PHP - Query In Php Not Working For Danish Characters

Evening everybody, hoping someone can see my error here!
$sql="INSERT INTO transactions (t_ref, m_affid, m_name, order_val, order_date, order_status, u_id, u_comm) VALUES('$t_ref', '$m_affid', '$m_name', '$order_val', '$order_date', '$order_status', '$u_id', '$u_comm')";
if (!mysql_query($sql))
   { 
    mysql_error();
   }
else

{
$newq = mysql_query("SELECT * FROM userbalance WHERE u_id = '$u_id'");
echo mysql_error();
$row = mysql_fetch_assoc($newq);
$cpending = $row['pending'];
$npending = $cpending + $u_comm;
$updatepending = "INSERT into userbalance (pending)
VALUES ('$npending')
WHERE u_id = '$u_id'";

if (!mysql_query($updatepending))
   { 
    mysql_error();
   }

}

Basically  the 1st query works and inserts the data correctly, however once it gets to $newq = mysql_query("SELECT * FROM userbalance WHERE u_id = '$u_id'"); this query does not insert anything, no errors.

ok Im building a search page.  The user can enter a name, county, category, address into my search field.  The results page will show the details based on that.  It works great so here is a snippet of the working code.




$result = "-1";
if (isset($_POST['searchField'])) {
  $result = $_POST['searchField'];
}



$result = sprintf("SELECT Clubs.clubID, Clubs.name, Clubs.county, Clubs.logo, Clubs.postcode, Clubs.intro, Clubs.thumbsup, Clubs.cat, Category.*, County.*FROM Clubs

INNER JOIN Category ON Clubs.cat = Category.catID  
INNER JOIN County ON Clubs.county = County.countyID 
WHERE
 ((Clubs.name Like %s) OR (Clubs.cat LIKE %s) OR (County.county LIKE %s) OR (Clubs.area LIKE %s) OR (Clubs.postcode LIKE %s) OR (Category.categorys LIKE %s)) "

, String($result . "%", "text"),
String($result . "%", "text"),
String($result . "%", "text"),
String($result . "%", "text"),
String($result . "%", "text"),
String($result . "%", "text"));


This code works great no matter what I enter.  the problem Im having is I only want to show results based on the logged in users county.  I have a variable stored under $user['county']  which is the ID found in the County table under countyID

I have tried writing the code so many different ways.  Here is my latest attemped but it doesnt work.  It breaks the whole query and nothing is displayed.



$result = "-1";
if (isset($_POST['searchField'])) {
  $result = $_POST['searchField'];
}

$County= "-1";
if (isset($user['county'])) {
  $County= $user['county'];
}



$result = sprintf("SELECT Clubs.clubID, Clubs.name, Clubs.county, Clubs.logo, Clubs.postcode, Clubs.intro, Clubs.thumbsup, Clubs.cat, Category.*, County.* FROM Clubs
INNER JOIN Category ON Clubs.cat = Category.catID  
INNER JOIN County ON Clubs.county = County.countyID 
WHERE
 ((Clubs.name Like %s) OR (Clubs.cat LIKE %s) OR (County.county LIKE %s) OR (Clubs.area LIKE %s) OR (Clubs.postcode LIKE %s) OR (Category.categorys LIKE %s)) AND County.countyID = %s"

, String($result . "%", "text"),
String($result . "%", "text"),
String($result . "%", "text"),
String($result . "%", "text"),
String($result . "%", "text"),
String($result . "%", "text")
String($County. "%", "int"));


Thanks

Danny

Can someone review this piece of code and advise what's missing that's preventing the SQL query from working?
Issue: no results being echoed out from $results.

Note: I confirmed the db connection  established (only the query not working - no results pulled from db and displayed via echo statements).



$db = mysql_connect($host,$user,$pw)
        or die("Cannot connect to MySQL.");

mysql_select_db($database,$db)
        or die("Cannot connect to database.");

echo "Success!  Connected to database ".$database."<br /><br />";

// $jsearch is variable sent in from form "title" name
$jsearch = $_POST['jobsearch'];

// I confirmed that all the field names are correct

  //-query  the database table
  $sql="SELECT id,title,details,status FROM jobs WHERE title = '$jsearch';";
  //-run  the query against the mysql query function
  $result=mysql_query($sql) or die("<br>Query string: $result<br>Returned error: " . mysql_error() );
  //-create  while loop and loop through result set
  while($row=mysql_fetch_array($result)){
          $title  =$row['title'];
  echo title."<br />";
          $details=$row['details'];
  echo $details."<br />";
  $status=$row['status'];
  echo $status."<br />";
          $ID=$row['id'];

I am using a query to update the db when a user logs in. It sets every value correctly but it doesnt set the colum u_online which is INT(1).


$link->query("UPDATE ".TBL_PREFIX."users
                                SET 
                                u_last_login = '".$config['time_now']."',
                                u_online = 1,
                                u_ip = '".$_SERVER['REMOTE_ADDR']."',
                                u_activity_time = '".$config['time_now']."',
                                u_points = u_points + '$login_points'
                                WHERE u_username = '$username'")
                                or die(print_link_error());


when i echo the query:


UPDATE asf_users 
SET u_last_login = '1300310822',
 u_online = 1, 
u_ip = '127.0.0.1', 
u_activity_time = '1300310822', 
u_points = u_points + '1' 
WHERE u_username = 'doddsey65'


which works because i tried it in phpmyadmin and it changed the column. But it doesnt work in the actual script. It changes every other value that its supposed to but the u_online.

Anyone have any ideas why?

Hello I have a general php insert function that I want to use to insert data into tables regulary, the problem is, each time I use it it creates a new row in the table instead of using certain columns to add information to, here is the function, can you tell me why does it insert a new row each time I use it?

class Query{
function add($table,$cols,$vals){
		include 'conx.php';
		$query = "
			INSERT INTO $table($cols) VALUES('$vals');
		";
		
		$res = mysqli_query($con,$query) ;
		
		if($res){
			echo "Inserted value";
		}else{
			echo "Did not insert value";
		}
	}
}

The following mysql query is not returning rows like I expect it to. '$update_field' is a variable,  matching an actual field name in table 'users'. 'user_task[1]' is an integer value. What am I missing here?
Code: [Select]
$query_update_user = "UPDATE users SET ".$update_field." = 'Y' WHERE user_no = '".$user_task[1]."'";

This is for a faux stock photo website. When I only use the $query string to search for $searchSubject, everything goes golden. Once I put everything else in there though, the search function will not work at all.

Any help will be greatly appreciated. I've been staring at this for too long and it's starting to give me a headache.

PS. This is my first post here. Seems like a great community!

<?php

$searchSubject = $_POST['searchSubject'];
$searchPhotoname = $_POST['searchPhotoname'];
$searchLocation = $_POST['searchLocation'];

if (isset($_POST['search'])) {
   
   //Select records from the database

$query = "SELECT * FROM `photos` WHERE `subject` LIKE '%" . $searchSubject . "%' OR `name` LIKE '%" . $searchPhotoname . "%' OR `location` LIKE '%" . $searchLocation . "%' ";






} else {
   
   $query = "SELECT * FROM photos";
   
}

$result = mysql_query($query);
// count the number of rows in the database
$num = mysql_num_rows($result);

?>



If I leave it at this:


$query = "SELECT * FROM `photos` WHERE `subject` LIKE '%" . $searchSubject . "%'";

Then everything works perfect.

Hi, i am currently designing a new website and am testing a few different compnents of it. One part gathers two variables of the url using the GET method and selects the relevant data from the database. It then performs an UPDATE of a field called balance before INSERTING a new record. Everything works and gets inserted except for the UPDATE.

Here is the code

Code: [Select]
<?php
include ("connect.php");

$id = "";
$offerid = "";

$id = $_GET['user_id'];
$offerid = $_GET['offer_id'];

$sql_user = mysql_query("SELECT * FROM user_info WHERE id = $id");

$sql_offer = mysql_query("SELECT * FROM offers WHERE offerid = $offerid");

$balance += $payout;

mysql_query("UPDATE user_info SET balance = $balance WHERE id = $id");

mysql_query ("INSERT INTO offer_credited (id, offerid, date_credited, payout_to_user) VALUES ('$id', '$offerid', CURDATE(), '$payout')") or die(mysql_error());

echo "Your Offer has been recorded and Updated"
?>
The balance field is a float that i want to increase by the amount of the payout variable which is from the offers table. However, the balance never gets updated and remains at 0.

Any ideas what it could be, i am thinking it is something to do with,

Code: [Select]
$balance += $payout;
But i can't think of what else to do.

Hopefully this isn't too confusing.

Thanks for the help

Hi there,

I have this query, which outputs the records properly in PhpMyAdmin.
But when using this query inside my PHP code, the result is incomplete.
Instead of retrieving the expected 15 results, I only get the first one twice.
Also, I don't get any errors.

Code: (php) [Select]
<?php
$get_files = "SELECT `id` FROM `files` WHERE `category` = 'W';";

include_once('connection.inc.php');
$con = mysql_connect(MYSQL_SERVER, MYSQL_USERNAME, MYSQL_PASSWORD) or die ("Connection failed: " . mysql_error());
$dbname = "myDB";
$select_db = mysql_select_db($dbname, $con) or die ("Can't open database: " . mysql_error());

$result_files = mysql_query($get_files) or die ("Query failed: " . mysql_error());
$files = mysql_fetch_array($result_files);
$W_files = "'".implode("', '", $files)."'";

// echo $W_files; outputs the first number twice, instead of the expected 15 different ones

$get_checklist = "SELECT * FROM `checklist` WHERE `id` IN ($W_files) ORDER BY `id` ASC;";
$result_checklist = mysql_query($get_checklist) or die ("Query failed: " . mysql_error());

// .... code for displaying
?>

I've got to be missing something pretty basic here.. considering the query is pretty basic.

I'm trying to figure out how to pull a query as an array so I can compare it against another array (array_diff)

I'm doing a mysql_fetch_array, and I'm getting an error ( mysql_fetch_array(): supplied argument is not a valid MySQL result resource):
Quote

$checker = "SELECT ID FROM edible_uses";

$result2 = mysql_fetch_array($checker) or die(mysql_error());
//echoing to see if I'm getting what I need.
echo $row['ID'];


I've done a mysql_query and I get results. The table name and all that is correct. I'm stumped. This seems like a pretty simple query? I tried mysql_fetch_assoc as well. Same result? I tried it with an extra set of parenthesis around it. nope.

my SQL Query wont execute on on following lines:

Code: [Select]
$result = mysql_query("INSERT INTO 'gallery' ('image', 'memberid', 'caption') VALUES ('$newFileName', '$member_id', '$caption')")
    or die (mysql_error());
i get the following error:

Code: [Select]
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 ''gallery' ('image', 'memberid', 'caption') VALUES ('gallery/9074849_1.jpg', '1',' at line 1
here is my full code:
Code: [Select]
<?php
require_once('connect.php');

$rand = mt_rand(1,9999999);
$rand2 = mt_rand(1,9999999);
$member_id = $_SESSION['SESS_MEMBER_ID'];
$caption = $_POST["caption"];

if(isset($_FILES['uploaded']['name']))
{
$allowed_filetypes = array('.jpg','.gif','.bmp','.png','.jpeg');
$max_filesize = 524288; // Maximum filesize in BYTES (currently 0.5MB) 
    $fileName = basename($_FILES['uploaded']['name']);
$errors = array();
$target = "gallery/";
$fileBaseName = substr($fileName, 0, strripos($fileName, '.'));

   // Get the extension from the filename.
   $ext = substr($fileName, strpos($fileName,'.'), strlen($fileName)-1); 
 
   //$newFileName = md5($fileBaseName) . $ext;
   $newFileName = $target . $rand . "_" . $member_id.$ext;
   
   // Check if filename already exists
   if(file_exists("gallery/" . $newFileName)) 
    {
     $errors[] = "The file you attempted to upload already exists, please try again.";
    }
   // Check if the filetype is allowed.
   if(!in_array($ext,$allowed_filetypes))
    {
         $errors[] = "The file you attempted to upload is not allowed.";
    }
    
// Now check the filesize.
   if(!filesize($_FILES['uploaded']['tmp_name']) > $max_filesize)
    {
         $errors[] = "The file you attempted to upload is too large.";
    }
  
    // Check if we can upload to the specified path.
   if(!is_writable($target))
    {
         $errors[] = "You cannot upload to the specified directory, please CHMOD it to 777.";
    }
 
    //Here we check that no validation errors have occured.
    if(count($errors)==0)
    {

//Try to upload it.
        if(!move_uploaded_file($_FILES['uploaded']['tmp_name'], $newFileName))
        {
            $errors[] = "Sorry, there was a problem uploading your file.";
        }
    }
    //Lets INSERT database information here
    //Here we check that no validation errors have occured.
    if(count($errors)==0)
    {
$result = mysql_query("INSERT INTO 'gallery' ('image', 'memberid', 'caption') VALUES ('$newFileName', '$member_id', '$caption')")
    or die (mysql_error()); 

{
            $errors[] = "SQL Error.";
        }
}
    //If no errors show confirmation message
    if(count($errors)==0)
    {
         echo "<div class='notification success png_bg'>
<a href='#' class='close'><img src='img/cross_grey_small.png' title='Close this notification' alt='close' /></a>
<div>
The file {$newFileName} has been uploaded<br>\n
</div>
</div>";


//echo "The file {$fileName} has been uploaded";
echo "<br>\n";
echo "<a href='gallery.php'>Go Back</a>\n";

   }
    else
    {
        //show error message
        echo "<div class='notification attention png_bg'>
<a href='#' class='close'><img src='img/cross_grey_small.png' title='Close this notification' alt='close' /></a>
<div>
Sorry your file was not uploaded due to the following errors:<br>\n
</div>
</div>";
//echo "Sorry your file was not uploaded due to the following errors:<br>\n";
        echo "<ul>\n";
        foreach($errors as $error)
        {
            echo "<li>{$error}</li>\n";
        }
        echo "</ul>\n";
echo "<br>\n";
echo "<a href='gallery.php'>Go Back</a>\n";
    }
    
}
else
{
    //Show the form
echo "Use the following form below to add a new image to your gallery;<br />\n";
    echo "<form enctype='multipart/form-data' action='' method='POST'>\n";
    echo "Please choose a file: <input name='uploaded' type='file' /><br />\n";
echo "Caption: <input name='caption' type='text' /><br />\n";
    echo "<input type='submit' value='Upload' />\n";
    echo "</form>\n";

//Echo Tests!
    echo "<br /><br />Random FileName: "; 
echo $rand;
echo "<br />";
echo "member ID: #";
echo $member_id;
}
?>
any help appreciated. its prob something simple.

my table has the following fields:

"gallery"
id (primary Key, AUTO_INC)
memberid (fetched from session)
image (will store image name including extension)
caption (from "caption" text field in form)

I have a basic query where I am retrieving project records. I basically want to show the records from newest to oldest, but I also want to show the Featured projects first. Each Featured project is marked by "1" in the mysql table column.  If it's not featured, then it's marked "0".

$find_records = $db->prepare("SELECT * FROM projects ORDER BY project_id DESC, featured DESC LIMIT 10");
$find_records->execute();
$result_records = $find_records->fetchAll(PDO::FETCH_ASSOC);
if(count($result_records) > 0) {
  foreach($result_records as $row) {
    
  }
}

The issue with above query is that the ORDER BY works only for the first component(project_id) and ignores the second component(featured). Do you see what's wrong with it?

Basically I need to input data into two tables.

I am running 3 different query's but only 2 of them work. The other one doesn't.

None working query:
mysql_query("INSERT INTO users(username, password, email, pin, key) VALUES('$username', '$password', '$email', '$key', '$pin')");

Working querys:

mysql_query("DELETE FROM beta_keys WHERE keys_new='$key'");

**and**

mysql_query("INSERT INTO beta_keys(keys_used) VALUES('$key')");

So any ideas why the top one doesn't work but the bottom two do?

Hello all, I am using the Lynda.com PHP course to further learn php.  On one of the examples, they are using a SQL Query that works for them but is erroring out for me.  It is the query below for the $page_set variable, particularly the use of the WHERE subject_id = {$subject["id"]}". 


Code: [Select]
<?php require_once ('includes/conf.php'); ?>
<?php include ('includes/header.php'); ?>
<?php require_once ('includes/functions.php'); ?>

<table id="stucture">
<tr>
<td id="navigation">
<ul class="subjects">
<?php 
$subject_set = mysql_query("SELECT * FROM subjects", $connection);
if (!$subject_set) {
die ("Database query failed: " . mysql_error());
}

while ($subject = mysql_fetch_array($subject_set)) {
echo "<li>{$subject["menu_name"]}</li>";
}

$page_set = mysql_query("SELECT * FROM pages WHERE subject_id = {$subject["id"]}");  // This appears to be the problem area, unless it is fooling me. It works with this exact statement in their code.
if (!$page_set) {
die ("Database query failed: " . mysql_error());
}
echo "<ul class='pages'>";
while ($page = mysql_fetch_array($page_set)) {
echo "<li>{$page['menu_name']}</li>";
}
echo "</ul>";
?>
</ul>
</td>
<td id="page"><h2>Content Area</h2>
</td>
</tr>
</table>
<?php require ('includes/footer.php'); ?>

The error listed is this:     Database query failed: 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 '' at line 1


Thanks,  you all are the best!   BtD

I'm having a problem with adding icon/avatar to my database..this is a submit form where I just submit the avatar url. I don't know how to change the submit link from avatar_add.php?mode=submit to index.php?mode=submit&x=avatar_add.
http://www.graphics.allmerk-noplay.com/index.php?x=avatar_add


Code: [Select]
<?php
$host =""; // host name
$user =""; // user
$pass =""; // password
$dbase =""; // database name

// connect to the server
mysql_connect("$host","$user","$pass");
// and select the database
mysql_select_db($dbase);

$mode=$_GET["mode"]; // get the mode
$QUERY_STRING = 'x=avatar_add';

if(!isset($mode) || empty($mode)) // if mode is empty, switch to case index
{
$mode='index';
}

switch($mode)
{

case 'index'; // displays the form
?>


<form method="post" action="avatar_add.php?mode=submit">
Avatar URL:<br>
<input type="text" name="avatar_url" size="30"><br><br>
<input type="submit" value="Submit">


<?php
break;

case 'submit'; // form processing mode

$avatar_url=$_POST["avatar_url"];

// check for empty fields
if(empty($avatar_url))

{
echo "Error! Please go back and fill in all the required fields!";
// if you have a footer include, insert it before the die statement
die; // stops the script from executing the rest of the code
}

// let's get the date in YYYY-MM-DD format
$date = date("Y-m-d");

$query="INSERT into icons_db (id,avatar_url) VALUES ('','$avatar_url')";
$result=mysql_query($query) or die ("Couldnot execute query: $query." . mysql_error());

if($result) // if mysql query successful
{ echo "Thank you! The avatar has been added to or database."; }

break;

}

?>