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need help w/mysqli update

Posted: Wed Dec 03, 2014 8:31 pm
by limited
[text]Hi,
My project is to: select record from database menu(working),
click(?) that selection to update field1, field2
and link to the associated location. I'd appreciate any guidance.[/text]
==================================================
this is update code:

Code: Select all

<?php
$servername = "localhost";$username = "root";$password = "xxxx";
$dbname = "xxxxDB";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn)
 {die("Connection failed: " . mysqli_connect_error());}
    $id = $_POST['id'];
    $lastused = $_POST['lastused']; 
    $visits = $_POST['visits']; 
    $name = $_POST['target'];
  $sql = "UPDATE emailtbl SET visits = visits + 1, lastused = NOW() WHERE target = '".$name."'";
if (mysqli_query($conn, $sql))
 {echo "you done good";}
else
 {echo "back up charlie: " . mysqli_error($conn);}
?> 
====================================================
this code connects, selects & displays the desired record:

<html><title>email menu</title>
<head></head>
<BODY><center>
<FORM name=form method="post" action="">

Code: Select all

<?php 
include ('gethomedb.php');
// ==========This creates the drop down box using records in the table
       echo "<select name = 'target'>";
    echo '<option value="">'.'---select email account ---'.'</option>';
    $query = mysqli_query($con,"SELECT target FROM emailtbl");
    $query_display = mysqli_query($con,"SELECT * FROM emailtbl");
    while($row=mysqli_fetch_array($query))
    { echo "<option class=highlight value='". $row['target']."'>".$row['target']
    .'</option>';}
    echo '</select>';
    ?>
<input type="submit" name="submit" value="Submit"/>
   </form>
      <?php
        if(isset($_POST['target']))
 {
    $id = $_POST['id'];
    $visits = $_POST['visits'];
    $lastused = $_POST['lastused'];        
    $name = $_POST['target'];
    $fetch="SELECT target, username, password, emailused, visits, lastused,  
    purpose, saved FROM emailtbl WHERE target = '".$name."'";
    $result = mysqli_query($con,$fetch);
    if(!$result)
    {echo "Error:".(mysqli_error($con));}   
// =============================== this displays the table
    echo '<table border="1">'.'<tr>'.'
    <td bgcolor="#FFD47F" align="center">'. 'email menu'. '</td>'.'</tr>';
    echo '<tr>'.'<td>'.'<table border="1">'.'<tr>'.'
<td bgcolor="#ccffff">'.'target'.'</td>'.'
<td bgcolor="#ccffff">'.'username'.'</td>'.'
<td bgcolor="#ccffff">'. 'password' .'</td>'.'
<td bgcolor="#ccffff">'. 'emailused'. '</td>'.'
<td bgcolor="#FFD47F">'.'visits' .'</td>'.'
<td bgcolor="#FFD47F">'.'lastused' .'</td>'.'
<td bgcolor="#ccffff">'. 'purpose'. '</td>'.'
<td bgcolor="#ccffff">'. 'saved' .'</td>'.'</tr>';
    
    while($data=mysqli_fetch_row($result))

    {echo ("<tr><td>$data[0]</td><td>$data[1]</td><td>$data[2]</td>
    <td>$data[3]</td><td>$data[4]</td><td>$data[5]</td><td>$data[6]</td>
    <td>$data[7]</td></tr>");}
    echo '</table>'.'</td>'.'</tr>'.'</table>';
    }
    ?>
</body></html>

Re: need help w/mysqli update

Posted: Thu Dec 04, 2014 6:36 am
by Celauran
Is there a question in here somewhere?