PHP - Syntax For Variable In Select Statement

I have a Select statement that relies on two session variables to complete the statement. One variable, $egroup, supplies a column name. I cannot make the SELECT work. It works fine if I substitute the real name of one of the columns  such as "egroup1", or "egroup6", but that's what the variable $egroup is for. Can anyone tell me where I'm wrong?

I guess the answer is in the syntax, but I'm not very good at this and I've been on it for too long. Any help gratefully taken.

Code: [Select]
<?php                 
$query1 = mysql_query("SELECT * FROM topics WHERE managerId='".$managerId."' AND "$egroup"= 1 ORDER BY title ASC");
?>

$sql=mysql_query("SELECT * FROM `buds` WHERE `level`<='$user_level' UNION SELECT * FROM `buds`, `unlocked_buds` WHERE buds.`id` = unlocked_buds.`bud_id` ORDER BY buds.`id` ASC") or die("A MySQL error has occurred.<br />Your Query: " . $sql . "<br /> Error: (" . mysql_errno() . ") " . mysql_error());



I have been trying to learn about UNION select statements. I ran the query above and got this response:

Quote

Error: (1222) The used SELECT statements have a different number of columns



I think I know what the problem is, but not sure how to fix it.

There is two columns, one is "buds" which holds the flowers seeds info. The seconds is "unlocked_buds" which just links the "id" from buds to "user_id" to the user table.

Both buds and unlocked_buds are both 8 columns.

What do I need to learn?

Hey guys, I'm trying to get this working... No errors right now, but I'm not returning any results :/ been messing with it for days.

Code: [Select]
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post" />
<label for="search_by">Search By</label>
<select name="search_by">
    <option value"player">Player</option>
    <option value"city">City</option>
    <option value"alliance">Alliance</option>
    <option value"browse">Browse</option>
    </select>
    <input type="text" name"search">
    <input type="submit" value="search" name="search">
 <?php
  $search_by = $_POST['search_by'];
$search = $_POST['search'];

  echo "<table><tr><td>Player</td><td>city</td><td>alliance</td><td>x</td><td>y</td><td>other</td><td>porters</td><td>conscripts</td><td>Spies</td><td>HBD</td><td>Minos</td><td>LBM</td><td>SSD</td><td>BD</td><td>AT</td><td>Giants</td><td>Mirrors</td><td>Fangs</td><td>ogres</td><td>banshee</td></tr>" ; 

$dbc = mysqli_connect('xx', 'xx', 'xx', 'xx') or die ('Error connecting to MySQL server');

$sql = "SELECT * FROM players WHERE ('$search_by') LIKE ('$search') ";
//problem is here^^?? 
$result = mysqli_query($dbc,$sql)
  or die("Error: " .mysqli_error($dbc));
Not sure, any help would be greatly appreciated.

Hi All, I have the following Query:

Code: [Select]
$query = 'SELECT `BookingID`,`OfferID`, `name`, `OfferCount`,`OfferPrice` FROM `Offers_Finance` JOIN offers_details ON `OfferID` =  `o_d_id` WHERE `BookingID` ="' .$bookerid. '"';
if ( !$result = $mysqli->query( $query ) ) { die( $mysqli->error ); }
$field = $result->fetch_object();

This will return anywhere between 0 and 50 Rows of Data.  For Each Row returned, I want to do the following:

Quote

        echo '<tr>';
   echo '<td>' . $field->name.'</td>';
   echo '<td></td>';
   echo '<td>&pound;' . $field->OfferPrice.'</td>';
   echo '<td></td>';
   echo '<td>' . $field->OfferCount . '</td>';
   echo '<td></td>';
   echo '<td>&pound;' . $field->OfferPrice * $field->OfferCount . '</td>';
   echo '</td>';
   echo '</tr>';



The problem I have is - how do I loop through the results returned from the query to output multiple rows?

Sorry if this doesn't make much snese, I'm new to PHP and the whole web development world, but have been landed with finishing a project someone else started!

If someone would be kind enough to turn this into a working example, it would help no end as I have about 20 of these things to figure out across the project!

[text]I created two tables in a database called 'members' and 'blogs'.  The blogs table uses the primary key of the members table (member_id), as a foreign key.  The member_id is an auto incremented column in the members table and when I query and print out the rows of this table, the member_id values for the two members I created, turn out to be 1 and 2 as expected.    Now when I use a subselect query in an insert statement, to input member_id values into the blogs table, and then query the rows of this table, both member_id values show up as 0.  I will display both tables and the insert query for the blogs table below.  Can anyone identify the problem?  I'm convinced there is something about the subselect query that I'm not getting right.  Ill also include the select query that displays the results of the blogs table just in case. [/text]



Code: [Select]
<?php
$query = "CREATE TABLE members (

member_id INT UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY ,
username VARCHAR( 50 ) NOT NULL UNIQUE,
firstname VARCHAR( 50 ) NOT NULL ,
lastname VARCHAR( 50 ) NOT NULL ,
title VARCHAR(10),
password VARCHAR( 50 ) NOT NULL ,
primary_email VARCHAR(100),
secondary_email VARCHAR(100),
register_date DATE,
ip VARCHAR( 50 ) NOT NULL ,
UNIQUE (username)
      
          )";
?>




<?php
$query = "CREATE TABLE blogs (

blog_id INT UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY ,
member_id INT UNSIGNED,
like_id INT UNSIGNED,
title VARCHAR( 500 ) NOT NULL,
entry VARCHAR( 2000 ) NOT NULL ,
blog_date DATE  

          )";
?>




<?php
$query = "INSERT INTO blogs ( member_id, title, entry, blog_date) VALUES ( 'SELECT member_id FROM members', '{$_POST['title']}', '{$_POST['entry']}', NOW())";
?>




<?php
$query= 'SELECT * FROM blogs';
if($r = mysql_query ($query)) {//Run the query.
//Retrieve and print every record.
     while ($row = mysql_fetch_array ($r)) {
          print " {$row['title']}" ;
          print " {$row['entry']}" ;
          print " {$row['blog_date']}" ;
          print " {$row['member_id']}" ;
    }
 } else {//Query didn't run.
        die (' Could not retrieve the data becasue: .mysql_error().
             ');
}//End of query IF.
mysql_close(); //Close the database connection.
?>



Any help is appreciated.

Hi Guys

I don't know if this is possible but can someone point me in the right direction. I have a php function which takes two inputs and returns an output. for simplicity's sake let's say it's an addition function. What I want to do is use a mysql select statement to show all the rows from a database where field1 and field2 equal '3'. Here's the sort of thing I mean.

function addNumbers($one,$two) {
return $one + $two;
}

mysql_query("SELECT * FROM table WHERE 'addNumbers(field1,field2)' = '3'");

What I actually want to do is a lot more complex than this but I am trying to understand how to make the syntax work in simple terms first.

Can anybody help?

Many Thanks
Dan

Hi everybody
i am beginnet with working with sql in php..so i confused.....i made one scripte and i am making the secound and now i am facing a problem....i hope u could help!!
i have a dat=abse of my chatroom....this is the code i typed...please have a look at it!
Code: [Select]
$usern=$_SESSION['etchat_'.$this->_prefix.'username'];
$con1 = mysql_connect("localhost","manosir_main","********");
if (!$con1)
  {
  die('Could not connect: ' . mysql_error());
  }


mysql_select_db("manosir_manoshos_main", $con1);
$users= mysql_query(" select etchat_user_online_user_sex  from db1_etchat_useronline where etchat_user_online_user_name = $usern ");


 
in my database, i ahve a table called db1_etchat_useronline that keeps record of online users in chatroom.
as u usee i qant to to get the user sex ( etchat_user_online_user_sex) of th current user!!! as u see i conected to database correctly but now i dont know how to echo the $users . before this i echo my sql results with mysql_fetch_array ..but  now i cant.......i got nothing!!!! can someone help me!!!!!

here is some simple code for getting and displaying fata from a database

Code: [Select]
$sql="SELECT * FROM messages WHERE m_id = '".$id."'";

$result = mysql_query($sql);

<table border='0' cellspacing="4">

while($row = mysql_fetch_array($result))
  {
  echo "<tr>";
  echo "<th>Message</th>";
  echo "</tr>";
  echo "<tr>";
  echo "<td>" . $row['message'] . "</td>";
  echo "</tr>";
  }

now ive used that while loop to display those results how can i do that again without having a new SQL statement.

i cant do 2 while loops because the first one has already got to the end of the amount of rows

basically how can i have another while loop displaying those same results again?

This topic has been moved to MySQL Help.

http://www.phpfreaks.com/forums/index.php?topic=327011.0

I have got a set of data for a country for each week of the year, but want to display the data a quarter at a time, using a form which posts a value into LIMIT, which was OK for the 1st quarter, but for subsequent quarters I needed to use LIMIT in the form LIMIT start, rows. I found by experiment that this had to be the last statement in the query. when I added the start I got an error: Parse error: syntax error, unexpected ',' in /homepages/43/d344817611/htdocs/Admin/Visitdata.php on line 10 Here is the SQL part of my code:

$Visit_data="SELECT WeekNo.WNo, WeekNo.WCom, Countries.Country, ctryvisits.CVisits FROM ctryvisits 
LEFT JOIN Countries
ON ctryvisits.country=Countries.CID
LEFT JOIN WeekNo
      ON ctryvisits.WNo=WeekNo.WNo
      WHERE Countries.Country = '".$_POST["Country"]."'
ORDER BY ctryvisits.WNo LIMIT '".$_POST["QUARTER"]."'" - 13, '".$_POST["QUARTER"]."'";

Is this the best way of doing what I require or is there an easier way of achieving what I want?

//DATABASE CONNECTION VARIABLES
$myserver ="localhost";
$myname = "myname";
$mypassword = "mypassword";
$mydb ="mygamedb";


/*SQL CONNECTION*/
// Create connection
$conn = new mysqli($myserver, $myname, $mypassword, $mydb);

// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}
else {

//variables
$username = $_POST["username"];
$password = $_POST["password"];
$ip = $_SERVER['REMOTE_ADDR'];

//INSERT USER
//prepare and bind
$stmt = $conn->prepare("INSERT INTO Players (Username, Password, IP) VALUES (?, ?, ?)");
//bind parameters
$stmt->bind_param("sss", $username, $password, $ip);
//set parameters and execute
$stmt->execute();
//close
$stmt->close();

//FETCH ID
$resultnews = mysql_query("SELECT * FROM Players WHERE Username ='$username'");
$rownews = mysql_fetch_array($resultnews);
$user_id = $rownews["ID"];


}

After having suffered an SQL injection into one of my sites, I figured it was time to overhaul it and use prepared statements. I am new to this. I figured out how to an INSERT with an example, but now I need to fetch an ID and cannot get it to work. Any help much obliged. All I need is just one good example. Looked all over the place, but all I get are insert examples, which is NOT what i need. Really need one with a select and fetch.

This topic has been moved to MySQL Help.

http://www.phpfreaks.com/forums/index.php?topic=322930.0

I need to convert the following select statement to a pdo->query but have no idea how to get it working:

SELECT t.id FROM 
  ( SELECT g.* 
    FROM location AS g
    WHERE g.start <=  16785408
    ORDER BY g.start DESC, g.end DESC
    LIMIT 1
  ) AS t
WHERE t.end >= 16785408;

Here's the code I'm trying:

<?php

$php_scripts = '../../php/';
require $php_scripts . 'PDO_Connection_Select.php';
require $php_scripts . 'GetUserIpAddr.php';
function mydloader($l_filename=NULL)

{
$ip = GetUserIpAddr();
if (!$pdo = PDOConnect("foxclone_data"))
{    
    exit;
}
    if( isset( $l_filename ) ) {  
        $ext = pathinfo($l_filename, PATHINFO_EXTENSION);
        $stmt = $pdo->prepare("INSERT INTO download (address, filename,ip_address) VALUES (?, ?, inet_aton('$ip'))");
        $stmt->execute([$ip, $ext]) ; 

        $test = $pdo->prepare("SELECT t.id FROM ( SELECT g.id FROM lookup AS g WHERE g.start <= inet_aton($ip) ORDER BY g.start DESC, g.end DESC ) AS  t  WHERE t.end >=inet_aton($ip)");
        $test ->execute() ; 
        $ref = $test->fetchColumn();
        $ref = intval($ref);

        $stmt = $pdo->prepare("UPDATE download SET ref = '$ref' WHERE address = '$ip'");
        $stmt->execute() ;         

        header('Content-Type: octet-stream');
        header("Content-Disposition: attachment; filename={$l_filename}");
        header('Pragma: no-cache');
        header('Expires: 0');        
        readfile($l_filename);
       }
        
    else {
        echo "isset failed";
        }  
}
mydloader($_GET["f"]);
exit;

 It gives the following error:

Quote

Fatal error: Uncaught PDOException: SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near '.144.181) ORDER BY g.start DESC, g.end DESC ) AS  t  WHERE t.end >=inet_aton(7' at line 1 in /home/foxclone/test.foxclone.com/download/mydloader.php:19 Stack trace: #0 /home/foxclone/test.foxclone.com/download/mydloader.php(19): PDO->prepare('SELECT t.id FRO...') #1 /home/foxclone/test.foxclone.com/download/mydloader.php(38): mydloader('foxclone40a_amd...') #2 {main} thrown in /home/foxclone/test.foxclone.com/download/mydloader.php on line 19

How do I fix this?

I'm able to use PHP-ODBC to read the tables in an MDB file, as well as the column names on the tables. But when I try to read data from a particular table, it fails (I don't get an error, instead the browser pops up a message asking if I want to save the file "mdbtest.php" which is 0 bytes in size)! Any idea what's wrong?

Code: [Select]
//THIS WORKS
$conn = odbc_connect( 'TestDB', '', '' );
if ( !$conn) exit("Connection Failed: " . $conn);

//STILL FINE HERE
$sql = 'SELECT * FROM Customers';

//INCLUDING THIS LINE OF CODE CAUSES THE PROBLEM
$rs = odbc_exec( $conn, $sql );

how do i handle single quotes in sql query

Code: [Select]
" SELECT name from phrase WHERE name='$stitle' ";this returns an error because the name contains single quotes like this:
Johnson's.