PHP - Anthing Wrong With This Query?

This topic has been moved to MySQL Help.

http://www.phpfreaks.com/forums/index.php?topic=326252.0

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!

This topic has been moved to MySQL Help.

http://www.phpfreaks.com/forums/index.php?topic=347365.0

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);
?>

This topic has been moved to MySQL Help.

http://www.phpfreaks.com/forums/index.php?topic=307477.0

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.

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?

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());

Here is my code:
// Start MySQL Query for Records
$query = "SELECT codes_update_no_join_1b" .
"SET orig_code_1 = new_code_1,
       orig_code_2 = new_code_2" .
"WHERE concat(orig_code_1, orig_code_2) = concat(old_code_1, old_code_2)";
$results = mysql_query($query) or die(mysql_error());
// End MySQL Query for Records


This query runs perfectly fine when run direct as SQL in phpMyAdmin, but throws this error when running in my script??? Why is this???

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 '= new_code_1, orig_code_2 = new_code_2WHERE concat(orig_code_1, orig_c' at line 1

Hello all,   Based on the suggestion of you wonderful folks here, I went away for a few days (to learn about PDO and Prepared Statements) in order to replace the MySQLi commands in my code. That's gone pretty well thus far...with me having learnt and successfully replaced most of my "bad" code with elegant, SQL-Injection-proof code (or so I hope).   The one-and-only problem I'm having (for now at least) is that I'm having trouble understanding how to execute an UPDATE query within the resultset of a SELECT query (using PDO and prepared statements, of course).   Let me explain (my scenario), and since a picture speaks a thousand words I've also inlcuded a screenshot to show you guys my setup:   In my table I have two columns (which are essentially flags i.e. Y/N), one for "items alreay purchased" and the other for "items to be purchased later". The first flag, if/when set ON (Y) will highlight row(s) in red...and the second flag will highlight row(s) in blue (when set ON).   I initially had four buttons, two each for setting the flags/columns to "Y", and another two to reverse the columns/flags to "N". That was when I had my delete functionality as a separate operation on a separate tab/list item, and that was fine.   Now that I've realized I can include both operations (update and delete) on just the one tab, I've also figured it would be better to pare down those four buttons (into just two), and set them up as a toggle feature i.e. if the value is currently "Y" then the button will set it to "N", and vice versa.   So, looking at my attached picture, if a person selects (using the checkboxes) the first four rows and clicks the first button (labeled "Toggle selected items as Purchased/Not Purchased") then the following must happen:   1. The purchased_flag for rows # 2 and 4 must be switched OFF (set to N)...so they will no longer be highlighted in red. 2. The purchased_flag for row # 3 must be switched ON (set to Y)...so that row will now be highlighted in red. 3. Nothing must be done to rows # 1 and 5 since: a) row 5 was not selected/checked to begin with, and b) row # 1 has its purchase_later_flag set ON (to Y), so it must be skipped over.   Looking at my code below, I'm guessing (and here's where I need the help) that there's something wrong in the code within the section that says "/*** loop through the results/collection of checked items ***/". I've probably made it more complex than it should be, and that's due to the fact that I have no idea what I'm doing (or rather, how I should be doing it), and this has driven me insane for the last 2 days...which prompted me to "throw in the towel" and seek the help of you very helpful and intellegent folks. BTW, I am a newbie at this, so if I could be provided the exact code, that would be most wonderful, and much highly appreciated.   Thanks to you folks, I'm feeling real good (with a great sense of achievement) after having come here and got the great advice to learn PDO and prepared statements.   Just this one nasty little hurdle is stopping me from getting to "end-of-job" on my very first WebApp. BTW, sorry about the long post...this is the best/only way I could clearly explaing my situation.   Cheers guys!

case "update-delete":
	if(isset($_POST['highlight-purchased'])) 
		{
			// ****** Setup customized query to obtain only items that are checked ******
			$sql = "SELECT * FROM shoplist WHERE";
					for($i=0; $i < count($_POST['checkboxes']); $i++)
						{
							$sql=$sql . " idnumber=" . $_POST['checkboxes'][$i] . " or";
						}
					$sql= rtrim($sql, "or");
											
			$statement = $conn->prepare($sql);
			$statement->execute();
																						
			// *** fetch results for all checked items (1st query) *** //
			$result = $statement->fetchAll();

			$statement->closeCursor();
											
			// Setup query that will change the purchased flag to "N", if it's currently set to "Y"
			$sqlSetToN = "UPDATE shoplist SET purchased = 'N' WHERE purchased = 'Y'";
											
			// Setup query that will change the purchased flag to "Y", if it's currently set to "N", "", or NULL
			$sqlSetToY = "UPDATE shoplist SET purchased = 'Y' WHERE purchased = 'N' OR purchased = '' OR purchased IS NULL";

			$statementSetToN = $conn->prepare($sqlSetToN);
			$statementSetToY = $conn->prepare($sqlSetToY);
											
			/*** loop through the results/collection of checked items ***/
			foreach($result as $row)
				{
					if ($row["purchased"] != "Y")
						{
							// *** fetch one row at a time pertaining to the 2nd query *** //
							$resultSetToY = $statementSetToY->fetch();
															
							foreach($resultSetToY as $row)
								{															
									$statementSetToY->execute();																
								}
						}
					else
						{
							// *** fetch one row at a time pertaining to the 2nd query *** //
							$resultSetToN = $statementSetToN->fetch();
															
							foreach($resultSetToN as $row)
								{															
									$statementSetToN->execute();																
								}
						}
				}
				break;
		}

 CRUD Queston.png   20.68KB   0 downloads    

If you also have any feedback on my code, please do tell me. I wish to improve my coding base.

Basically when you fill out the register form, it will check for data, then execute the insert query. But for some reason, the query will NOT insert into the database. In the following code below, I left out the field ID. Doesn't work with it anyways, and I'm not sure it makes a difference.

Code:

  Code: [Select]
mysql_query("INSERT INTO servers (username, password, name, type, description, ip, votes, beta) VALUES ($username, $password, $name, $server_type, $description, $ip, 0, 1)");
Full code:

Code: [Select]
<?php
include_once("includes/config.php");
 
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title><? $title; ?></title>
<meta http-equiv="Content-Language" content="English" />
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<link rel="stylesheet" type="text/css" href="style.css" media="screen" />
</head>
<body>
 
<div id="wrap">
 
<div id="header">
<h1><? $title; ?></h1>
<h2><? $description; ?></h2>
</div>
 
<? include_once("includes/navigation.php"); ?>
 
<div id="content">
<div id="right">
<h2>Create</h2>
<div id="artlicles">
<?php
 
if(!$_SESSION['user'])
{
 
        $username = mysql_real_escape_string($_POST['username']);
        $password = mysql_real_escape_string($_POST['password']);
        $name = mysql_real_escape_string($_POST['name']);
        $server_type = mysql_real_escape_string($_POST['type']);
        $description = mysql_real_escape_string($_POST['description']);
 
        if(!$username || !$password || !$server_type || !$description || !$name)
        {

        echo "Note: Descriptions allow HTML. Any abuse of this will result in an IP and account ban. No warnings!<br/>All forms are required to be filled out.<br><form action='create.php' method='POST'><table><tr><td>Username</td><td><input type='text' name='username'></td></tr><tr><td>Password</td><td><input type='password' name='password'></td></tr>";
        echo "<tr><td>Sever Name</td><td><input type='text' name='name' maxlength='35'></td></tr><tr><td>Type of Server</td><td><select name='type'>
       
        <option value='Any'>Any</option>
        <option value='PvP'>PvP</option>
        <option value='Creative'>Creative</option>
        <option value='Survival'>Survival</option>
        <option value='Roleplay'>RolePlay</option>
       
        </select></td></tr>
       
        <tr><td>Description</td><td><textarea maxlength='1500' rows='18' cols='40' name='description'></textarea></td></tr>";
        echo "<tr><td>Submit</td><td><input type='submit'></td></tr></table></form>";
        }
        elseif(strlen($password) < 8)
        {
        echo "Password needs to be higher than 8 characters!";
        }
        elseif(strlen($username) > 13)
        {
                echo "Username can't be greater than 13 characters!";
        }
        else
        {
                $check1 = mysql_query("SELECT username,name FROM servers WHERE username = '$username' OR name = '$name' LIMIT 1");
               
                if(mysql_num_rows($check1) < 0)
                {
                        echo "Sorry, there is already an account with this username and/or server name!";
                }
                else
                {
                        $ip = $_SERVER['REMOTE_ADDR'];

                        mysql_query("INSERT INTO servers (username, password, name, type, description, ip, votes, beta) VALUES ($username, $password, $name, $server_type, $description, $ip, 0, 1)");
                        echo "Server has been succesfully created!";
                }
        }
       
}
else
{
        echo "You are currently logged in!";
}
 
?>
</div>
</div>
 
<div style="clear: both;"> </div>
</div>
 
<div id="footer">
<a href="http://www.templatesold.com/" target="_blank">Website Templates</a> by <a href="http://www.free-css-templates.com/" target="_blank">Free CSS Templates</a> - Site Copyright MCTop
</div>
</div>
 
</body>
</html>

I'm trying to update every record where one field in a row is less than the other. The code gets each row i'm looking for and sets up the query right, I hope I combined the entire query into one string each query seperated by a ; so it's like
UPDATE `table` SET field2= '1' WHERE field1= '1';UPDATE `table` SET field2= '1' WHERE field1= '2';UPDATE `table` SET field2= '1' WHERE field1= '3';UPDATE `table` SET field2= '1' WHERE field1= '4';UPDATE `table` SET field2= '1' WHERE field1= '5';
this executes properly if i run the query in phpMyAdmin, however when I run the query in PHP, it does nothing... Any advice?

I was just wondering if it's possible to run a query on data that has been returned from a previous query?  For example, if I do

Code: [Select]
$sql = 'My query';
$rs = mysql_query($sql, $mysql_conn);

Is it then possible to run a second query on this data such as

Code: [Select]
$sql = 'My query';
$secondrs = mysql_query($sql, $rs, $mysql_conn);

Thanks for any help

Say I have this query:

site.com?var=1

..I have a form with 'var2' field which submits via get.

Is there a way to produce:

site.com?var=1&var2=formdata

I was hoping there would be a quick way to affix, but can't find any info. Also, the query could sometimes be:

site.com?var2=formdata&var=1

I would have to produce:

site.com?var2=updatedformdata&var=1

Is my only option to further parse the query?

here's the code:

Code: [Select]
$companyName = 'big company';
$address1 = 'big bay #8';
$address2 = 'some big warehouse';
$city = 'big city';
$province = 'AB';
$postalCode = 'T1T0N0';
$phone = '0123456789';
$email2 = 'bigKahuna@bigKahuna.edu';

$query = "INSERT INTO clients (
                   companyName, address1, address2, city, province, postalCode, phone, email)
                 VALUES ( ".
                    $companyName.",".$address1.",".$address2.",".$city.",".$postalCode.",".$phone.",".$email2.")";

$result = mysql_query($query, $connexion);

if ($result)
{
// Success!
echo "Fabulous!  check the DB, we did it! :D<br>";
?>

<pre>
<?php
print_r($result);
?>
</pre>

<?php
} else {
// Fail!
echo"CRAAAAAPP!  something went wrong.  FIX IT!  :P<br>";
echo mysql_error();
}

if (isset($connexion)) { mysql_close($connexion); }
i copied it over from an old *working* file to illustrate how a simple INSERT works.

this is the error i get:

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 'company,big bay #8,some big warehouse,big city,T1T0N0,0123456789,bigKahuna@bigKa' at  line 4
looks completely valid to me.  all the database table elements are set to VARCHAR(80), so it can't be a space/type issue...

halp! 

WR!