PHP - Database Access By Table Prefix

Hi everyone! I'm getting the error "Access denied for user ''@'localhost' to database 'crystalair'" (crystalair is the name of my database) whenever I submit my form which inserts a new row to the database table "order". The same happens when I try to retrieve data from the table using select query.

However, I can successfully perform operations (select, insert) in my other pages which involves another database table "user". If this is a connection error, how can I successfully perform operations with another table? Also, my connection query does not return an error. I have also checked to make sure my user has been granted "All Privileges" in the phpmyadmin. I've run out of ideas. Can you please help? Thanks a lot.

Hi,
I have the below code to show the suggestion text, now it’s showing all the words which contain the letter when I first type in the text field. For example, If I typed the letter “a”, it should show only the words which start with the letter “a”, now it’s showing all the words which contain the letter “a” in the suggestion. Below is my code. Thanks in advance.

 

<?php
$xmlDoc=new DOMDocument();
$xmlDoc->load("links.xml");

$x=$xmlDoc->getElementsByTagName('link');

//get the q parameter from URL
$q=$_GET["q"];

//lookup all links from the xml file if length of q>0
if (strlen($q)>0) {
  $hint="";
  for($i=0; $i<($x->length); $i++) {
    $y=$x->item($i)->getElementsByTagName('title');
    $z=$x->item($i)->getElementsByTagName('url');
    if ($y->item(0)->nodeType==1) {
      //find a link matching the search text
      if (stristr($y->item(0)->childNodes->item(0)->nodeValue,$q)) {
        if ($hint=="") {
          $hint="<a href='" .
          $z->item(0)->childNodes->item(0)->nodeValue .
          "' target='_blank'>" .
          $y->item(0)->childNodes->item(0)->nodeValue . "</a>";
        } else {
          $hint=$hint . "<br /><a href='" .
          $z->item(0)->childNodes->item(0)->nodeValue .
          "' target='_blank'>" .
          $y->item(0)->childNodes->item(0)->nodeValue . "</a>";
        }
      }
    }
  }
}

// Set output to "no suggestion" if no hint was found
// or to the correct values
if ($hint=="") {
  $response="no suggestion";
} else {
  $response=$hint;
}

//output the response
echo $response;
?> 

 

hey guys trying to find out what the short name of FireFox2 is. (for example the "IE 6" in this code) What would it be for firefox versions 2?

Code: [Select]
<!--[if IE 6]>
<meta http-equiv="refresh" content="0; url=http://example.com/">
<![endif]-->
THANKS!!!

This topic has been moved to HTML Help.

http://www.phpfreaks.com/forums/index.php?topic=342336.0

this is my sql code
Code: [Select]
<?php
$query = "SELECT Player, 
                 SUM(GLI) AS 'gli', 
                 SUM(Goals) AS 'goals', 
                 SUM(Saves) AS 'saves', 
                 SUM(SOG) AS 'sog', 
                 SUM(Assists) AS'assists',  
                 SUM(CK) AS 'ck',  
                 SUM(YC) AS 'yc', 
                 SUM(RC) AS 'rc'  
          FROM pinkpanther_stats 
          GROUP BY Player; ";
?>
their are 11 separate players and i want to set variables $adam_glie and $tyler_glie equal to the glie column for that player(the player being the first part of the variable) and i dont have any idea how to go about doing this 

Good morning:

This is the output of my test database table:

2011-05-30 Chevrolet Monte Carol 1975 9000 15000 James Bond's 007-1212 james@bond.com B I am setting the rocord for bond's most active movies.
2011-05-31 Dodge Stratus 2008 5000 7500 James Cagney 555-1818 jc@hollywood.com S This is not going to be the best of time's. Ain't it?

Within my table I have two important fields--the record date-created field shown on the left and the (hidden) EXPIRED field--on which the UPDATE function depends.  This is my code for (wherein lies the problem) UPDATING the record within my table and setting the EXPIRED field to 'Y' for filtering and easy deletion of the record(s).
Code: [Select]
<?php
// Connect to database.
include("connect.php");
// Access the database 

@mysql_select_db($database,$con) or die( "Unable to select database");

$result = mysql_query("SELECT * FROM Autos"); //read all records into array

while($row = mysql_fetch_array($result)) //loop through array 
 {
  if (1 - strcmp(date("Y-m-d"),$row['DateCreated']) == 0) //compare current date with record date and act when result is zero 
  {
    mysql_query("UPDATE Autos SET Expired = 'Y' WHERE Expired == 'N'"); //if ready to expire then update table EXPIRED field  
  } else {
     if ($row['BuyerSeller'] == 'B') {
     echo "Contact buyer below";
   } elseif ($row['BuyerSeller'] == 'S') {
     echo "Contact seller below";
   }
   echo "<br />";
    echo $row['DateCreated'] . " " . $row['Make'] . " " . $row['Model'] . " " . $row['Year'] . " " . $row['MinPrice'] . " " . $row['MaxPrice'] . " " . $row['POC'] . " " . $row['POCPhone'] . " " . $row['POCEmail'] . " " . $row['Notes'];
    echo "<br />";
  }
 }
mysql_close($con)
?>
Lastly, does the mysql_close function with the single parameter $con also close the database?

MOD EDIT: code tags added.

Code: [Select]
mysql_query("SELECT $field FROM iNettuts WHERE user_id='1'") or die (mysql_error());
And I have a table named iNettuts.... in my database, my db is called "forum"

But I get the error...


Code: [Select]
Table 'forum.iNettuts' doesn't exist
Makes no sense? This works on localhost, but when I transfer to a life server on my VPS, it seems to bring this error. any idea? ty

im trying to write a script takes an xml files with tv show info, splits it into the show and the eppisode info and the place it into a table, i have got it to proccess all the info and print it out on a web page, but i cant, for the life of me, get it to insert said data into the table, it seems to be just ignoring the code and prints out the data as if nothing happens, no errors or anything. here is my code (abit messy but im only just starting and its my test.php)

Code: [Select]
<?php
$tvdb_mirror = "http://www.thetvdb.com/api/";
$tvdb_time = "http://www.thetvdb.com/api/Updates.php?type=none";
$dbname = "mediadb";
$dbuser = "root";
$dbpass = "";
$dbserv = "127.0.0.1";

$rss = simplexml_load_file('sample.xml');

$showName = "Show Name = ".$rss->Series->SeriesName;
print $showName;
print "<br />Show Discription = ".$rss->Series->Overview;
print "<br />";

mysql_connect('127.0.0.1', 'root', '');
@mysql_select_db('mediadb') or die("Unable to select database");

foreach ($rss->Episode as $item) {
$seasonnum = $item->Combined_season;
$EpisodeNumber = $item->EpisodeNumber;
 if($EpisodeNumber < 10){
$EpisodeNumber = "0".$EpisodeNumber;
 };
$EpisodeName = $item->EpisodeName;
$Overview = $item->Overview;
$airdate = $item->FirstAired;
$tvdbid = $item ->id;

$query = "INSERT INTO eppisodes VALUES('', '1', ".$EpisodeName.", ".$Overview.", ".$airdate.", '1', ".$tvdbid.", '-1', ".$seasonnum.", ".$EpisodeNumber.")";
mysql_query($query);

print "<br />".$showName." - ".$seasonnum."x".$EpisodeNumber." - ".$EpisodeName." Overview:<br />".$Overview;
}
mysql_close();
?>
i am a noob @ php and mysql, but i have doubke and triple checked the names of the db and table.

here is an sql dump of my db
Code: [Select]
-- phpMyAdmin SQL Dump
-- version 3.3.5
-- http://www.phpmyadmin.net
--
-- Host: 127.0.0.1
-- Generation Time: Nov 13, 2010 at 12:48 PM
-- Server version: 5.1.49
-- PHP Version: 5.3.3

SET SQL_MODE="NO_AUTO_VALUE_ON_ZERO";


/*!40101 SET @OLD_CHARACTER_SET_CLIENT=@@CHARACTER_SET_CLIENT */;
/*!40101 SET @OLD_CHARACTER_SET_RESULTS=@@CHARACTER_SET_RESULTS */;
/*!40101 SET @OLD_COLLATION_CONNECTION=@@COLLATION_CONNECTION */;
/*!40101 SET NAMES utf8 */;

--
-- Database: `mediadb`
--

-- --------------------------------------------------------

--
-- Table structure for table `eppisodes`
--

CREATE TABLE IF NOT EXISTS `eppisodes` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `showID` int(11) NOT NULL,
  `eppname` varchar(255) NOT NULL,
  `eppdesc` longtext NOT NULL,
  `airdate` date NOT NULL,
  `format` int(11) NOT NULL,
  `tvdbid` varchar(20) NOT NULL,
  `dohave` tinyint(1) NOT NULL,
  `season` varchar(2) NOT NULL,
  `eppisode` varchar(3) NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=1 ;

--
-- Dumping data for table `eppisodes`
--


-- --------------------------------------------------------

--
-- Table structure for table `shows`
--

CREATE TABLE IF NOT EXISTS `shows` (
  `id` int(100) NOT NULL AUTO_INCREMENT,
  `name` varchar(255) NOT NULL,
  `description` longtext NOT NULL,
  `TVDBID` int(100) NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=MyISAM  DEFAULT CHARSET=latin1 AUTO_INCREMENT=2 ;

--
-- Dumping data for table `shows`
--

INSERT INTO `shows` (`id`, `name`, `description`, `TVDBID`) VALUES
(1, 'higogo', 'some info', 67546);

any help would be very much appericiated.

fyi im running win7 with easyPHP 5.3.3 with php 5.3.3, mysql 5.1.49 apache 2.2.16

Ok, I got someone to help me fix this but he had no idea what the error was...

I have 2 tables, one called points and the other called members.

In members i have got:
id
name

In points i have got:
id
memberid
promo

I have the following code:
Code: [Select]
<?php
$con = mysql_connect("localhost","slay2day_User","slay2day");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("slay2day_database",$con);
$sqlquery="SELECT Sum(points.promo) AS score, members.name, members.id = points.memberid Order By members.name ASC";
$result=mysql_query($sqlquery,$con);
while ($row = mysql_fetch_array($result))
{

//get data
$id = $row['id'];
$name = $row['name'];
$score = $row['score'];

echo "<b>Name:</b> $name<br />";
echo "<b>Points: </b> $score<br />" ;
echo "<b>Rank: </b>";
if  ($name == 'Kcroto1'):
  echo 'The Awesome Leader';
else:
if ($points >= '50'):
  echo 'General';
elseif ($points >= '20'):
  echo 'Captain!';
elseif ($points >= '10'):
  echo 'lieutenant';
elseif ($points >= '5'):
  echo 'Sergeant';
elseif ($points >= '2'):
  echo 'Corporal';
else:
  echo 'Recruit';
endif;
endif;
echo '<br /><br />';

}

?>
I am getting the following error when i do the query in mysql:
Code: [Select]
#1109 - Unknown table 'points' in field list
And when i open the webpage i get the following error:
Code: [Select]
Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /home/slay2day/public_html/points/members.php on line 18
Please Help me?

Hi,

I am trying to store a sql query in a database, but every time I try to read it, it shows all the code as text. eg just print ' . $parts_row['part_number'] . '

Here is the code on the page:       

    $parts_sql = "SELECT * FROM parts WHERE part_cat = " . $category_row['cat_id'] . " ORDER BY CAST(part_a AS DECIMAL(10,2))";
  			$parts_result = mysql_query($parts_sql);
  
  			if(!$parts_result)
  			{
  				echo '<tr><td>The parts could not be displayed, please try again later.</tr></td></table>';
  			}
  			else
 			{
 
 				while($parts_row = mysql_fetch_array($parts_result))
 				{
                                               $pageformat_sql = "SELECT * FROM category_pages WHERE catpage_number = " . $category_row['cat_page'] . "";
 
                                               $pageformat_result = mysql_query($pageformat_sql);
 
 
                                                if(!$pageformat_result)
                                                {
                                                echo 'Error';
                                                }
                                                 else
                                                 {
                                                 while($pageformat_row = mysql_fetch_assoc($pageformat_result))
                                                 {
                                                   $data = $pageformat_row['catpage_table'];
                                                 echo '<table class="table2 centre" style="width:750px">
 	                                              ' . $pageformat_row['catpage_tabletitle'] . '' . $data . '';
 
 
 
                                                 }
                                                 }
 
     }}

And this is what is stored in the database table:

 <tr>
 <td class="cell left">' . $parts_row['part_number'] . '</td>
 <td class="cell centre">' . $parts_row['part_a'] . ' mm</td>
 <td class="cell centre">' . $parts_row['part_b'] . ' mm</td>
 <td class="cell centre">' . $parts_row['part_c'] . ' mm</td>
 <td class="cell centre">' . $parts_row['part_d'] . ' mm</td>
 <td class="cell centre">' . $parts_row['part_e'] . ' mm</td>
 <td class="cell centre">' . $parts_row['part_imped100'] . '</td>
 <td class="cell centre"><a href="datasheets/' . $parts_row['part_datasheet'] . '.pdf"><img border="0" src="images/pdf.jpg" alt="Download"</a></td></tr>

I would be very grateful to anyone who can help me with this.

Hello again!   I'm here with another PHP/Mysqli related question, I want to output the data i receive from the Query:

SHOW tables; 

Into html code, more specific: links. This is what i try'd to do:

<div class="container">
  <div class="header" align="center">
   <nav>
		<ul>
			<li><a href="#">Home</a></li>
			<li><a href="#">Databases</a>
				<ul>
<?php
$mysqli = new mysqli('localhost','*****','*****','*****');
if ($mysqli->connect_error) {
    die("Connection failed: " . $mysqli->connect_error);
} 
$sql = 'show tables from osiris1q_mtg';
$result = $mysqli->query($sql);
if(!$result = $mysqli->query($sql)){
    die('There was an error running the query [' . $mysqli->error . ']');
}
    // output data of each row
    while($row = $result->fetch_assoc()){
    echo "<li><a href='#'></a>"$row"</li>";
}
} else {
    echo "<li><a href='#'></a>0 results</li>";
}				</ul>
			</li>
		</ul>
	</nav>
</div>    <!-- end .header -->
?>

This is the header.php file which is included in the actual index page. This doesn't return anything not even the html parts of it. Even all the files i have 'included' aren't showing up anymore. I hope i explained myself good enough for anyone to understand   How Anyone can help me out!   Thank you in Advance!

Hi everyone, wondering if any one could help me, probably simple to most. I am using an API from a video game and I am trying to insert the xml data I get back in to a MYSQL database table. but can't seem to figure it out.

I created a simple echo foreach row which works, but I cannot seem to figure it out how to insert in to DB table. also the rows are attributes don't know if that makes a difference?
Code: [Select]
xml = simplexml_load_file('file.xml');

foreach ($xml->result->rowset->row as $row) {
echo '<tr class="' . $row['ownerName2'] . '">' . '<td>' . $row['date'] . '</td>' . '<td>' . $row['ownerName1'] . '</td>' . '<td>' . $row['ownerName2'] . '</td>' . '<td>' . $row['amount'] . '<b>ISK</b>' . '</td>' . '<td>' . $row['balance'] . '<b>ISK</b></td>' . '<td>' . $row['reason'] . '&nbsp;</td>' . '</tr>';
};

This is the structure of the xml document if you need to have a look.

Code: [Select]
<result>
<rowset name="entries" key="refID" columns="date,refID,refTypeID,ownerName1,ownerID1,ownerName2,ownerID2,argName1,argID1,amount,balance,reason">
<row date="2011-09-16 14:19:00" refID="4873095949" refTypeID="37" ownerName1="name" ownerID1="435353535" ownerName2="name" ownerID2="2342523523" argName1="name" argID1="325235252" amount="-1101525.00" balance="0.00" reason="DESC:"/>
<row date="2011-09-16 14:19:00" refID="4873095949" refTypeID="37" ownerName1="name" ownerID1="435353535" ownerName2="name" ownerID2="2342523523" argName1="name" argID1="325235252" amount="-1101525.00" balance="0.00" reason="DESC:"/>
<row date="2011-09-16 14:19:00" refID="4873095949" refTypeID="37" ownerName1="name" ownerID1="435353535" ownerName2="name" ownerID2="2342523523" argName1="name" argID1="325235252" amount="-1101525.00" balance="0.00" reason="DESC:"/>
</rowset>
</result>

Thanks in advance.

 

Hello forum.
I'm new here, but I've been reading and finding useful things for a while now.

I'm still new to PHP and I need a little help.

I'm doing a school project and I have some things I want to do, but do not know how to write it down in PHP.

I think I'll ask a lot of questions this week, and I hope I will get some help..

For start I want to ask this:

I've been using Code: [Select]
mysql_fetch_array() for doing loops and populating check-boxes. And everything's working fine.. but what I want is to control the actual loop by clicking buttons. Let's say first time a while-do is run, my check-boxes get populated from the database and every other loop the next data from the table is added.. pretty straightforward.
I want to be able to populate once, then click "next" and the new data to be added and so on..
Code: [Select]
<?php
$tema = mysql_query("SELECT * from questions")or die(mysql_error());

function answer1($string)
{
$string1 = explode("/", $string);
echo $string1[0];
}
function answer2($string)
{
$string1 = explode("/", $string);
echo $string1[1];
}

while ($row=mysql_fetch_array($tema))
{
echo mysql_fetch_array($tema);
$tip=$row["tip"];
if ($tip==2)
{
$id=$row['prasanje_id'];
$question=$row['question'];
$answer=$row['answer'];
?>
   <label> <?php echo $question?></label><br>
   <input type="checkbox"  name="CheckboxGroup1" value="checkbox" id="CheckboxGroup1_0" />
      <?php answer1($tekst) ?></label>
    <label>
      <input type="checkbox"  name="CheckboxGroup1" value="checkbox" id="CheckboxGroup1_1" />
      <?php answer2($tekst) ?></label>
<?php
}
}
?>I want an alternative to the while-do loop..
Is it possible to do this?

Thanks!!