_POST in tear_done.php

tear_done.php

$Name = $row ['name'];
$Dept = $row ['dept'];
$DZ_M = $_POST ['Dz_M'.$no]; 
$DZ_T = $_POST ['Dz_T'.$no];
$DZ_W = $_POST ['Dz_W'.$no];
$DZ_F = $_POST ['Dz_F'.$no];
$DZ_S = $_POST ['Dz_S'.$no];

This solved my problem and now in the echo (again, only for testing) I see the sql looks correct. Thanks Much for your help everyone.