PHP - How To Php Loop With Multiple Databases

I have a code that I am using to update multiple products at one time instead of one at a time. It works great except that my processing page is not updating the very last product. My process page is receiving all of my products as I know how many products that I am updating and I have my process page tell me how many products are updated as well as listing the name of each product updated. It is always the last product that is not updated.
Here is my process update code.
Code: [Select]
<?php

$size = count($_POST['casketid']);
//$size = 306;
$i = 1;
while ($i < $size) {

$product=$_POST['productid'][$i];//getting case #
$price=$_POST['price'][$i];
$model=$_POST['model'][$i];
$active=$_POST['active'][$i];

//Getting pricerange
if(($price > "0") && ($price <= "1000.00"))
$pricerange = "1";
elseif(($price >= "1001.00") && ($price <= "1500.00"))
$pricerange = "2";
elseif(($price >= "1501.00") && ($price <= "2000.00"))
$pricerange = "3";
elseif(($price <= "2001.00") && ($price <= "2500.00"))
$pricerange = "4";
elseif(($price <= "2501.00") && ($price <= "3000.00"))
$pricerange = "5";
elseif(($price >= "3001.00") && ($price <= "3500.00"))
$pricerange = "6";
elseif(($price >= "3501.00") && ($price <= "4000.00"))
$pricerange = "7";
elseif(($price >= "4001.00") && ($price <= "4500.00"))
$pricerange = "8";
elseif(($price >= "4501.00") && ($price <= "4500.00"))
$pricerange = "9";
elseif($price >= "5001.00")
$pricerange = "10";

$query = "UPDATE products SET price='$price', pricerange='$pricerange', active='$active' WHERE productid='$product'";
mysql_query($query) or die ("Error in query: $query");
echo "$model = $price & active = $active</em><br />$size<br />";
++$i;
}


?>

So here is what I a trying to do:

Code: [Select]

<?php

/**
 * @author 
 * @copyright 2010
 */
$res = mysql_query("SELECT * FROM `planetsTable` WHERE `section` = '1' LIMIT 9 ASC");
while($row = mysql_fetch_array($res)){ ?>
 
 
            <li>
            <span id="sun">Solar System Star</span>
            <p>The Sun is a star, a hot ball of glowing gases at the heart of our solar system. Its influence extends far beyond the orbits of distant Neptune and Pluto. Without the Sun's intense energy and heat, there would be no life on Earth. And though it is special to us, there are billions of stars like our Sun scattered across the Milky Way galaxy.</p>
            </li>
 
        <li>
            <span id="mercury"><?php echo $row[0]['name']; ?></span>
            <p>Sun-scorched Mercury is only slightly larger than Earth's Moon. Like the Moon, Mercury has very little atmosphere to stop impacts and it is covered with craters. Mercury's dayside is super heated by the Sun, but at night temperatures drop hundreds of degrees below freezing. Ice may even exist in craters. Mercury's egg-shaped orbit takes it around the Sun every 88 days.</p>

            </li>
           
            <li>
            <span id="venus"><?php echo $row[1]['name']; ?></span>
            <p>Venus is a dim world of intense heat and volcanic activity. Similar in structure and size to Earth, Venus' thick, toxic atmosphere traps heat in a runaway 'greenhouse effect.' The scorched world has temperatures hot enough to melt lead. Glimpses below the clouds reveal volcanoes and deformed mountains. Venus spins slowly in the opposite direction of most planets.</p>
            </li>
           
            <li>
            <span id="earth"><?php echo $row[3]['name']; ?></span>

            <p>Earth is an ocean planet. Our home world's abundance of water - and life - makes it unique in our solar system. Other planets, plus a few moons, have ice, atmospheres, seasons and even weather, but only on Earth does the whole complicated mix come together in a way that encourages life - and lots of it.</p>
            </li>
           
            <li>
            <span id="mars"><?php echo $row[4]['name']; ?></span>
            <p>Though details of Mars' surface are difficult to see from Earth, telescope observations show seasonally changing features and white patches at the poles. For decades, people speculated that bright and dark areas on Mars were patches of vegetation, that Mars could be a likely place for life-forms, and that water might exist in the polar caps. When the Mariner 4 spacecraft flew by Mars in 1965, many were shocked to see photographs of a bleak, cratered surface. Mars seemed to be a dead planet. Later missions, however, have shown that Mars is a complex member of the solar system and holds many mysteries yet to be solved.</p>
            </li>
           
            <li>

            <span id="jupiter"><?php echo $row[5]['name']; ?></span>
            <p>The most massive planet in our solar system, with four large moons and many smaller moons, Jupiter forms a kind of miniature solar system. Jupiter resembles a star in composition. In fact, if it had been about 80 times more massive, it would have become a star rather than a planet.</p>
            </li>
           
            <li>
            <span id="saturn">Saturn</span>
            <p>Saturn was the most distant of the five planets known to the ancients. Like Jupiter, Saturn is made mostly of hydrogen and helium. Its volume is 755 times greater than that of Earth. Winds in the upper atmosphere reach 500 meters (1,600 feet) per second in the equatorial region. These super-fast winds, combined with heat rising from within the planet's interior, cause the yellow and gold bands visible in the atmosphere.</p>
            </li>

           
            <li>
            <span id="uranus">Uranus</span>
            <p>The first planet found with the aid of a telescope, Uranus was discovered in 1781 by astronomer William Herschel. The seventh planet from the Sun is so distant that it takes 84 years to complete one orbit.</p>
            </li>
           
            <li>
            <span id="neptune">Neptune</span>
            <p>Nearly 4.5 billion kilometers (2.8 billion miles) from the Sun, Neptune orbits the Sun once every 165 years. It is invisible to the naked eye because of its extreme distance from Earth. Interestingly, the unusual elliptical orbit of the dwarf planet Pluto brings Pluto inside Neptune's orbit for a 20-year period out of every 248 Earth years</p>

            </li>
           
            <li>
            <span id="pluto">Pluto</span>
            <p>Tiny, cold and incredibly distant, Pluto was discovered in 1930 and long considered to be the ninth planet. But after the discoveries of similar intriguing worlds even farther out, Pluto was reclassified as a dwarf planet. This new class of worlds may offer some of the best evidence of the origins of our solar system.</p>
            </li>
        </ul>
   
<?}

?>

What I need to know is if this is possible and if so how would I got about doing this?
Code: [Select]
<?php 
$row['0']['name']; // Main Issue.
?>
McBryver

Let me explain my problem. I have an array with dates and numbers in format ($cronograma)

Ex:

Array ( [2020-09-21] => Array ( [0] => 2020-09-21 [1] => 2 [2] => 2 [3] => 2 ) [2020-09-28] => Array ( [0] => 2020-09-28 [1] => 2 [2] => 2 [3] => 4 )

Then i have another array with 2 ids (in this case 58,60) ($id)

Finally i have a third array with numbers only (in this case 34,34) $tot

So what i want is cross information beween them, for example for id 58 I must get dates (first element and last element when $tot = 34) for id 60 I must get dates (first element after $tot =34 and last element of array)

Whath i have so far is this

foreach ($id as $idPlan) { foreach ($cronograma as $c) { $t1 = 0; foreach ($tot as $d) { $t1 += (int)$d['tempos']; if ($c[3] == $t1) { $newAr[] =$idPlan; $newAr[] = $c[0]; } } } }

My response

array(8) { [0]=> string(2) "58" [1]=> string(10) "2021-02-01" [2]=> string(2) "58" [3]=> string(10) "2021-06-14" [4]=> string(2) "60" [5]=> string(10) "2021-02-01" [6]=> string(2) "60" [7]=> string(10) "2021-06-14" } null

So it's clear that i have all repeated

I should have a line like:

58 - 2020-09-21 -2021-02-01

Any help?

Hi iam currently writing some code for my website and i have a number of records that i simply want to display in a table format, i have done this in a while loop and it works well, however i only want to display a maximum of 10 records per page as my div needs to be a fixed height and was wondering how i could include a next button that would act as like a page number. The thing is at any given point there could be from 1 record to display to 500, so i need a way of allowing the user to quickly navigate to the other records. How can i do this without using mulitple pages.

Hope this makes sense.

Heres my code

Code: [Select]
<?php echo"<table width='100%' border='1' align='center'><tr><th>Referal Name</th><th>Amount Received</tr>";
while ($row = mysql_fetch_array($sql_info))
{
extract($row);
echo "<tr><td align='center'>".$username."</td><td align='center'>".$cumulative_referral."</td></tr>";
}
echo "</table>";


?>
Thanks for your help

Hello,

I'm working on a script which should read lines from a file keyword.txt create a new file with that variable name, insert the content close and loop through all the lines in the keyword file. The loop itself works but now that i'm adding the fopen fwrite and fclose i'm running into trouble. I'm not that great a coder and have pieced this together so far. I'd appreciate any suggestions on how to fix the fopen and other file functions to create unique files using the keywords from the file

Thanks in advance


<?php
$read_file = file('keywords.txt');
foreach($read_file as $var)
    {
    $file = fopen($var, "x");
$content = "<!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>This is the title</title>
</head>
<body>
<h2>The keyword is $var</h2>
<p><p>
</body>
</html>";
fwrite($file, $content);
fclose($file);
    }
?>

Just wondering, which is better php databases or sql? I have phpMyAdmin and I've that you can convert the database into php scripts?

What would be the fastest way to search 2+ databases with the same search information?

Each database is different, and may return different information.

ok , here is my mysql code to get all posts from the posts table .
Code: [Select]
$query = mysql_query("SELECT id,to_id,from_id,post,type,state,date FROM posts WHERE state='0' ORDER BY id DESC LIMIT 50"); and here is the code to display the users friends...
Code: [Select]
$sqlArray = mysql_query("SELECT friend_array FROM myMembers WHERE id='" . $logOptions_id ."' LIMIT 1");
while($row=mysql_fetch_array($sqlArray)) { $iFriend_array = $row["friend_array"]; }
$iFriend_array = explode(",", $iFriend_array);
if (in_array($id, $iFriend_array))see now i got as far as , if(in_array($id, $iFriend_array))
How would i put these togeather to where it would get the posts from the posts table that there friends posted?

I'm trying to connect to two databases and I'm having problems with the following code.  I googled to come up with this but can't figure out the errors I'm getting.
Code: [Select]
$connection="localhost";
$username="user";
$password="password";
$database1="dbone";
$database2="dbtwo";

$db1 = mysql_connect($connection,$username,$password) or die(mysql_error());
$sel1 = mysql_select_db($database1, $db1);

$query1 = "SELECT * FROM TBLUSERS";
$result1 = mysql_query($query1, $db1);
while($nt1 = mysql_fetch_array($result1, $db1))
{

}

$db2 = mysql_connect($connection,$username,$password) or die(mysql_error());
$sel2 = mysql_select_db($database2, $db2);
$query2 = "SELECT * FROM TBLPD20101101";
$result2 = mysql_query($query2, $db2) or die(mysql_error());
while($nt2 = mysql_fetch_array($result2, $db2))
{

}
The error I get is
Quote

Warning: mysql_fetch_array() expects parameter 2 to be long, resource given in C:\xampp\htdocs\HighVisibility\DashBoard2.php on line 13

Warning: mysql_fetch_array() expects parameter 2 to be long, resource given in C:\xampp\htdocs\HighVisibility\DashBoard2.php on line 22

I have a site where I need to have lets call it image1 displayed, then I want to change this image based on a php if statement, for instance:

if $var == $var2
change the image
....blah blah

so I was also going to have the names of my images stored in my database, i.e. image1.jpg and image2.jpg in my database.

the image is in its own div tag set as the background image of the div tag if that makes any diference.

Thanks

I have just read my upcoming modules for my final year at uni and 'multimedia databases' is one of them. I am just wondering if any of you had any clue on what a multimedia database is? I am guessing it's a database populated with directory data, but that would be to simple...

Hi,

I am in the procress of creating discussion system however I am a bit puzzled about the best way to go about it.

I am starting the discussion by creating an ID number and then match the answer to the initial ID number.  However, I dont know whether if is best to put the responses into a different database.

I'm a bit puzzled how ID matching systems works.

Lets say:

Question 1 =    ID1
Question 2 =    ID2
Question 3 =    ID3
Question 1 Answer 1    =   ID4   (How is this matched to ID1)
Question 2 Answer 1    =   ID5   (How is this matched to ID2)

is this based on preg_match?

Below is some code I have that I would like to connect to databases that reside on different physical machines. The problem I can't seem to solve is how to ignore machines that are unreachable.
Code: [Select]
<html>
<head>

</head>

<body>
<?php
 $mysql_conn=new mysqli('192.168.1.67','admin','password','monkey');
$result=$mysql_conn->query("SELECT * from testtable");
echo "<table>";
echo "<th>Test Query</th>";
while ($row=$result->fetch_array(MYSQL_ASSOC)){
echo "<tr><td>";
echo $row['test1'];
echo "</td></tr>";
}
echo "</table>";

$mysql_conn=new mysqli('192.168.1.126','admin','','monkey1');
$result=$mysql_conn->query("SELECT * from demo");
echo "<table>";
echo "<th>Test Processor</th>";
while ($row=$result->fetch_array(MYSQL_ASSOC)){
echo "<tr><td>";
echo $row['name'];
echo "</td></tr>";
}
echo "</table>";
$mysql_conn=new mysqli('192.168.1.127','admin','','monkey2');
$result=$mysql_conn->query("SELECT * from demo");
echo "<table>";
echo "<th>Test Processor</th>";
while ($row=$result->fetch_array(MYSQL_ASSOC)){
echo "<tr><td>";
echo $row['processor'];
echo "</td></tr>";
}
echo "</table>";
?>

</body>
</html>
If 192.168.1.126 is offline, the webpage doesn't go any further and display 192.168.1.127's info...Appreciate any insight...

Dear All,

I am having 2 different DB on 2 different hosts.

I am running MySQL Server on my local PC where the user is entering data in the tables. I have a website which has the identical DB on the web. I am able to connect to the database by using the codes on the server.

I want to update the server DB with the local system DB by running one update command. I get the error "-SELECT command denied to user 'localusername'@'localhost' for table 'pst_data'"

Given below is the code used for the process :

//connecting the remote system DB
$link = mysql_connect('IPAddress:3306', 'remoteusername', 'Password');
if (!$link) {
    die('Not connected : ' . mysql_error());
}

$db_selected = mysql_select_db('remotedb', $link);
if (!$db_selected) {
    die ('Can\'t use Remote System DB: ' . mysql_error());
}

//connecting the local database on the webstite
$weblink = mysql_connect("localhost","localusername","password");
if (!$weblink) {
    die('Not connected : ' . mysql_error());
}
$webdb_selected = mysql_select_db('localwebdb', $weblink);
if (!$webdb_selected) {
    die ('Can\'t use WebServer Database : ' . mysql_error());
}

//query to fetch the records from remote ystem and insert into local website database
$upd_Query=mysql_query("INSERT INTO  localwebdb.`table` SELECT * FROM  remotedb.`table` where field=' some condition' ");
---------------------------------------------------

I have tested that I have connected the remote DB by running queries on the webserver.

Could anyone bail me out so that i can copy the DB from remote to local

Any help would be appreciated

right i did this code
<table width="400" border="1" align="center" cellpadding="2" cellspacing="0" bordercolor="#000000" class=thinline>
        <tr class="header" background="includes/grad.jpg">
          <td height="20" colspan="3" background="includes/grad.jpg"  class=header><div align="center" class="header">Last 10 Kills</div></td>
        </tr>
        <tr class="header" background="includes/grad.jpg"> 
          <td width="166" height="20" background="includes/gradgrey.jpg">Name</td>
          <td width="157" height="20" background="includes/gradgrey.jpg">Rank</td>
          <td width="157" height="20" background="includes/gradgrey.jpg">Killed Time</td>
        </tr>
      <?
   $c=mysql_query("SELECT * FROM attempts  WHERE outcome='Dead' ORDER BY id DESC LIMIT 10");
   while($d=mysql_fetch_object($c)){
   echo "<tr><td><a href='profile.php?viewuser=$d->target'>$d->target</a></td><td>$d->rank</td><td>$d->date</td></tr>";
   
   
   
   }
   ?>

but you see the $d->rank

is in another database called users

How can i link the databases so the last 10 kills shows the users rank