CODEWITHSUNDEEP
phpmysqlhtmlsql
Michael Harvey
Michael Harvey

Reputation: 148

previous/next record button php/mysql woes

My question deals with my next/previous buttons. I can get my update/delete buttons to work, but I'm so ready to tear out my hair when dealing with the next/previous buttons. Any help would be spectacular! Here's my code. Also, I'm pretty new to PHP so if this is bad coding, please let me know and point me in the right direction so I can fix my mistakes. Thanks!!!

session_start();

include "connectionfile.php";

if (isset($_POST['fname']) &&
isset($_POST['lname']) &&
isset($_POST['email']) &&
isset($_POST['login']) &&
isset($_POST['password']) &&
isset($_POST['super']) &&
isset($_POST['foldername']))
{
$id = get_post('id');
$fname = get_post('fname');
$lname = get_post('lname');
$email = get_post('email');
$login = get_post('login');
$password = hash('sha256', get_post('password'));
$super = get_post('super');
$foldername = get_post('foldername');

if ($_POST['submit']==0){

$query = mysql_query("SELECT * FROM `Logins` WHERE ID < '".$id."'  ORDER BY ID DESC LIMIT 1;");
while($row = mysql_fetch_array($query)){
   $id = $row['ID'];
   $fname = $row['fname'];
   $lname = $row['lname'];
   $email = $row['email'];
   $login = $row['login'];
   $password = $row['password'];
   $super = $row['super'];
   $foldername = $row['foldername'];

}
}else if ($_POST['submit']==1){
$query = "UPDATE Logins SET fname = '$fname', lname='$lname', email='$email".'@carouselclinical.com'."', login='$login', password='$password', super='$super', foldername='$foldername'";
$query .= "WHERE ID = '$id';";
if (!mysql_query($query, $connect))
echo "INSERT failed: $query<br />" .
mysql_error() . "<br /><br />";

}else if($_POST['submit']==2){

$delete_query = "DELETE FROM Logins WHERE ID = '".$id."';";
mysql_query($delete_query);
$rc = mysql_affected_rows();
echo "Rows Affected " . $rc;

}


if ($_POST['submit']==3){

$query = mysql_query("SELECT * FROM `Logins` WHERE ID= '". $id ."' ORDER BY ID ASC LIMIT 1;");
while($row = mysql_fetch_array($query)){
   $id = $row['ID'];
   $fname = $row['fname'];
   $lname = $row['lname'];
   $email = $row['email'];
   $login = $row['login'];
   $password = $row['password'];
   $super = $row['super'];
   $foldername = $row['foldername'];

}

}
}

mysql_close($connect);

function get_post($var)
{
return mysql_real_escape_string($_POST[$var]);
}
?>

<form action="" method="post"><pre>
id <input type="text" readonly="readonly" name="id" value="<?php echo "$id"; ?>" />
First Name <input type="text" name="fname" value="<?php echo "$fname"; ?>" />
Last Name <input type="text" name="lname" value="<?php echo "$lname"; ?>" />
Email <input type="text" name="email" value="<?php echo "$email"; ?>" /> There's no need to put @carouselclinical.com.
Login <input type="text" name="login" value="<?php echo "$login"; ?>"/>
Password <input type="text" name="password" value="<?php echo "$password"; ?>"/>
Super? <input type="text" name="super" value="<?php echo "$super"; ?>" />
foldername <input type="text" name="foldername" value="<?php echo "$foldername"; ?>" />
<button name="submit" value="0">Previous</button>
<button name="submit" value="1">UPDATE</button>
<button name="submit" value="2">Delete</button>
<button name="submit" value="3">Next</button>
</pre>
<a href="super.php">Home</a> <br />
<a href="logout.php">Log out</a>
</form>

Upvotes: 0

Views: 3855

Answers (1)

Sean
Sean

Reputation: 12433

Try adding an else right above mysql_close($connect);. My guess is that on the initial page load you are not posting any values, so no action is taken. This will create a default ID if none is defined in your top if.

else{
$query = mysql_query("SELECT * FROM `Logins` ORDER BY ID ASC LIMIT 1;");
while($row = mysql_fetch_array($query)){
  $id = $row['ID'];
  $fname = $row['fname'];
  $lname = $row['lname'];
  $email = $row['email'];
  $login = $row['login'];
  $password = $row['password'];
  $super = $row['super'];
  $foldername = $row['foldername'];
}

Also, on your if ($_POST['submit']==3), you need to change the = to > in your $query so you can get the next record. Currently you would be selecting the same ID, not the next higher.

$query = mysql_query("SELECT * FROM `Logins` WHERE ID > '". $id ."' ORDER BY ID ASC LIMIT 1;");

Finally, when doing Previous/Next you also need to take into consideration how you will deal with Previous when you are on the first ID, and Next when you are on the last id, as you will return an empty result set from MySQL.

Upvotes: 1

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