如何在单击行特定按钮时将从数据库获取的记录值传递到另一个PHP页面

Question

I m trying to pass the email fetched from database to next page based on the row button clicked. But i m not able to use inside the value attribute of tag.

i have created a button for each row. for any button clicked i want to move to a common page with row email. the email varies based on selected button.

while($row= mysqli_fetch_assoc($results))
{
   $_email=$row["email"];
   echo  "<tr><td>". $row["email"] ."</td><td>" . $row["name"] .
   "</td><td>". $row["credit"] ."</td><td>".
   '<form method="post" action="user.php">
    <input type="hidden" name="data" value="<?php echo $_email?>"/>
    <button><b>Home<b></button>
    </form>'. "</td></tr>";
}

the email corresponding to the row button should be avialable in user.php file using $_POST method. i m not getting email value when i m doing $data=$_POST["data"] in the user.php page

Answer 1

If you use a double quote outer wrapper and then single quotes inside that wrapper for the attributes, it easier to read and debug.

Also remember that if you have an assoc array and you want to place $row["email"] inside the double quoted string literal, you can leave the " off like this $row[email]

See your edited code below

   $_email=$row["email"];
   echo 
    "<tr>
        <td>$row[email]</td>
        <td>$row[name]</td>
       <td>$row[credit]</td>
       <td>
            <form method='post' action='user.php'>
                <input type='Hidden' name='data' value='$_email'/>
                <button><b>Home<b></button>
            </form>
        </td>
    </tr>";