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aebstract

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    [SOLVED] login script issue

    in PHP Coding Help

    Posted

    It's all in one file, account.php. The form redirects back to the same page and if submit is hit that is when the top part goes in to play. I'll post all the php coding below so you can see. I have tried to echo out the id from the selection but it doesn't seem to be displaying a single thing. I really think the id isn't getting passed.

     

    <?php
session_start();  
header("Cache-control: private");   
if(isset($_SESSION["id"]))  
{  
header("Location: acchome.php");  
}

if(!isset($_SESSION["id"])) 
{

if (isset ($_POST['submit']))
{

$problem = FALSE;


      if (empty ($_POST['password'])) {
$problem = TRUE;
$error .= 'You must fill in a password <br />';
      }

$echoedid = $_POST['id'];
echo "$echoedid";


if (!$problem) {

mysql_connect("localhost","berryequipment","gU8Kso8Y") or die(mysql_error());
mysql_select_db("berryequipment_net");


$result = mysql_query("SELECT id, plantloc, password, address FROM plants WHERE id=('$_POST[id]') AND password=('".md5($_POST['password'])."')") or die ("error");
            if (mysql_num_rows($result) == 0)
              {
echo 'The pasword you entered did not match the plant location you chose.';
} else {

                      $worked = mysql_fetch_array($result);
    			    $_SESSION["id"] = $worked[id];

                      header("Location: acchome.php");
} 




}

}

}



?>

     

     

    and the form area

    
<?php



echo "$error";


$content .= '

<form action="account.php" method="post">
   plant loc<br />

<select name="dropdown">';

mysql_connect("localhost","berryequipment","gU8Kso8Y") or die(mysql_error());
mysql_select_db("berryequipment_net");

$result = mysql_query("SELECT * FROM plants ORDER BY plantloc ASC") or DIE(mysql_error());
while ($r=mysql_fetch_array($result)) {

$id=$r["id"];
$plantloc=$r["plantloc"];

$content .= "<option value=\"$id\">$plantloc</option>\n";

} 


$content .= '</select>


             <br /><br />
   password <br />
   <input type="password" maxlength="6" class="textfield" name="password" value="' . $_POST[password] . '" size="6" /><br /><br />
   <input type="submit" name="submit" class="textfield" value="login" />


</form>';


echo "$content";



?>

    [SOLVED] login script issue

    in PHP Coding Help

    Posted

    I have been working around looking for the problem still and have come to a more solid conclusion that the id variable isn't being passed from that drop down to my other part of the script. I changed the error coding like you said, and it IS displaying the error of "password doesn't match". Which proves that the ID isn't passing correctly even more.

    [SOLVED] login script issue

    in PHP Coding Help

    Posted

    Still has an issue with logging in, was just putting an update on how it is displaying the drop down in case someone wanted to point the for loop as an issue.

     

    Here is the code that is trying to connect and match up the variable, maybe it will help out:

     

    $result = mysql_query("SELECT id, plantloc, password, address FROM plants WHERE id=('$_POST[id]') AND password=('".md5($_POST['password'])."')") or die ("error");
            if (mysql_num_rows($result) == 0)
              {
  $echo = 'The pasword you entered did not match the plant location you chose. <br />';
} else {

                      $worked = mysql_fetch_array($result);
    			    $_SESSION["id"] = $worked[id];

                      header("Location: acchome.php");
}

    [SOLVED] login script issue

    in PHP Coding Help

    Posted

    Trying to a get login function to work. I have gone through several areas of the code to try and limit the problem to one specific area. I'm passing up the password fine, but I think the 'id' from the database isn't getting passed up from the drop down. I put an echo in the top to display the id and when I hit submit, nothing displays.  ???

     

    the code:

    <select name="dropdown">';

mysql_connect("localhost","berryequipment","gU8Kso8Y") or die(mysql_error());
mysql_select_db("berryequipment_net");

$result = mysql_query("SELECT * FROM plants ORDER BY plantloc ASC") or DIE(mysql_error());
$allplants = mysql_num_rows($result); 
while ($r=mysql_fetch_array($result)) {

$id=$r["id"];
$plantloc=$r["plantloc"];
for ($op = 1; $op <= $allplants; $op++) {
$content .= "<option value=\"$id\">$plantloc</option>\n";


}
} 


$content .= '</select>

    form passing variable

    in PHP Coding Help

    Posted

    Okay, sorry I keep having these issues. I think my whole drop down code is pretty messed up right now. So basically I should be able to login with this script, which is pulling the plant locations and then checking with what password they put in. I don't think my drop down is passing the information through when you hit submit. It needs to let me know what id is being chosen from the drop down and then I am going to compare that to that plants id/password in the database.

     

    mysql_connect("localhost","berryequipment","gU8Kso8Y") or die(mysql_error());
mysql_select_db("berryequipment_net");
$result = mysql_query("SELECT * FROM plants ORDER BY plantloc ASC") or DIE(mysql_error());

$options="";

while ($r=mysql_fetch_array($result)) {

$id=$r["id"];
$plantloc=$r["plantloc"];
    $options .= "<OPTION VALUE=\"$id\">".$plantloc;
} 

?>



<form action="account.php" method="post">
   plant loc<br />

<SELECT NAME=dropdown>
<OPTION VALUE=0>
<?=$options?>
</SELECT>

    [SOLVED] trouble generating dropdown from mysql

    in PHP Coding Help

    Posted

    Alright, I am generating content for a drop down based on the information that I have stored in a database. It is very basic information, it is a table with "id", "plantloc", "password", and "address". What I want to do is generate a drop down that contains every plant. It should reference by id and should display as "plantloc". The plantloc is a nice looking description of the location, this way the user chooses their location and continues to login.

     

    Here is what I have:

     

    <?php
mysql_connect("localhost","berryequipment","gU8Kso8Y") or die(mysql_error());
mysql_select_db("berryequipment_net");
$result = mysql_query("SELECT * FROM plants ORDER BY plantloc ASC") or DIE(mysql_error());

$options="";

while ($row=mysql_fetch_array($result)) {

$id=$r["id"];
$plantloc=$r["plantloc"];
    $options .= "<OPTION VALUE=\"$id\">".$plantloc;
} 

?>



<form action="account.php?menu=login" method="post">
   plant loc<br />

<SELECT NAME=dropdown>
<OPTION VALUE=0>
<?=$options?>
</SELECT>
<br /><br />
   password <br />
   <input type="password" maxlength="6" class="textfield" name="password" size="6" /><br /><br />
   <input type="submit" name="submit" class="textfield" value="login" />


</form>

    [SOLVED] info not going to db, email not sending

    in PHP Coding Help

    Posted

    I'm back on the phone right now, will ask and I found this in my php.ini:

    [mail function]
; For Win32 only.
SMTP = localhost

; For Win32 only.
sendmail_from = me@localhost.com

; For Unix only.  You may supply arguments as well (default: "sendmail -t -i").
sendmail_path = /usr/sbin/sendmail -t -i

     

    I'm thinking it may be wrong "me@localhost.com" and I'm on a .net domain. So maybe it isn't sending out correctly? I'll come back with another update soon.

    [SOLVED] info not going to db, email not sending

    in PHP Coding Help

    Posted

    Alright I called in because I looked at my error log and had nothing dealing with mail in it. Couldn't find my php.ini file so he placed one in my html directory. I don't know anything about the php.ini file so I have no clue how to find what I need or check what it should be. He did a simple one line mail function test and said it was working so it may be my syntax?

    [SOLVED] info not going to db, email not sending

    in PHP Coding Help

    Posted

    Hi again, back so soon! I put a else die on the end of my connecting to DB.. so I am 99% sure that isn't my problem here. No information is being placed in the database and no email is being sent out. Any ideas on why not?

     

    <?php
session_start();
header("Cache-control: private");

if (isset ($_POST['submit'])) {
$problem = FALSE;



if (empty ($_POST['plantloc'])) {
$problem = TRUE;
$error .= 'Must enter a plant location<br />';
}







mysql_connect("*","*","*");
mysql_select_db("*");


if (!$problem) {

$plantloc  = $_POST['plantloc'];

$length = 6;
$password = "";
$possible = "0123456789bcdfghjkmnpqrstvwxyz"; 
$i = 0; 
    
while ($i < $length) { 
    $char = substr($possible, mt_rand(0, strlen($possible)-1), 1);
    if (!strstr($password, $char)) { 
      $password .= $char;
      $i++;
    }

  }

$password2 = md5($_POST['password1']);
$address = $_POST['address'];


$result = MYSQL_QUERY("INSERT INTO plants (plantloc,password,address)".
"VALUES ('$plantloc, '$password2', '$address,')");


$sendmail = "
$password2
";

mail ('tcantwell@berryequipment.net', 'Thank You', $sendmail, 'From: aebstract@gmail.com');



} else {
$content .= "$error";
}
}
?>

    [SOLVED] form isnt passing my variable?

    in PHP Coding Help

    Posted

    Hey guys, been awhile since I have been around these parts, but I am having a little trouble.

    I got a form with two simple inputs. I put a check at the top of my page to see if the field has information entered or not. Well it is returning false when it has information and when it doesn't.

     

    Here is what I have:  (file - account2.php)

     

    <?php
session_start();
header("Cache-control: private");

if (isset ($_POST['submit'])) {
$problem = FALSE;



if (empty ($_POST['plantloc'])) {
$problem = TRUE;
$error .= 'Must enter a plant location<br />';
}







mysql_connect("*","*","*");
mysql_select_db("*");


if (!$problem) {

$plantloc  = $_POST['plantloc'];

$length = 6;
$password = "";
$possible = "0123456789bcdfghjkmnpqrstvwxyz"; 
$i = 0; 
    
while ($i < $length) { 
    $char = substr($possible, mt_rand(0, strlen($possible)-1), 1);
    if (!strstr($password, $char)) { 
      $password .= $char;
      $i++;
    }

  }

$password2 = md5($_POST['password1']);
$address = $_POST['address'];


$result = MYSQL_QUERY("INSERT INTO plants (plantloc,password,address)".
"VALUES ('$plantloc, '$password2', '$address,')");


$sendmail = "
$password2
";

mail ('tcantwell@berryequipment.net', 'Thank You', $sendmail, 'From: aebstract@gmail.com');



} else {
$content .= "$error";
}
}
?>

     

     

     

     

    and the form area:

     

    
<?php
$content .= '<form action="account2.php?menu=create" method="post">
<br />
Plant Location: <br /><input type="text" maxlength="100" class="textfield" name="username" size="40" value="' . $_POST[plantloc] . '" /><br /><br />

Address:<br />
<textarea cols="50" rows="4" name="address" value="' . $_POST[address] . '"></textarea>
<br /><br />

<input type="submit" name="submit" value="submit" class="textfield" /></form>';



echo "$content";
?>

    [SOLVED] 403 forbidden error

    in PHP Coding Help

    Posted 

    <?php



if (isset($work)) {

echo "

<img src=\"/print/$work.png\" />

";

} else {

echo "
<div id=\"thumbs\">
<a href=\"/print/1/\"><img src=\"/thumb.png\" border=\"0\" /></a><img src=\"/spacerw.png\" /><img src=\"/thumb.png\" /><img src=\"/spacerw.png\" /><img src=\"/thumb.png\" /><img src=\"/spacerw.png\" /><img src=\"/thumb.png\" /><img src=\"/spacerw.png\" /><img src=\"/thumb.png\" /><img src=\"/spacerw.png\" /><img src=\"/thumb.png\" /><br /><br />

<img src=\"/thumb.png\" /><img src=\"/spacerw.png\" /><img src=\"/thumb.png\" /><img src=\"/spacerw.png\" /><img src=\"/thumb.png\" /><img src=\"/spacerw.png\" /><img src=\"/thumb.png\" /><img src=\"/spacerw.png\" /><img src=\"/thumb.png\" /><img src=\"/spacerw.png\" /><img src=\"/thumb.png\" /><br /><br />

<img src=\"/thumb.png\" /><img src=\"/spacerw.png\" /><img src=\"/thumb.png\" /><img src=\"/spacerw.png\" /><img src=\"/thumb.png\" /><img src=\"/spacerw.png\" /><img src=\"/thumb.png\" /><img src=\"/spacerw.png\" /><img src=\"/thumb.png\" /><img src=\"/spacerw.png\" /><img src=\"/thumb.png\" /><br /><br />
</div>
";

}


?>

     

    This code is included in to an index.php file and whenever I go to /print/ it gives the forbidden error (that is the bottom half of the script) and whenever I go to /print/1/ the top half of the script works and it all displays correctly.

    www.carbenco.com and click print, the first thumbnail is clickable (though it prob. won't show the page without hitting that error)

    www.carbenco.com/print/ - problem

    www.carbenco.com/print/1/ - no problem

     

    Thanks!

    [SOLVED] simple php code displaying white page

    in PHP Coding Help

    Posted

    Eh, I don't see a problem with it and I have isolated it down to this chunk of code, so maybe someone can see what is causing it. I know there are probably shorter ways to do this same code, but for now just help getting the error out if you don't mind. Thanks!

     

    <?php


if ($page == home) {
echo '<div id="nav1"><a href="/print/"><img src="/printx.png" border="0" /></a></div>
<div id="nav2"><a href="/identity/"><img src="/identityxx.png" border="0" /></a></div>
<div id="nav3"><a href="/web/"><img src="/webx.png" border="0" /></a></div>
<div id="nav4"><a href="/home/"><img src="/home.png" border="0" /></a></div>';
} elseif ($page == web) {
echo '<div id="nav1"><a href="/print/"><img src="/printx.png" border="0" /></a></div>
<div id="nav2"><a href="/identity/"><img src="/identityx.png" border="0" /></a></div>
<div id="nav3"><a href="/web/"><img src="/web.png" border="0" /></a></div>
<div id="nav4"><a href="/home/"><img src="/homex.png" border="0" /></a></div>';
} elseif ($page == identity) {
echo '<div id="nav1"><a href="/print/"><img src="/printxx.png" border="0" /></a></div>
<div id="nav2"><a href="/identity/"><img src="/identity.png" border="0" /></a></div>
<div id="nav3"><a href="/web/"><img src="/webxx.png" border="0" /></a></div>
<div id="nav4"><a href="/home/"><img src="/homexx.png" border="0" /></a></div>';
} elseif ($page == print) {
echo '<div id="nav1"><a href="/print/"><img src="/print.png" border="0" /></a></div>
<div id="nav2"><a href="/identity/"><img src="/identityxxx.png" border="0" /></a></div>
<div id="nav3"><a href="/web/"><img src="/webxxx.png" border="0" /></a></div>
<div id="nav4"><a href="/home/"><img src="/homexx.png" border="0" /></a></div>';
} else {
echo '<div id="nav1"><a href="/print/"><img src="/printx.png" border="0" /></a></div>
<div id="nav2"><a href="/identity/"><img src="/identityxx.png" border="0" /></a></div>
<div id="nav3"><a href="/web/"><img src="/webx.png" border="0" /></a></div>
<div id="nav4"><a href="/home/"><img src="/home.png" border="0" /></a></div>';
}
?>

    [SOLVED] php pages & htaccess

    in PHP Coding Help

    Posted

    Put it in, index page loads fine, click a link and it 404's.

    Just so we're on the same page with the index.php file, here is what I'm using in that php section:

    <?php

$page = (isset($_GET['page'])) ? $_GET['page'] : 'home';

$page_path = $_SERVER['DOCUMENT_ROOT'] . '/' . $page . '.php';

if(file_exists($page_path))
{
    include $page_path;
}
else
{
    die("$page_path does not exist");
}

?>

     

     

     

    EDIT:

    Okay error was in my linking, had a / on the end, which I can easily put in to the .htaccess, what about the /1/ part, adding another variable to the url, is that hard?

    [SOLVED] php pages & htaccess

    in PHP Coding Help

    Posted

    somehow my index file just now got corrupted to:

    +ADwAIQ-DOCTYPE html
PUBLIC +ACI--//W3C//DTD XHTML 1.0 Strict//EN+ACI
+ACI-http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd+ACIAPg
+ADw-html xmlns+AD0AIg-http://www.w3.org/1999/xhtml+ACI xml:lang+AD0AIg-en+ACIAPg
+ADw-head+AD4
+ADw-title+AD4-Carbenco Designs+ADw-/title+AD4

+ADw-link href+AD0AIg-/stylesheet.css+ACI rel+AD0AIg-stylesheet+ACI type+AD0AIg-text/css+ACI title+AD0AIg-default+ACI /+AD4
+ADw-/head+AD4
+ADw-body+AD4

+ADw-div id+AD0AIg-container+ACIAPg

+ADw-div id+AD0AIg-top+ACIAPgA8-img src+AD0AIg-logo.png+ACI /+AD4APA-/div+AD4
+ADw-div id+AD0AIg-middle+ACIAPgA8-div id+AD0AIg-midleft+ACIAPgA8-img src+AD0AIg-midleft.png+ACI /+AD4APA-/div+AD4

+ADw-div id+AD0AIg-midmid+ACIAPg
+ADw-div id+AD0AIg-midmidtop+ACIAPgA8-img src+AD0AIg-midtop.png+ACI /+AD4APA-/div+AD4
+ADw-div id+AD0AIg-midmidcontent+ACIAPg

+ADw-div id+AD0AIg-nav+ACIAPg
+ADw-div id+AD0AIg-nav1+ACIAPgA8-a href+AD0AIg-/print/+ACIAPgA8-img src+AD0AIg-print.png+ACI border+AD0AIg-0+ACI /+AD4APA-/a+AD4APA-/div+AD4
+ADw-div id+AD0AIg-nav2+ACIAPgA8-a href+AD0AIg-/indentity/+ACIAPgA8-img src+AD0AIg-identityxxx.png+ACI border+AD0AIg-0+ACI /+AD4APA-/a+AD4APA-/div+AD4
+ADw-div id+AD0AIg-nav3+ACIAPgA8-a href+AD0AIg-/web/+ACIAPgA8-img src+AD0AIg-webxxx.png+ACI border+AD0AIg-0+ACI /+AD4APA-/a+AD4APA-/div+AD4
+ADw-div id+AD0AIg-nav4+ACIAPgA8-a href+AD0AIg-/home/+ACIAPgA8-img src+AD0AIg-homexx.png+ACI border+AD0AIg-0+ACI /+AD4APA-/a+AD4APA-/div+AD4
+ADw-/div
+ADw-div id+AD0AIg-mmiddle+ACIAPg
+ADw-img src+AD0AIg-middle.png+ACI /+AD4
+ADw-/div+AD4
+ADw-div id+AD0AIg-art+ACIAPg


hi


+ADw-/div+AD4
+ADw-/div+AD4
+ADw-div id+AD0AIg-midmidbottom+ACIAPgA8-img src+AD0AIg-midbottom.png+ACI /+AD4APA-/div+AD4
+ADw-/div+AD4

+ADw-div id+AD0AIg-midright+ACIAPgA8-img src+AD0AIg-midright.png+ACI /+AD4APA-/div+AD4APA-/div+AD4
+ADw-div id+AD0AIg-bottom+ACIAPgA8-/div+AD4

+ADw-/div+AD4

+ADw-/body+AD4
+ADw-/html+AD4

    What is up with that?

     

     

     

    EDIT:  I went back and redid/saved everything. I think it was transfering it over to a weird style when saving so I made it save and not change a thing, its working now. Could you explain the part I was wondering about doing something like carbenco.com/home/1/ having a second variable if I wanted? Thanks for your help!

     

    EDIT2: Go to carbenco.com and look at the links, it's still bugging up. Not as bad, but a 404. So it isn't finding the variable or something?

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