PHP - Moved: What's Wrong With My Sql Query?

Hey guys,

I have a problem that I find rather strange, but maybe that's just me. I want to output a list of news that is stored in a MySQL database. I have coded paging ($start in the example below), but below I have narrowed my code down and removed code that is not related to the problem. The problem is that my script does not output the first row. That is, if I limit the query below from 0 to the next 10 (LIMIT 0, 10), the result will start at row 2. The same happens if $start is 10; news 12 will be the first one rather than 11. I have tried to put the numbers directly into the SQL query instead of $start, but with the same result. I have tested the query in phpMyAdmin, and the query works fine there. But for some reason, why PHP script does not output the first result. I guess I have made some silly mistake, but at the moment, I do not see it. 


<?php

// Connect to MySQL & select db

$start = 0;

$result = mysql_query("SELECT * FROM Content ORDER BY Time ASC LIMIT $start, 10");

if ($result) {
if (mysql_num_rows($result) > 0) {
$row = mysql_fetch_assoc($result);

while ($row = mysql_fetch_array($result)) {
echo '<div class="someClass">' . $row['title'] . '</div>';
}
}
}

?>

Thanks in advance!

My variable `users`.`id` isn't being passed through correctly.


$ms = 'SELECT `messages_inbox`.`from_id`, `messages_inbox`.`subject`, `messages_inbox`.`body`, `messages_inbox`.`files`, `messages_inbox`.`inbox` AS `box`';
$ms.= 'SELECT LEFT `users`.`firstname`, `users`.`lastname`';
$ms.= 'SELECT FROM `messages_inbox` INNER JOIN `users` ON `messages_inbox`.`from_id` = `users`.`id`' or die(mysql(error)); 

I am having a warning which indicates there is a not valid mysql result, I think the problems lay down at the WHERE clause, but I am not sure.

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /storeprueba/sidebar.php on line 21

Code: [Select]

$categoryurl = $_GET['categoryurl'];
$sql= mysql_query("SELECT * FROM products, categories WHERE products.category = '$categoryurl' DESC LIMIT 6");
$productCount = mysql_num_rows($sql); // line 21
if ($productCount>0 )
     {
     while($row = mysql_fetch_array($sql))
          {
            $id= $row["id"];
$product_name= $row["product_name"];
$price = $row["price"];
$category = $row["category"];
$subcategory = $row["subcategory"];
$location = $row["location"];
$date_added = strftime("%b %d, %Y", strtotime($row["date_added"]));


thanks.

ive looked at this for a while now and im not sure whats wrong, the error occured when i entered $cat and catagory into it, without that the query works perfectly fine, any help appreciated.
<?php
INSERT INTO newnotes (uid, name, catagory, location) VALUES ('1', 'test', '1001 Laws - Polceing', '/home/mikeh/public_html/uzEr Upl0ds/Alyssa O'Leary.doc')
 
the <php tag is only to give color and make it less dull the query is from a die($query) statment

I have a need to do a special sorting for a bunch of leads. The person who sees the leads (l.user) is looking at leads that are for a few different people (l.team_id). When I run my query in sqlyog, it gives me the results I am looking for, where the rank is used to sort. When I run my query through phpMyAdmin, php, or the MySQL command line, all of the ranks are NULL, so nothing gets sorted. All of these attempts are on the same database, on the same computer (Windows 7). Any ideas or help is appreciated.

SELECT 
  x.lead_id,
  x.rank,
  x.team_id,
  x.status 
FROM
  (SELECT 
    l.lead_id,
    l.team_id,
    l.status,
    CASE
      WHEN @ps != l.team_id 
      THEN @rownum := 0 
      WHEN @ps = l.team_id 
      THEN @rownum := @rownum + 1 
      ELSE @rownum 
    END AS rank,
    @ps := l.team_id 
  FROM
    `system-leads` l 
  WHERE (l.user = 189905706)  
    AND l.status != "closed" 
  LIMIT 100000000 OFFSET 0) `x`
ORDER BY x.status ASC,
  x.rank ASC,
  x.lead_id ASC 

I wonder why it works in sqlyog and not the others, but I need it to work regardless of what client I am using.

Hi everyone.

I'm stuck on the following query. I need to display all the fields listed below on a page, but linked via communications.CommID.  I'd appreciate any assistance you can provide. thank you.

Code: [Select]

<?php
 $result = mysql_query("SELECT records.NameFirst_1, records.NameLast_1, records.CompanyName, records.CompanyBranch, records.CompanyReferenceNumber, records.CaseOwnerSelect, communications.ConversionType, communications.Contact, communications.ContactFrom, communications.CommID, communications.ContactPosition, communications.ContactTelephone, communications.ContactEmail, communications.ContactFax, communications.CallDate, communications.CallTime, communications.ActionTextField
FROM records INNER JOIN communications ON records.IDNumber = '$IDNumber'")

 or die(mysql_error()); 
 $row = mysql_fetch_array($result);
?>

Ok, I'm going start off simple. If I have to provide more code I will.

I am doing an update on a table called countries. Yet my query just will not update the db. Is there anything wrong with this query?

mysql_query("UPDATE countries SET country_id = '{$_POST['update_value']}' WHERE country_id = '{$_POST['original_html']}'") or die(mysql_error());

I have a table korisnici in SQLite with INTEGER field aktivan that can have only 0 or 1 value (CHECK constraint).

Field aktivan has value 0 but PHP returns value 1, why? Is this a bug?

This is PHP code that I am running:

 $sql = "SELECT ime, aktivan FROM korisnici WHERE lower(ime) = '" . $ime . "'" . " AND sifra = '" . $_POST["sifra"] . "'";
                
$result = $db->query($sql);
$row = $result->fetchArray(SQLITE3_ASSOC);

$row['aktivan'] = 1 but in table the value is 0.

When I run same query in DB Browser for SQLite I get correct value 0.

Is this a bug?

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