PHP - Mysql, Php - Turn Result Into Variable

Hello,
So I have a content management site set up, but I'm stuck on one part. I have a query. Code: [Select]
Select `email` FROM author WHERE author.id IN(SELECT authorid FROM job WHERE id = '$_POST[id]')
I'm a beginner in PHP and SQL. Basically I need the query to run, it will come out with one result and I need that to be where the email is sent to.

This is the existing code I have for sending out the email $subject = " Job Application Confirmation";

$message = "Hello $_POST[name] , \r\rYou, or someone using your email address, has applied for the job: $text Using the Resource Locator at Prahan.com. Keep this as future reference. \r\r Full Name: $_POST[name] \r Location: $_POST[location] \r Email: $_POST[email] \r Additional Information: $_POST[info] \r Job ID: $id  \r Job Name: $text  \r \r Expect a response shortly. \r\r Regards, Prahan.com Team";

$headers = 'From: Prahan.com Team <noreply@prahan.com>';

    'Reply-To: noreply@prahan.com' . "\r\n" .

    'X-Mailer: PHP/' . phpversion();

$to1      = 'pratik@prahan.com';

mail($to1, $result, $message, $headers);


Thanks in Advance!

Hello Guys,
I have a column named message_status it can only have the values "read" or "unread" in it. I would like to show an image in the column next to it. If the message is "read" then I would like to display a green dot, and if it is unread I would like to display a red dot. I would also like to be able to have anything in the row that's "unread" to appear in bold. I completely stumped by this! I have written the following code, which doesn't seem to work, presumably because it is run after the table has been displayed? I am very new to this so please be gentle!

Regards,

AJLX



$result = mysql_query("SELECT * FROM messages where username ='$username'ORDER BY {$_GET['orderbycol']} $sort");

echo "<table border='1'>
<tr width='200'>
<th>View</th>
// code to display table removed
</tr>"; 
$green_dot = $message_status; 
$red_dot = $message_status;
$green = "";
$red = "";
$bold = "";
while ($row = mysql_fetch_array($result))
{
  echo  "<tr>";
  $id = $row['ID'];
  echo  "<td><a href='/view_message.php/?view_message=$id'>View</a></td>";
  echo "<td>"  .$green . $red."</td>";
  echo  "<td>" .$bold . $message_status = $row['message_status'] . "</td>";
  echo  "<td>" .$bold. $row['date_sent'] . "</td>";
  echo  "<td>" .$bold. $row['contact_name'] . "</td>";
  echo  "<td>" .$bold. $row['subject'] . "</td>";
  echo  "</tr>";
 }
 if ($green_dot = 'read'){
 $green ='<img src=\"assets/red.jpg\"/>';
 echo $green;
 } else {
  $red ='<img src=\"assets/red.jpg\"/>';
  $bold = "<b>";
  echo $red;
 

I want to save the results of a loop as a variable ie $output.  I have tried encasing the php within quotes but it does not work.  Is there a way to save the complete results as a variable?

$result = mysql_query("SELECT * FROM $table2 WHERE $db_item_1 OR $db_item_2", $connection);   if (!mysql_num_rows($result)) {   echo "Error 13424 - not working";   exit();   }    while ($row = mysql_fetch_array($result))  { echo "<tr><td scope=\"col\" style=\"font-size: 16px; color: #333333; padding: 10px 8px; border-bottom: 1px solid #919191;\"><div align=\"left\" style=\"font-size: 20px;\">" . $row['date'] . "</div></td>            <td scope=\"col\" style=\"font-size: 16px; color: #333333; padding: 10px 8px; border-bottom: 1px solid #919191;\"><div align=\"left\"></div></td>            <td scope=\"col\" style=\"font-size: 16px; color: #333333; padding: 10px 8px; border-bottom: 1px solid #919191;\"><div align=\"left\" style=\"font-size: 20px;\">" . $row['title'] . "</div></td>            <td scope=\"col\" style=\"font-size: 16px; color: #333333; padding: 10px 8px; border-bottom: 1px solid #919191;\"><div align=\"left\"></div></td>            <td scope=\"col\" style=\"font-size: 16px; color: #333333; padding: 10px 8px; border-bottom: 1px solid #919191;\"><div align=\"left\" style=\"font-size: 20px;\">" . $row['cost'] . "</div></td> </tr>"; }echo $complete;()

The part I want as one result (ie. $complete) is the result of the while loop.  There is always at least one result but sometimes 10 which means that it creates 10 table rows.  I need to do it this way as later on in the page I use a pdf converter which does not allow loop checks within it otherwise I would just place it within the converter.

Hello,

I am creating a class which contains one private function that deals with connecting to a SQL Server and executing the sql that is passed in to that function.
I have defined a class variable which is assigned the value of sqlsrv_query like so.

$this->QueryResult = sqlsrv_query($conn,$sql);
I have placed private $QueryResult at the top of my class.

The other public methods call this private function to assign the result set to the class variable, the private method returns and then the public methods loop through the results array like so.

while($row = sqlsrv_fetch_array($this->QueryResult))

.. but the while loop never gets entered.

If I declare everything in the same method then it works, however there are going to be several public methods that will use this private method and I don't want to duplicate all the database coding.

Hope someone can help

How can I check if a returned mysql value is equal to '' i.e. nothing?

I keep getting an error where the page won't load because the returned value is '' so i need to check for it

I am trying to allow the user to update a variable he chooses by radio buttons, which they will then input text into a box, and submit, to change some attributes. I really need some help here.  It works just fine until I add the second layer of variables on top of it, and I can't find the answer to this question anywhere.

<?PHP require('connect.php'); ?>

<form action ='' method='post'> <select name="id">

<?php
$extract = mysql_query("SELECT * FROM cars");
   
   while($row=mysql_fetch_assoc($extract)){
   $id = $row['id'];
   $make= $row['make'];
   $model= $row['model'];
   $year= $row['year'];
   $color= $row['color'];

echo "<option value=$id>$color $year $make $model</option>
";}?>

</select>

Which attribute would you like to change?<br />
<input type="radio" name="getchanged" value="make"/>Make<br />
<input type="radio" name="getchanged" value="model"/>Model<br />
<input type="radio" name="getchanged" value="year" />Year<br />
<input type="radio" name="getchanged" value="color"   />Color<br /><br />

   <br /><input type='text' value='' name='tochange'>
   <input type='submit' value='Change' name='submit'>

</form>

//This is where I need help...

<?PHP
if(isset($_POST['submit'])&&($_POST['tochange'])){
mysql_query("
   UPDATE cars
   SET '$_POST[getchanged]'='$_POST[tochange]'
   where id = '$_POST[id]'
         ");}?>

hello
I'm using this code:
Code: [Select]
$query="SELECT * FROM `second_content` WHERE CHANGED =0 limit 0,1";to query first row data when I want see it and echo it I recived :
Code: [Select]
Resource id #2can anyone help me ?


thank you

Hello, i created this script for a client and have ran into an annoying error with the results displaying on a new line for each result instead of side by side, any help is welcome

Cheers

Code: [Select]
  <?php
  $subcat = mysql_real_escape_string(strip_tags(htmlspecialchars(protect($_GET['subcat']))));
  $cat =  mysql_real_escape_string(strip_tags(htmlspecialchars(protect($_GET['cat']))));
$sql = @mysql_query("SELECT * FROM cakes WHERE category =\"$cat\" AND sub_cat=\"$subcat\" ORDER BY id DESC");
while ($row = mysql_fetch_array($sql)) {
$reference = $row['reference'];
$image = $row['image'];
echo ("<p><img src='./images/cakes/$image' height='289px' width='177px' alt='IMAGE OF CAKE'></img><br />");
echo ("<b>Reference:</b>$reference"."</p>");
}
if (!$reference) {
echo 'There are no cakes in this category yet.';
}
?>
I know the errors only going to be something small i'm missing, but i've been coding all day

Hi,

I don't know anything about php and Mysql but I found a tutorial for a shopping cart and everything is working.
The only thing is they don't have the email part in the totorial so I'm kind of stuck with my file.
Anyone know how to email the resul to me via email with this code?

<?
   include("includes/db.php");
   include("includes/functions.php");
   
   if($_REQUEST['command']=='update'){
      $name=$_REQUEST['name'];
      $email=$_REQUEST['email'];
      $address=$_REQUEST['address'];
      $phone=$_REQUEST['phone'];
      
      $result=mysql_query("insert into customers values('','$name','$email','$address','$phone')");
      $customerid=mysql_insert_id();
      $date=date('Y-m-d');
      $result=mysql_query("insert into orders values('','$date','$customerid')");
      $orderid=mysql_insert_id();
      
      $max=count($_SESSION['cart']);
      for($i=0;$i<$max;$i++){
         $pid=$_SESSION['cart'][$i]['productid'];
         $q=$_SESSION['cart'][$i]['qty'];
         $price=get_price($pid);
         mysql_query("insert into order_detail values ($orderid,$pid,$q,$price)");
      }
      die('Thank You! your order has been placed!');
   }
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Billing Info</title>
<script language="javascript">
   function validate(){
      var f=document.form1;
      if(f.name.value==''){
         alert('Your name is required');
         f.name.focus();
         return false;
      }
      f.command.value='update';
      f.submit();
   }
</script>
</head>


<body>
<form name="form1" onsubmit="return validate()">
    <input type="hidden" name="command" />
   <div align="center">
        <h1 align="center">Billing information</h1>
        <table border="0" cellpadding="2px">
           <tr><td>Total:</td><td><?=get_order_total()?></td></tr>
            <tr><td>Name :</td><td><input type="text" name="name" /></td></tr>
            <tr><td>Address :</td><td><input type="text" name="address" /></td></tr>
            <tr><td>Email :</td><td><input type="text" name="email" /></td></tr>
            <tr><td>Phone :</td><td><input type="text" name="phone" /></td></tr>
            <tr><td>&nbsp;</td><td><input type="submit" value="Place Order" /></td></tr>
        </table>
   </div>
</form>
</body>
</html>

Im not sure where to post this but since it includes php il post it here instead of in the mysql forum.

ok so, i have a table and i get the values using while($row = mysql_fetch_array($result)){ and then echo them in rows.

that works fine but i need to add a class to the last row of my table. I would need somehow to fetch the last row of the array and make it echo something different.

Any help is appreciated

Thank you

I am seeking to learn more about the noted subject, how to use PHP to allow a user to enter search terms and search a database.  I have experimented with this with little results save for errors.  Please see code listed below:

search.php
<?
//// filename = search.php
<form method="post" action="result.php3">
<select name="metode" size="1">
<option value="row_name1">metode1</option>
<option value="row_name2">metode2</option>
</select>
<input type="text" name="search" size="25">
<input type="submit" value="Begin Searching!!">
</form>
?> 

results.php
//// filename = result.php3
<?
$hostname = "mysql7.000webhost.com"; // Usually localhost.
$username = "a4542527_root"; // If you have no username, leave this space empty.
$password = "*******"; // The same applies here.
$usertable = "people"; // This is the table you made.
$dbName = "a4542527_test1"; // This is the main database you connect to.
MYSQL_CONNECT($hostname, $username, $password) OR DIE("Unable to connect to database");
@mysql_select_db( "$dbName") or die( "Unable to select database");
?>
<?
//error message (not found message)
$XX = "No Record Found";
$query = mysql_query("SELECT * FROM $usertable WHERE $metode LIKE '%$search%' LIMIT 0, 30 ");
while ($row = mysql_fetch_array($query))
{
$variable1=$row["row_name1"];
$variable2=$row["row_name2"];
$variable3=$row["row_name3"];
print ("this is for $variable1, and this print the variable2 end so on...");
}

//below this is the function for no record!!
if (!$variable1)
{
print ("$XX");
}
//end
?>

Upon viewing search.php I receive the error message:
Parse error: syntax error, unexpected '<' in /home/a4542527/public_html/search.php on line 3

I believe I may be missing something and am a bit lost.
Thank-you in in advance for any help or suggestions.
~Matty

I am having a little bit of trouble with this piece of code.  I'm sure it's something simple, but I have been working on this thing all day and want to get it finally finished.

Here's the troublesome code:


function rrmdir($dir) {
   if (is_dir($dir)) {
     $objects = scandir($dir);
     foreach ($objects as $object) {
       if ($object != "." && $object != "..") {
         if (filetype($dir."/".$object) == "dir") rrmdir($dir."/".$object); else unlink($dir."/".$object);
       }
     }
     reset($objects);
     rmdir($dir);
   }
 }

$sql_clean = "SELECT * complete WHERE createdate < date_sub(current_date, interval 1 minute)";
$sql_list = mysql_query($sql_clean);

while($row = mysql_fetch_assoc($sql_list)) {
    $directory = "complete/" . $row['fileurl'];
    rrmdir($directory);
    }

The purpose of this particular bit is to run on a cron every few days.  It gets "createdate" and other info from the "complete" table in order to know how old the record is.  If the record is older than (in the example, 1 minute; it will be set to several days on public) the defined max age, it removes that directory and everything within it to keep the directory clean and the disk usage down.

The error returned is
Quote

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home1/latenit2/public_html/kindleprocessor/process/garbagecleaner.php on line 46


Line 46 is
Quote

while($row = mysql_fetch_assoc($sql_list)) {


I may be doing the look-up on the MySQL database incorrectly, too.  I haven't discounted that, and I'd be thankful if someone could help me out with this issue.

hey guys,

i need some help with my php/mysql iplogger. My code:

<?php

    //finds out ip
    $ip = $_SERVER['REMOTE_ADDR'];
    //conects to the mysql server
    $connection = mysql_connect('localhost', 'root', '');
    //sellects the database
    mysql_select_db('iplog', $connection);
    //looks for duplacute ips
    $dup = mysql_query("SELECT COUNT(number) FROM logged_ips WHERE ip_address = '$ip'",$connection);
    $count = mysql_result($dup, 0);
    //checks to see if there is a duplecate name
    if ($count == 0){
        //inserts the ip in to the database
        $string = 'INSERT INTO `logged_ips` (`aid`, `ip_address`, `ip_visits`) VALUES (\'' . null . '\', \'' . $ip . '\', \'0\')';
        mysql_query($string, $connection);
    }else{
        //adds a visit to the database
        $string2 = "UPDATE `logged_ips` SET `ip_visits` =  '++1' WHERE `ip_address` = $ip LIMIT 0,1";
        mysql_query($string2, $connection);
    }
    //outputs the ip
    echo $ip;
?>

error:

Warning: mysql_result() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\randoms\logger\test.php on line 11
127.0.0.1

It also doesnt put anything in to the mysql database. Please help.

Thanks jragon

Hey guys,

Currently Im using:

$row = mysql_fetch_array($result) or die(mysql_error());
echo $row['user_family']. " - ". $row['user_registered'];
$row['user_family'] = $fam;
$_SESSION['family'] = $fam;

to take data from a mysql table & set it as SESSION family. However, I cant seem to get this to set.

The information IS being taken from mysql because its being echo'd earlier up in the code, but its just not passing to the session.

Any ideas?