Question 8362 in PHP Versions Question posted in
You can find a list of all versions here.

How to update multiple rows in mysql and php with a different values using variables and different files? – PHP Versions

I am trying to make a discord bot in PHP (which works, i know not the best choice of language but that is what i am most comfortable with for my own needs for now) with a dashboard.
The idea was to be able to add and edit the commands on the website.

I am selecting my current commands using mysql for the form like this:

if ($result->num_rows > 0) {
                // output data of each row
                while($row = $result->fetch_assoc()) {
                  $disable = $row['Disabled'];
                  echo "<input type='hidden' name='id[]' value='";
                  echo $row['Id'];
                  echo "'>";
                  echo '<tr>';
                  echo '<td class="tg-515c"><input type="text" name="command[]" value="';
                  echo $row['Command'];
                  echo '" class="form-control"></td>';
                  echo '<td class="tg-wp8o"><input type="text" name="text[]" value="';
                  echo $row['Text'];
                  echo '" class="form-control"></td>';
                  echo '<td class="tg-wp8o"><select class="form-control" name="disabled[]">';
                  if($disable == '0')
                  {
                    echo "<option value='1'>Yes</option><option value='0' selected>No</option>";
                  }
                  else
                  {
                    echo "<option value='1' selected>Yes</option><option value='0'>No</option>";
                  }
                  echo "</select>";
                  echo '</td>';
                echo '</tr>';
                }

and i have the form outside of my php syntax.
That seems to be working fine.

// Create connection
$command = $_POST['command'];
$text = $_POST['text'];
$id = $_POST['id'];
$disable = $_POST['disabled'];
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
  die("Connection failed: " . $conn->connect_error);
}

$sql="UPDATE dashboard SET Command='$command[0]', Text='$text[0]', Disabled='$disable[0]' WHERE Id='$id[0]'";

if ($conn->query($sql) === TRUE) {
    echo "Record updated successfully";
  } else {
    echo "Error updating record: " . $conn->error;
  }

header("Location: /");
$conn->close();

I have just copy/pasted this code few times and changed the array from [0] to [1], etc.

This now works, however, i have another file which adds more commands to database, so i need it to be automatized each time i add a new command and not manually type for each input.

Edited the code to updated version.

Tags:

2

Answers


  1. Chosen as BEST ANSWER
    0

    I was able to figure out a solution by doing this:

    <?php
$servername = "xxxt";
$username = "xxx";
$password = "xxx";
$dbname = "xxx";
$command = $_POST['command'];
$text = $_POST['text'];
$id = $_POST['id'];
$disable = $_POST['disabled'];
$count = count($id);
for ($x = 0; $x <= $count; $x++) {
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
  die("Connection failed: " . $conn->connect_error);
}

$sql = "UPDATE dashboard SET Command='$command[$x]', Text='$text[$x]', Disabled='$disable[$x]' WHERE Id='$id[$x]'";

if ($conn->query($sql) === TRUE) {
    echo "Record updated successfully";
  } else {
    echo "Error updating record: " . $conn->error;
  }

$conn->close();
}
header("Location: /");
?>

    So i decided to do a for loop and count how many inputs i have, so for each input, it automatically creates a new query updating it.


  2. 0

    I made an example for you that I think works for you and it works now and I tried it

    <?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "createde";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
  die("Connection failed: " . $conn->connect_error);
}


// When you press the change button, it will completely change the data
        if(isset($_POST["sub"])){
            $command = $_POST['command'];
            $text =  $_POST['text'];
            $id = "1";
            $disable = "0";
        
            $sql="UPDATE deew SET Command='$command', Text='$text', Disabled='$disable' WHERE id='$id'";
            $conn->query($sql);
        
        
        }

$conn->close();

?>

    The second page is where the values are entered

    <?php
include("sql.php");
?>
<!DOCTYPE html>
<html lang="en">
<head>
    <meta charset="UTF-8">
    <meta http-equiv="X-UA-Compatible" content="IE=edge">
    <meta name="viewport" content="width=device-width, initial-scale=1.0">
    <title>Document</title>
</head>
<body>

<form action="#" method="POST">
<textarea name="command" id="command" cols="30" rows="10"></textarea>
<input type="text" name="text" >
<input type="submit" name="sub">

</form>
    
</body>
</html>

    I know that your code calls all the data and determines through them what data you want to change

    Login or Signup to reply.
Please signup or login to give your own answer.

Back To Top
Search