CODEWITHSUNDEEP
phphtmlmysql
user2884789
user2884789

Reputation: 533

Getting data to display in a form

I'm trying to use HTML, PHP and MYSQL to pull data from a database and display it in a form (to later be edited). At this point I'm only trying to pull that data and display it in a form. (I'll worry about updating later). I pull the data but nothing displays in my textboxes:

<?php 
$con = mysqli_connect("XXXXX");   //removed for privacy

if (mysqli_connect_errno())
{
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$query="select * from VOLUNTEER";
echo '$query';
$result = mysqli_query($con, $query);

echo "<table>";

if ($result)                                   
{                        
    while($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) 
{
    echo '<form method = "post" action="insertvolunteer.php">';
    echo '<tr>';
    echo '<td>First Name:</td>';
    echo '<td>' . '<input type=text name=FirstName' . $row["FirstName"] . '</td>';
    echo '<td>' . '<input type=hidden name=VolunteerId' . $row["VolunteerId"] .  '</td>';
echo '</tr>';
}
}
 echo "</form>";
 echo "</table>";                             
 mysqli_close($con);
 ?>

Upvotes: 0

Views: 75

Answers (2)

rafaeldefazio
rafaeldefazio

Reputation: 449

connect.php

<?php

$server = "server";
$user = "user";
$password = "password";
$bd = "yourbd";

$connect = mysql_connect($server, $user, $password);
$connect = mysql_select_db("$bd", $connect);

if  (!$connect){
echo mysql_error(); exit;
}

?>

namefile.php

<?php
include('connect.php');

    $select = mysql_query("select * from VOLUNTEER");

    while ($show  = mysql_fetch_assoc($select)):

    echo "<table>";
    echo "<form method = 'post' action='insertvolunteer.php'>";
     echo '<tr>';
    echo '<td>First Name:</td>';
    echo '<td><input type="text" name="FirstName" value="'.$show["FirstName"].'"></td>';
    echo '<td><input type="text" name="FirstName" value="'.$show["VolunteerId"].'"></td>';
    echo '</tr>';
    echo "</form";
    echo "</table>";

    endwhile;

?>

When you create an MySQL query, you need to declare this. How?

$var = "SELECT * FROM SOMEWHERE"; wrong

$var = mysql_query("SELECT * FROM SOMEWHERE"); right

'n in echo '<td>' . '<input type=text name=FirstName' . $row["FirstName"] . '</td>';, you need to close the tag. And also have no need of separate <td> of <input>.

Try something like :)

@update I realized that you've got what was wished. Cheers.

Upvotes: 0

Abhik Chakraborty
Abhik Chakraborty

Reputation: 44844

Text box data needs to be displayed on value as

echo '<td><input type="text" name="FirstName" value="'.$row["FirstName"].'"></td>';

Upvotes: 2

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