PHP - Simple Script To Insert Punctuation Not Working....

Hi i have this upload script which works fine it uploads image to a specified folder and sends the the details to the database.
but now i am trying to instead make a modify script which is Update set so i tried to change insert to update but didnt work can someone help me out please
this my insert image script which works fine but want to change to modify instead
Code: [Select]
<?php
mysql_connect("localhost", "root", "") or die(mysql_error()) ;
mysql_select_db("upload") or die(mysql_error()) ;

// my file the name of the input area on the form type is the extension of the file
//echo $_FILES["myfile"]["type"];

//myfile is the name of the input area on the form 
$name = $_FILES["image"] ["name"]; // name of the file
$type = $_FILES["image"]["type"]; //type of the file
$size = $_FILES["image"]["size"]; //the size of the file
$temp = $_FILES["image"]["tmp_name"];//temporary file location when click upload it temporary stores on the computer and gives it a temporary name
$error =array(); // this an empty array where you can then call on all of the error messages
$allowed_exts = array('jpg', 'jpeg', 'png', 'gif'); // array with the following extension name values
$image_type = array('image/jpg', 'image/jpeg', 'image/png', 'image/gif'); // array with the following image type values
$location = 'images/'; //location of the file or directory where the file will be stored
$appendic_name = "news".$name;//this append the word [news] before the name so the image would be news[nameofimage].gif

// substr counts the number of carachters and then you the specify how how many you letters you want to cut off from the beginning of the word example drivers.jpg it would cut off dri, and would display vers.jpg
//echo $extension = substr($name, 3);

//using both substr and strpos, strpos it will delete anything before the dot in this case it finds the dot on the $name file deletes and + 1 says read after the last letter you delete because you want to display the letters after the dot. if remove the +1 it will display .gif which what we want is just gif
$extension = strtolower(substr($name, strpos ($name, '.') +1));//strlower turn the extension non capital in case extension is capital example JPG will strtolower will make jpg
// another way of doing is with explode
// $image_ext strtolower(end(explode('.',$name))); will explode from where you want in this case from the dot adn end will display from the end after the explode

$myfile = $_POST["myfile"];

     if (isset($image)) // if you choose a file name do the if bellow
       {
       
       // if extension is not equal to any of the variables in the array $allowed_exts error appears
        if(in_array($extension, $allowed_exts) === false )
       {
         $error[] = 'Extension not allowed! gif, jpg, jpeg, png only<br />'; // if no errror read next if line
       }
        // if file type is not equal to any of the variables in array $image_type error appears
        if(in_array($type, $image_type) === false)
       {
          $error[] = 'Type of file not allowed! only images allowed<br />';     
       }
       
        // if file bigger than the number bellow error message
        if($size > 2097152)
       {
          $error[] = 'File size must be under 2MB!';     
       }
       
       // check if  folder exist in the server
        if(!file_exists ($location))
       {
          $error[] = 'No directory ' . $location. ' on the server Please create a folder ' .$location;     
       }
            
     }
     // if no error found do the move upload function
       if (empty($error)){
           if (move_uploaded_file($temp, $location .$appendic_name))
           {
             // insert data into database first are the field name teh values are the variables you want to insert into those fields appendic is the new name of the image
mysql_query("INSERT INTO image (myfile ,image)
 VALUES ('$myfile', '$appendic_name')") ;
           exit();
           }
         }
       else
          {
        foreach ($error as $error)
           {
               echo $error;
           }
        }
      
     
  
//echo $type;
?>

Hello, im very green to php and I am having trouble creating a simple log in script. Not sure why this is not working, maybe a mysql_query mistake? I am not receiving any errors but nothing gets updated in the members table and my error message to the user displays.  any help is appreciated! here is my php:


<?php
session_start();

$errorMsg = '';
$email = '';
$pass = '';

if (isset($_POST['email'])) {
   $email = ($_POST['email']);
   $pass = ($_POST['password']);
   
   $email = stripslashes($email);
    $pass = stripslashes($pass);
   $email = strip_tags($email);
   $pass = strip_tags($pass);
   
if ((!$email) || (!$pass)) {
   $errorMsg = '<font color="#FF0000">Please fill in both fields</font>';
}else {
   include 'scripts/connect_db.php';
   $email = mysql_real_escape_string ($email);
   $pass = md5($pass);
   
   $sql = mysql_query("SELECT * FROM members WHERE email='$email' AND password='$pass'");
   
   $log_check = mysql_num_rows($sql);
   
if ($log_check > 0) {
    while($row = mysql_fetch_array($sql)) {
      
      $id = $row["id"];
      $_SESSION['id'];
      $email = $row["email"];
      $_SESSION['email'];
      $username = $row["username"];
      $_session['username'];
      
      mysql_query("UPDATE members SET last_logged=now() WHERE id='$id' LIMIT 1");
      
   }//Close while loop
   echo "You are logged in";
             exit();
}
else {
   $errorMsg = '<font color="#FF0000">Incorrect login data, please try again</font>';
}


     }
}
?>



and the form:

<?php echo $errorMsg; ?>
<form action="log_in.php" method="post">
Email:<br />
<input name="email" type="text" /><br /><br />
Password:<br />
<input name="password" type="password" /><br /><br />
<input name="myBtn" type="submit" value="Log In" />
</form>

Alright so I am trying to code a very simple form that adds a user to my SQL database with certain information filled in and certain information kept hidden with default values in a form.

My index.php file is as follows (not all of the code but just the form part):

<?PHP
$submitMessage = $_GET['value'];
if ($submitMessage == success)
{
echo '<div class="congratulations"><font color="#84d20c"><strong>Congratulations:</strong></font> The user has been added.</div>';
}else if($submitMessage == NULL){
}else if($submitMessage == fail){
echo '<div class="error"><font color="#d3430c"><strong>Error:</strong></font> An error occurred please check the error details below:</div>';
echo 'ERROR DETAILS:';
echo mysql_error();
}
?>
<h1><i>Add User</i></h1>
<p>This page is where you would manually add a user to the donations databases (for example you or a member of the community with elevated status that does not need to pay).</p>
            <br/>

<form enctype="multipart/form-data" action="scripts/admin_save_sm_db.php" method="post">

<input name="authtype" value="steam"/><br /><br />

<h3><i>STEAM ID:</i></h3>
<input name="identity" value="STEAM_0:X:XXXXXXXX"/><br />
<div class="form_help">Steam ID of the player you wish to add. Must be in the format STEAM_0:X:XXXXXXXX</div><br />

<input name="flags" value="o"/><br /><br />

<input name="name" value="Donator"/><br /><br />

<input name="immunity" value="0"/><br /><br />

<h3><i>Your Email Address</i></h3>
<input name="user_email"/><br />
<div class="form_help">Your Email Address for security purposes.</div><br />
                
<input name="Submit" type="submit" value="Submit" class="form_submit" />
</form>



My admin_save_sm_db.php file is as follows:

<?php
session_start();
if(!isset($_SESSION["username"]))
{
   header('Location: login.php');
   exit;
}
@require("../sm_admin_db.php");
$authtype = $_POST['authtype'];
$identity = $_POST['identity'];
$flags = $_POST['flags'];
$name = $_POST['name'];
$immunity = $_POST['immunity'];
$user_email = $_POST['user_email'];
$link = @mysql_connect(_HOST,_USER,_PASS);
@mysql_select_db(_DB);
$sql = @mysql_db_query(_DB,"INSERT INTO "._TBL." (`authtype`, `identity`, `flags`, `name`, `immunity`, `user_email`) VALUES (NULL, '$authtype', '$identity', '$flags', '$name', '$immunity', '$user_email')");
$okay = @mysql_affected_rows();
if($okay > 0) {





header( "Location: ../index.php?value=success" );



} else {





header( "Location: ../index.php?value=fail" );
}
@mysql_close($link);
?>



And finally my sm_admin_db.php file is as follows:
<?PHP
//host loaction
define("_HOST", "localhost", TRUE);
//database username
define("_USER", "thatsact_admin", TRUE);
//database username`s password
define("_PASS", "*********", TRUE);
//database name
define("_DB", "thatsact_sadadmins", TRUE);
//database table name change this to install multiple version on same database
define("_TBL", "sm_admins", TRUE);
//This is the email that you will recieve emails when someone signs up.
define("_EMAIL", "admin@tag-clan.com", TRUE);
?>

All of this together does not work and gives the echo error code, however the mysql_error() string does not return anything. Nothing is inserted into the database. Can I get some help on where I went wrong and please keep in mind I am VERY new to php and didn't write most of this script so just tell me what to edit and exactly how to do it please.

Thanks in advance.      

Im sure this is simple, but I cannot see what my problem is!

I am hitting an error on my insert query

Code: [Select]
Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '1' at line 1
I know the output of $user_id is 1, so my error is on $mysavepath

Code: [Select]
$mysavepath = $folder.'/'.$worldname.'_'.date("dMjY");
echo $mysavepath;
$savepath = mysql_query("INSERT INTO saves (user_id,savepath) VALUES ('$user_id','$mysavepath')");
echo '<br>'.$savepath;
if(!mysql_query($savepath))
{
die('<br>Error: ' . mysql_error());
}

however it all echos out ok?

Code: [Select]
188ea678f0dcdc8252aeb15e3c910408/world_15Jan152012
1
Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '1' at line 1

Can anyone see the problem?

Cheers
Dave

Hello, can any1 show me how to insert the following in db
Code: [Select]

<?php
include('dbconnect.php');
include('simple_html_dom.php');
$html = file_get_dom('test3.php');
foreach($html->find('h2') as $heading)
{
//echo $heading;
foreach ($heading->find('a') as $link)
{
                                       $item_title = md5($link);
echo $link."<br/>";
$sql = "INSERT INTO articles(article_id, article_link) VALUES($item_title, $link)";
$result = mysql_query($sql);
}
}
?>

what i want to do is insert every link in the db, but this is not working. The output is ok but the insert doesnt work. Any help please??

Hi Guys, in code below i dont see any reason why it doesnt add anything into my mysql, can u please tell me where is wrong?


<?php
session_start();
include ("includes/db.php");
include ("includes/function.php");
//authentication
if (loggedin()==FALSE)
{
Header("Location: login.php");
exit();
}
//username session
$_SESSION['username']=='$username';
$username=$_SESSION['username'];
echo "Welcome, " .$_SESSION['username']."!<p>";
//get details from user table to be passed to page(text fields)

$getdetails=mysql_query("SELECT * FROM users WHERE username='$username'");
while($row = mysql_fetch_array($getdetails))
{
$users_id=$row['id'];
$first_name=$row['first_name'];
$last_name=$row['last_name'];
$email=$row['email'];

}

///////////////////////////////////////
//****************************//


//post starts here
if (isset($_POST['next']) && $_POST['next'])
    {


$myfirstname = addslashes(strip_tags($_POST['firstname']));
$mylastname = addslashes(strip_tags($_POST['lastname']));
$myemail = addslashes(strip_tags($_POST['email']));
$mybookingdate = addslashes(strip_tags($_POST['datetime']));
$mytime = addslashes(strip_tags($_POST['time']));
$myseatnumber = addslashes(strip_tags($_POST['seatnumber']));
$ride_id=1;

//check if booking exists

$checkbooking=mysql_query("SELECT * FROM booking WHERE date='$mybookingdate' AND time='$mytime' AND seat_no='$myseatnumber'");
   if (mysql_num_rows($checkbooking)>=1)
   {
           echo "Seat No $myseatnumber For $mybookingdate at $mytime has been already taken, Please try a different seat, time or date";
   }
else
{
$addbooking=mysql_query("INSERT INTO booking (ride_id, users_id, time, date, seat_no) VALUES ('$ride_id','$users_id','$mytime','$mybookingdate','$myseatnumber'");
}
}



?>
<html>
<head>

<link type="text/css" href="css/smoothness/jquery-ui-1.8.6.custom.css" rel="stylesheet" />

<script type="text/javascript" src="includes/jquery-1.4.2.min.js"></script>
<script type="text/javascript" src="includes/jquery-ui-1.8.6.custom.min.js"></script>


<script type="text/javascript">  


 $(function() {  
    $('#datetime').datepicker({  
dateFormat: 'dd.mm.yy'
 });  
});  
</script>

</head>

<body>

<form name="booking" method="post" action="">

  <p>First Name:<br />
    <input type="text" name='firstname' value="<? echo "$first_name";?>">
    
<p>Last Name:<br />
  <input type="text" name='lastname' value="<? echo "$last_name";?>">
    
<p>Email:<br />
  <input type="text" name='email' value="<? echo "$email";?>">


<p>Choose a Date:<br />
    <input type="text" name='datetime' id="datetime">
        </p>
<p>
        Select a Time:<br/>
  <select name='time'>
    <option value="9:00">9:00
    <option value="9:07">9:07
    <option value="9:14">9:14
    <option value="9:21">9:21
    <option value="9:28">9:28
    <option value="9:35">9:35
    <option value="9:42">9:42
    <option value="9:49">9:49
    <option value="9:56">9:56
    <option value="10:03">10:03
    <option value="10:10">10:10
    <option value="10:17">10:17
    <option value="10:24">10:24
    <option value="10:31">10:31
    <option value="10:38">10:38
    <option value="10:45">10:45
    <option value="10:52">10:52
    <option value="10:59">10:59
    <option value="11:02">11:02
    <option value="11:09">11:09
    <option value="11:16">11:16
    <option value="11:23">11:23
    <option value="11:30">11:30
    <option value="11:37">11:37
    <option value="11:44">11:44
    <option value="11:51">11:51
    <option value="11:58">11:58         
    </select>
        </p>
<p>
          
  Select Seat Number:<br/>
  <select name='seatnumber'>
    
    <option value="1">Seat 1
      <option value="2">Seat 2
        <option value="3">Seat 3
          <option value="4">Seat 4
          <option value="5">Seat 5
          <option value="6">Seat 6
          <option value="7">Seat 7
          <option value="8">Seat 8
          <option value="9">Seat 9
          <option value="10">Seat 10
          <option value="11">Seat 11
          <option value="12">Seat 12
          <option value="13">Seat 13
          <option value="14">Seat 14
          <option value="15">Seat 15
          <option value="16">Seat 16
          <option value="17">Seat 17
          <option value="18">Seat 18
          <option value="19">Seat 19
          <option value="21">Seat 20
          <option value="22">Seat 22
          <option value="23">Seat 23
          <option value="24">Seat 24
          <option value="25">Seat 25
          <option value="26">Seat 26
          <option value="27">Seat 27
          <option value="28">Seat 28
          <option value="20">Seat 29
          <option value="30">Seat 30
          <option value="31">Seat 31
          <option value="32">Seat 32
          <option value="33">Seat 33
          <option value="34">Seat 34
          <option value="35">Seat 35
          <option value="36">Seat 36
          <option value="37">Seat 37
          <option value="38">Seat 38
          <option value="39">Seat 39
           <option value="40">Seat 40
           <option value="41">Seat 41
           <option value="42">Seat 42
           <option value="43">Seat 43
           <option value="44">Seat 44
           <option value="45">Seat 45
           <option value="46">Seat 46
           <option value="47">Seat 47
           <option value="48">Seat 48                                                                                              
                 </select>                                                          
                                                                                            

<p>
  <input type='submit' name='next' value='Next Step'>
</form>
<p>
    </body>
    
</html>

I have a form that submits a record and saves it to the database, I've got that working already, I'm trying to figure out if there is a cleaner way to delete a record with a button that gets a value from the unique key, in this case 'id'
here is how the delete button looks like:

   // get info from table
$query  = "SELECT id, Title, Message FROM table_name";
$result = mysql_query($query);

//displaying all data
while($row = mysql_fetch_assoc($result))
{
echo"
<div class='status'><h3>{$row['Title']} <br></h3>" . "
         <h5>{$row['Message']} <br><br></h5>"; 
 $id = $row['id']; //trying to get unique key from database
//delete button
 echo "</div>
 <form action='delete.php' method='post' />
 <input type='hidden' name='delete' value='yes' /> 
 <input type='hidden' name='id' value='$id' />
 <input type='submit' value='remove' /></form>";
}

As you can see, i'm trying to give $id a unique value but for some reason i'm not getting it.
The delete.php code looks like this:

if (isset($_POST['delete']))// check if delete was clicked
        {
         $query = "DELETE FROM table_name WHERE id='$id'";
echo "$id Deleted successfully";
}
elseif(!mysql_query($query, $db_server))
{
echo "DELETE failed: $query<br />" .
myql_error() . "<br /><br />";
         }
mysql_close();

I'm new to php and still learning so, if you think there are other ways to do this, please let me know, the button won't do anything to the data.

Thanks in advance

I am trying to do an INSERT on my "bio_answer" table which is a junction table in between the "member" and "bio_question" tables.

My PHP just tries to do the INSERT into "bio_answer" and doesn't touch the parent tables, which I assume is okay?!

Here is a snippet of my code...

if ($_SERVER['REQUEST_METHOD']=='POST'){
// Form was Submitted (Post).

// Initialize Errors Array.
$errors = array();

// Trim all form data.
$trimmed = array_map('trim', $_POST);


// ************************
// Validate Form Data. *
// ************************

// Validate Answer1.
if (strlen($trimmed['answer01']) >= 2 && strlen($trimmed['answer01']) <= 1024){
// Valid Answer1.
$answerArray[0] = $trimmed['answer01'];
$questionID = 1;


echo '<p>$memberID = ' . $memberID .  '</p>'; // Resolves to 19 which exists in the "member" table
echo '<p>$questionID = ' . $questionID .  '</p>'; // Resolves to 1 which exists in the "bio_question" table
echo '<p>$answerArray[0] = ' . $answerArray[0] . '</p>'; // Resolves to whatever I type in my form, e.g. "This is a test..."

}else{
// Invalid Answer1.
$errors['question01'] = 'Answer must be 2-1024 characters.';
}//End of VALIDATE ANSWER1


// ******************************
// Attempt to Create Thoughts. *
// ******************************
if (empty($errors)){
// Valid form data.

// Build query.
$q1 = "INSERT INTO bio_answer(member_id, question_id, response, created_on)
VALUES(?, ?, ?, NOW())";

// Prepare statement.
$stmt1 = mysqli_prepare($dbc, $q1);

// Bind variables to query.
mysqli_stmt_bind_param($stmt1, 'iis', $memberID, $questionID, $answerArray[0]);

// Execute query.
mysqli_stmt_execute($stmt1);

// Verify Insert.
if (mysqli_stmt_affected_rows($stmt1)==1){
// Insert Succeeded.
$_SESSION['resultsCode'] = 'THOUGHTS_NEW_THOUGHTS_CREATED_2138';

}else{
// Insert Failed.
$_SESSION['resultsCode'] = 'THOUGHTS_NEW_THOUGHTS_FAILED_2139';
}//End of UPDATE MEMBER RECORD

// Close prepared statement.
mysqli_stmt_close($stmt1);

// Set Error Source.
$_SESSION['errorPage'] = $_SERVER['SCRIPT_NAME'];

// Redirect to Display Outcome.
header("Location: " . BASE_URL . "/members/results.php");

// End script.
exit();
}//End of ATTEMPT TO CREATE THOUGHTS



My script keeps failing and errors to 'THOUGHTS_NEW_THOUGHTS_FAILED_2139'

What is wrong with my Script/SQL??

Thanks,


Debbie

EvilSpider
123456
FakeEmail@yahoo.com
Connected successfullyYou have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '@yahoo.com)' at line 2
mysql_query("INSERT INTO user
(username, password,email) VALUES($user, $pass, $email) ")

and when i dont do the @ sign i still get the error
EvilSpider
123456
FakeEmail
Connected successfully
Unknown column 'EvilSpider' in 'field list'

hello, i am returning the error in the following code.  im i doing something wrong?

Code: [Select]
{
$sql = "UPDATE jobs SET contact = ' . mysql_real_escape_string{$_POST['contact']} .', contactphone = ' . mysql_real_escape_string{$_POST['contactphone']} .', customer = ' . mysql_real_escape_string{$_POST['customer']} .', initial = ' . mysql_real_escape_string{$_POST['initial']} .', address = ' . mysql_real_escape_string{$_POST['address']} .', city = ' . mysql_real_escape_string{$_POST['city']} .', postal = ' . mysql_real_escape_string{$_POST['postal']} .', province = ' . mysql_real_escape_string{$_POST['province']} .', description = ' . mysql_real_escape_string{$_POST['description']} .', mechanic = ' . mysql_real_escape_string{$_POST['mechanic']} .', ponumber = ' . mysql_real_escape_string{$_POST['ponumber']} .', status = ' . mysql_real_escape_string{$_POST['status']} .' WHERE id = '$id";
 
$result = mysql_query($sql) or die('Error, updating job failed.  Check you fields and try again.');

echo "<center> You have successfully updated the new job.  You can see your changes below.";

}

Hello PHP freaks.  Having a few issues with my PHP scripts, specifically the $query = sprintf and mysql_real_escape_string functions.  Kind of new to this, so if you do reply, explain it to me like I am a complete moron...

Oh, before I forget.  My specific problem is that I can click the "Post" button and follow the header to "Location: view.php", but the actual text in the Subject and Message fields is not being sent to the database.  Finally managed to get rid of all the error messages I was getting, and now I get this...

Thanks in advance!

Here is my script for the entire page:

_________________________________________________ ____________

<?php
require_once('auth.php');
?>
<?php
mysql_connect('xxxxxx', 'xxxxxx', 'xxxxxx');

$subject = $_POST['subject'];
$message_text = $_POST['message_text'];

// add entry to the databse if the form was submitted and
// the necessary information was supplied in the form
if (isset($_POST['submitted']) && $subject && $message_text)
{
$query = sprintf('INSERT INTO FORUM_MESSAGE (SUBJECT, MSG_TEXT) VALUES ($subject, $message_text)', Ryan_iframe,
mysql_real_escape_string($subject),
mysql_real_escape_string($message_text));
mysql_query($query);

// redirect user to list of forums after new record has been stored
header('Location: view.php');
}

// form was submitted but not all the information was correctly filled in
else if (isset($_POST['submitted']))
{
$message = '<p>Not all information was provided.  Please correct ' .
'and resubmit.</p>';
}
?>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Member Index</title>
<link href="loginmodule.css" rel="stylesheet" type="text/css" />
</head>
<body>
<h1>Welcome <?php echo $_SESSION['SESS_FIRST_NAME'];?></h1>
Home | <a href="member-profile.php">My Profile</a> | <a href="logout.php">Logout</a>

<br /><br />

<form method="post">
<div>
<label for="subject">Subject: </label>
<input type="text" id="subject" name="subject" value="<?php echo htmlspecialchars($subject); ?>" /><br />
<label for="message_text">Message: </label>
<input type="text" id="message_text" name="message_text" value="<?php echo htmlspecialchars($message_text); ?>" /><br />
<input type="hidden" name="submitted" value="true" />
<input type="submit" value="Post" />
</div>
</form>
   
</body>
</html>

Hi can someone pls help, im tryin a tutorial but keep getting errors, this is the first one i get after registering.

You Are Registered And Can Now Login
Warning: Cannot modify header information - headers already sent by (output started at /home/aretheyh/public_html/nealeweb.com/regcheck.php:43) in /home/aretheyh/public_html/nealeweb.com/regcheck.php on line 46

I'd like to write - or acquire - code that displays a simple table (name, phone number, email address, plus a comments field) on a web page in a password-protected page and allows a user to add his own information, update it or delete it.

I figure hundreds of people and companies have written something like this so I'd like to find either an example I can imitate or even an existing package that I can simply customize to the specifics for my own table.

Can anyone help me with that? Or am I going to have to reinvent the wheel for the gazillionth time and write it myself?

I'm new to PHP/MySQL and have been searching high and low for a solution to this...

When using my PHP script to insert a row into the database it errors out but does not give me a specific error. When I use the MySQL command line client I can insert the first, second, third, or any number of rows without a problem.

What I don't understand is, once I have inserted a row using the command line client, my PHP script will insert rows without issue.

So my question is, what is it about the first row being inserted that makes it so special? One of the columns in the table is an AUTO_INCREMENT field, so i'm guessing that has something to do with it?