PHP - Need Help For Email Mysql Result

Hi,

I am new to php but not to programming.

I created a script on a windows platform which connects to the mysql database and returns the results of a table.  A very basic script which I wrote to simply test my connection worked.  The script works fine on my windows machine but not on my new mac.  On the mac it simply does not display any records at all.

I know that the database connection has been established because there is no error but I can not see why the result set is not being displayed on screen, as I said it worked fine on my windows machine.

The Mac has mysql (with data) and apache running for php.

Please could someone help as I have no idea what to do now?

Script below:

Code: [Select]
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = 'root';

$conn = mysql_connect($dbhost, $dbuser, $dbpass) or die ('Error connecting to mysql');

$dbname = 'test';

mysql_select_db($dbname);

mysql_select_db("test", $conn);

$result = mysql_query("SELECT * FROM new_table");

while($row = mysql_fetch_array($result))
  {
  echo $row['test1'] . " " . $row['test2'] . " " . $row['test3'];
  echo "<br />";
  }

mysql_close($con);

Hi,

I'm trying to make a mysql output to a link so the name will be a link so when you hit this link you will get the full information of this mysql input.

Can someone point me in the correct direction?

Here is how i get the output from my mysql database into my table.
Code: [Select]
<td>";
echo $row['name'];
echo "</td>

I am getting  the row results from mysql as called, but they appear as a straight line instead of new table row when I echo like:

echo "TABLETABLETABLETABLE";

I am looking for:

echo "TABLE
          TABLE
          TABLE
          TABLE";

Here is my code:
Code: [Select]

$brand = $_POST['brand'];
$city = stripslashes($_POST['city']);
$state = $_POST['state'];
$zip_code = stripslashes($_POST['zip_code']);


if($city !=""){
$query = "SELECT name, bus_type, street, city, state, zip_code, brand, quantity, price1, price2, date FROM `prices` WHERE city ='".$city."' AND state ='".$state."' AND brand = '".$brand."'";
}else{
$query = "SELECT name, bus_type, street, city, state, zip_code, brand, quantity, price1, price2, date FROM `prices` WHERE zip_code ='".$zip_code."'";
}
$result = mysql_query($query);
 
     $count=mysql_num_rows($result);
if($count==0 && $zip_code !=""){?>

    <td width="100%" class="style9">Sorry! There are no results for that city, state and brand.</td>
 
<?
}else{
   while($row = mysql_fetch_array($result, MYSQL_ASSOC)){
?>
  <td width="100%" class="style9"><?php 
  echo "<table width=\"100%\" border=\"1\" cellspacing=\"0\" cellpadding=\"0\">
  <tr>
     <td{$typ_image}</td>
<td><b><a href=\"\" title=\"{$row['street']} {$row['city']}, {$row['state']}. {$row['zip_code']}\" target=\"_blank\">{$row['name']}</a></b><br>Last Updated:{$row['date']}</td>
    <td><b>{$row['brand']} - {$row['quantity']}</b><br>Before Tax:\${$row['price1']} After Tax:\${$row['price2']}</td>
<td{$pago}</td>
  </tr>
</table><BR>";
  ?></td>
<?php
 }
} 
?></table>   
    <p>&nbsp;</p>
    </td>

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

Hello Everyone,
I have a quick question for you all, I think its fairly simple...
I have created a database and I am using PHP to grab the data:


$usera = $_SESSION['username'];
$query2 = "SELECT * FROM tracker WHERE id = '$usera', hidden = yes";
mysql_query($query2) or die('Error, query failed : ' . mysql_error()); 


This hopefully will return multiple rows which look like this in the database.

id username date reps hidden 1 supremebeing 2011-01-02 30 yes 4 supremebeing 2011-04-02 46 yes
How would i turn each result into a variable eg:


$date1 =  2011-01-02;
$date2 =  2011-04-02;
$reps1 = 30;
$reps2 = 46;


I think i have explained that well enough for you to understand, please reply if not though and i will provide more information.

Thanks in Advance

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.

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

im getting an error:

Warning: mysql_fetch_array() [function.mysql-fetch-array]: The result type should be either MYSQL_NUM, MYSQL_ASSOC or MYSQL_BOTH. in C:\www\library\mysql.class.php on line 353


$result_type = "MYSQL_ASSOC";
$row = mysql_fetch_array($result, $result_type)


but if i use (below) it works fine...is the constant() function the right way to what im doing?


$result_type = "MYSQL_ASSOC";
$row = mysql_fetch_array($result, constant($result_type))

thank you

Hi Everyone.

I get the following data from a result from database search

<?php echo $rsjobs['CompanyURL'];?>  It prints out the url but when I try to turn it into a clickable link it does not work

Here is the coding I have so far.


<a href="<?php echo $rsjobs['CompanyURL'];?>"><?php echo $rsjobs['CompanyURL'];?></a>


 What am I doing wrong ? please help.

Now i use a very complex normal for, how i can use a foreach?

soo i get info from a form and i want to search for it in my db .. if i find user_name AND user_password the $result returns true .. otherwise false ..

<?php require_once("../includes/connection.php"); ?>
<?php
$user_name=$_POST["user_name"];
$user_password=$_POST["user_password"];
?>
<?php
$query = "SELECT *
          FROM users
          WHERE  '{$user_name}' = user_name
          AND '{$user_password}'= user_password
          ";
$result = mysql_query($query,$connection);
if($result==false){ 
redirect("../index.php");}
elseif($result==true){
redirect("../welcome.php");
}
?>
<?php //close connection!
mysql_close($connection);
?>


but the problem is that it always returns TRUE !!!!
whats the problem here exactly ???

I am trying to update a mysql table called AvItems with the value 'Torso' in the Equip "section?" I have been through the forums and cannot see anything to match.
I dont mind if the page looses the onsubmit() and has a button instead.
Though I would like to update the database and link back to the same page: There is a display that shows the item that is currently equiped, I have put this in to show it works, or doesn't as the case may be.
Hope I got the code /code right this time.

many thanks in advance

Andy Curtis

Code: [Select]
create table Items(
ItemID integer unsigned auto_increment primary key,
ItemName varchar(20) not null,
Type varchar(10),
UsedOn varchar(10),
);

create table AvItems(
AvItemID integer unsigned auto_increment primary key,
AvID integer unsigned,
ItemID integer unsigned,
Equip varchar(8));


<?php

$username="root";
$password="MyPassword";
$database="MyDataBase";
$AvName = "AndyJCurtis";

mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");

$AvAccR = mysql_query( " SELECT AvID FROM AvAcc WHERE AvName = '$AvName' " );
$AvID = mysql_result($AvAccR, 0, 'AvID');
///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
$Torso = mysql_query(" select ItemName from AvItems, Items where AvItems.itemID = Items.itemID AND AvItems.AvID = '$AvID' AND UsedON = 'Body' ");
$TorsoE = mysql_query(" select ItemName from AvItems, Items where AvItems.itemID = Items.itemID AND AvItems.AvID = '$AvID' And UsedON = 'Body' AND Equip = 'Body' ");
if(mysql_num_rows($TorsoE) != 0)
{
$TorsoItem = mysql_result($TorsoE ,0,"ItemName");
//mysql_close();
?>

<title></title>
<head></head>
<body>
<form action="http://localhost/CI/Equip2.php" method="post">

<table border=1>
<tbody>
<tr>
<td>Torso<BR>
<?PHP
echo "$TorsoItem <BR>";
?>
<select name="Torso" onchange="submit();" value =" Update">
<?PHP
while($TorsoRow = mysql_fetch_array($Torso))
{
echo "<option value=\"".$TorsoRow['ItemName']."\">".$TorsoRow['ItemName']."\n </option>";
}
?>
</select>
</td>
</tr>

////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
<?php
if($_POST['Torso'] == 'Update')
{
mysql_query("update AvItems set Equip = '' where Equip='Torso'") or die("cant update unequip"); 
mysql_query("update AvItems set Equip = 'Torso' where ItemID='{$_POST['ItemName']}'") or die("cant update equip"); 
}
?>
///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

</tbody>
</table>
</.form>
</body>
</html>

Hello -

Suppose I had a MySQL result set that was something like this:
ITEM: COLOR: ball red ball blue book red book green book black
If I were to use the code:
Code: [Select]
do {
   echo $row_myQuery['item'] . " - " . echo $row_myQuery['color'] . "<br />";
} while ($row_myQuery= mysql_fetch_assoc($myQuery));

Then I end up with this:

Code: [Select]
ball - red
ball - blue
book - red
book - green
book - black
MY QUESTION:

What I'd like to know is, how could I re-write that php code so that I could get an output like this?:

Code: [Select]
ball - red, blue
book - red, green, black
Thanks!