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PHP - Results From Echo Not Showing
Code: [Select]
if (isset($_POST['update'])) { $signature = $_POST['signature']; $type = $_POST['type']; $name = $_POST['name']; $b=0; While ($b < $i) { echo $signature[$b]."<br>"; echo $type[$b]."<br>"; echo $name[$b]."<br><br>"; $b++; }} I did echo '<pre>'; print_r($_POST); echo '</pre>'; The results are Code: [Select] Array ( [sname] => [notes] => [signature] => Array ( [0] => 1-1 [1] => 1-2 [2] => 1-3 ) [type] => Array ( [0] => Test1 [1] => Test2 [2] => Test3 ) [name] => Array ( [0] => Testa [1] => Testb [2] => Testc ) [update] => Update ) But for some reason nothing is showing up on the screen. Similar Tutorialsi have built pages that paginate with 10 rows per page (some pages show more but for the moment i want to focus on this particular page)
//Define Some Options for Pagination
$num_rec_per_page=10;
if (isset($_GET["page"])) { $page = $_GET["page"]; } else { $page=1; };
$start_from = ($page-1) * $num_rec_per_page;
$results = mysql_query("SELECT * FROM `ecmt_memberlist` WHERE toonCategory='Capital' AND oldMember = 0 ORDER BY CONCAT(MainToon, Name) LIMIT $start_from, $num_rec_per_page") or die(mysql_error());
$results_array = array();
while ($row = mysql_fetch_array($results)) {
$results_array[$row['characterID']] = $row;
}
The above sets the variables for the pagination and below the results are echo with 10 rows per page
then i show the pagination links:
<?php
$sql = "SELECT * FROM `ecmt_memberlist` WHERE toonCategory='Capital' AND oldMember = 0 ORDER BY CONCAT(MainToon, Name)";
$rs_result = mysql_query($sql); //run the query
$total_records = mysql_num_rows($rs_result); //count number of records
$total_pages = ceil($total_records / $num_rec_per_page);
?>
<table width="100%" border="0" cellspacing="0" cellpadding="0">
<tr>
<td align="center"><div style="width:100%; text-align:center;">
<ul class="pagination">
<li class="selected">
<?php
echo "<a href='capitalmember.php?page=1'>".'«'."</a> ";?>
</li>
<?
for ($i=1; $i<=$total_pages; $i++) {
echo "<li><a href='capitalmember.php?page=".$i."'>".$i."</a></li> ";
};
?>
<li class="selected">
<?
echo "<a href='capitalmember.php?page=$total_pages'>".'»'."</a> "; // Goto last page
?></li>
</ul></div>
<?php
$pageNr = $page; // Get Current Page Number
$from = $pageNr * $rowsPerPage; // 3 * 10 = 30 // 3 * 10 = 30
$to = $from + $rowsPerPage; // 30 + 10 = 40
echo $pageNr;
/* Result: From page 30 to 40 */
?></td>
</tr>
</table>
this works great and shows me 10 results per page,
what i need to work out and work on next is:
echo the number of results above the records (for example: "showing records 1 to 10" and then on page 2 "showing records 11 to 21" and page 3 "showing records 22 to 32"
how can i work out the maths for this echo?
i was thinking along the lines of;
<?php
$pageNr = $page; // Gets Current Page Number
$from = $pageNr * $rowsPerPage; // 3 * 10 = 30
$to = $from + $rowsPerPage; // 30 + 10 = 40
// Now Show Results
echo $from; // Echo from
echo $to // Echo to
?>
but i'm still working on this..
then i need to look at shortening the amount of page links on the page if for example i have 500 records it shows me links 1 to 50 and eats up the page....
Appreciate any help and light onto my problems.
Many thanks
Edited by jacko_162, 11 January 2015 - 05:43 PM. How do I present these documentaries like this:
0-9
10 Things You Need to Know About Sleep
20 Animals That Will Kill You
A
Ant Kingdom
Atkins Diet
B
Battle of the Brains
Body Talk
I don't even know where to start!
I'm not asking anyone to do this for me; I just need a push in the right direction.
Any help will be appreciated.
Ok, here's my code: How do i make it so that it outputs a maximum of only 5 results from the query where the img file exists. When I add LIMIT to the sql query it doesnt work so I guess its something else, but I have no clue. Can anyone help? $query = mysql_query("SELECT * FROM table WHERE date >= '$now' ORDER BY max(date) desc"); while ($row = mysql_fetch_assoc($query)) { $cid = $row['cat_id']; $title = $row['name']; $seoname = $row['seourl']; $img = 'images/'.$seoname.'.jpg'; if (!file_exists($img)) { $img = ''; } else { $img = 'images/'.$seoname.'.jpg'; } if (!empty($img)) { echo '<a href="images/'.$seoname.'/"><img src="'.$img .'" style="width:528px;" alt="" /></a>' . "\n"; } } Thanks. Hi, I have a searchbar which looks for a value in my tables, it works great but i want to know how i can make the results of the search clickable; go to a page relevent to that search, let me explain in more depth. User searches for a postcode, script finds and displays the postcode on the next page, user clicks the result postcode to goto the next page. What im stuck with is how to take that result and automatically display it on the clicked "final" page? Search script: Code: [Select] <?php // Get the search variable from URL $var = @$_GET['q'] ; $trimmed = trim($var); //trim whitespace from the stored variable // rows to return $limit=10; // check for an empty string and display a message. if ($trimmed == "") { echo "<p>Please enter a search...</p>"; exit; } // check for a search parameter if (!isset($var)) { echo "<p>We dont seem to have a search parameter!</p>"; exit; } //connect to your database ** EDIT REQUIRED HERE ** mysql_connect("xxxxx","xxxx","xxxx"); //(host, username, password) //specify database ** EDIT REQUIRED HERE ** mysql_select_db("removalspacelogin") or die("Unable to select database"); //select which database we're using // Build SQL Query $query = "(SELECT postcode FROM freelistings WHERE postcode like '%$trimmed%') UNION (SELECT postcode FROM basicpackage WHERE postcode like '%$trimmed%') UNION (SELECT postcode FROM premiumuser WHERE postcode like '%$trimmed%') ORDER BY postcode"; // EDIT HERE and specify your table and field names for the SQL query $numresults=mysql_query($query); $numrows=mysql_num_rows($numresults); $result = mysql_query($query) or die("<b>Query:</b> $query<br><b>Error</b>: " . mysql_error()); // If we have no results, offer a google search as an alternative if ($numrows == 0) { echo "<h4>Results</h4>"; echo "<p>Sorry, your search: "" . $trimmed . "" returned zero results</p>"; echo "<p><a href=\"http://www.google.com/search?q=" . $trimmed . "\" target=\"_blank\" title=\"Look up " . $trimmed . " on Google\">Click here</a> to try the search on google</p>"; } // next determine if s has been passed to script, if not use 0 if (empty($s)) { $s=0; } // get results $query .= " limit $s,$limit"; $result = mysql_query($query) or die("Couldn't execute query"); // display what the person searched for echo "<p>You searched for: "" . $var . ""</p>"; // begin to show results set echo "Results for $q"; $count = 1 + $s ; // now you can display the results returned while ($row= mysql_fetch_array($result)) { $title = $row['postcode']; echo " $title" ; $count++ ; } $currPage = (($s/$limit) + 1); //break before paging echo "<br />"; // next we need to do the links to other results if ($s>=1) { // bypass PREV link if s is 0 $prevs=($s-$limit); print " <a href=\"$PHP_SELF?s=$prevs&q=$var\"><< Prev 10</a>  "; } // calculate number of pages needing links $pages=intval($numrows/$limit); // $pages now contains int of pages needed unless there is a remainder from division if ($numrows%$limit) { // has remainder so add one page $pages++; } // check to see if last page if (!((($s+$limit)/$limit)==$pages) && $pages!=1) { // not last page so give NEXT link $news=$s+$limit; echo " <a href=\"$PHP_SELF?s=$news&q=$var\">Next 10 >></a>"; } $a = $s + ($limit) ; if ($a > $numrows) { $a = $numrows ; } $b = $s + 1 ; echo "<p>Showing results $b to $a of $numrows</p>"; ?> <form method="post" action="localarea.php"> <a href="localarea.php"><?php echo """ .$var. ""</p>";?></a> </form> [\code] The form at the bottom im guessing is where i should take the $var from but how exactly? or if not where should it be pulled from to display on the next page? I want this "next" page because ideally there would be lots of results in postcode, so the user can choose which one is more relevant to them and go to that specific page with all the companies within that postcode. Hope this helps for a better understanding of what im trying to achieve, its just implementing it in? I am working on building a code that will check the results returned from and xml pull to see if they match the database and print back pass or fail on the screen, it works for the first result but then fails to display anything else. any help on where I went wrong would be helpful. Code: [Select] <?php function pass($name, $level) { $sql1 = "SELECT * FROM `reqskills` WHERE `level` > 0 and `name` = '$name'"; $result1 = mysql_query($sql1); while($row1 = mysql_fetch_array($result1)) { if($level >= $row1['level']) { echo "Pass"; } Else { echo "Fail"; } } } $API = "address removed"; $xml = simplexml_load_file($API); $sql = "SELECT * FROM `reqskills` WHERE `level` > 0"; $result = mysql_query($sql); echo "<table width=100%"; echo "<tr><td>Skill</td><td>Level</td><td>Pass</td></tr>"; while($row = mysql_fetch_array($result)) { ?> <tr> <td><?php echo $row['name'] ?></td> <td><?php echo $row['level'] ?></td> <td><?php foreach ($xml->result->rowset[0] as $value) { $typeID = $value['typeID']; $sql = "select * from `invTypes` WHERE `typeID` = $typeID"; $result = mysql_query($sql); while ($row = mysql_fetch_array($result)) { echo pass($row['typeName'],$value['level']); }} echo "</td></tr>"; } echo"</table>"; ?> Hello, im new here, and i have little experience to php and mysql as i started for 2 weeks ago. I started out with some tutorials and feeling im getting the hang of it. Enough of me, lets get to the point: <?php $con = mysql_connect('localhost',$user,$pass)or die(mysql_error()); $selectdb = mysql_select_db($selectdb)or die(mysql_error()); $sql = "SELECT * From table"; $result = mysql_query($sql); $num = mysql_num_rows($result); $myarray = array($result); $i =0; while ($i < $num){ echo $myarray[$i]; $i++; } ?> Here i have written a dummyscript that does what the original script does, it tries to fetch the keys from the table and then trying to loop it and echo out the results. The output in the browser is this: Resource id #3 I know this probably is a simple fix but i cant seem to get it sorted out. Hope some of you could help me get this baby work, or maybe have another way of doing it more "simple". Thanks in advance! Dan-Levi Hello community, I am working on a database of specialties in the hospital I work. The doctor's referral requests are sent to a mySQL database and I have, with the help of online guidance, produced a working php script that displays the information I need it to. However, I need it a little bit more specific. I intend to make multiple copies of this file for each specialty, so that when they open the file they only have the requests for that particular specialty. My question is, with reference to my code below, can I make echo information so that online, for instance, if 'specialty1 = gastroenterology' (as in, that particular specialty that that referral request is for), then only the rows on the database that have that particular text are displayed only? Hope that makes sense. Code below for your reference and assistance is highly appreciated.
<!DOCTYPE html>
<html>
<head>
<title>Specialty Referral Form</title>
<style>
table {
border-collapse: collapse;
width: 100%;
color: #000000;
font-family: arial;
font-size: 10px;
text-align: center;
}
th {
background-color: #588c7e;
color: white;
}
tr:nth-child(even) {background-color: #f2f2f2}
</style>
</head>
<body>
<table>
<tr>
<th>Patient Details</th>
<th>Hospital Number</th>
<th>Date of Birth</th>
<th>Referred by:</th>
<th>New/Repeat Visit to Patient</th>
<th>Specialty</th>
<th>Admission Date</th>
<th>Too Ill for Clinic?</th>
<th>Diagnosis Aware?</th>
<th>Question</th>
<th>History</th>
<th>Medications</th>
<th>Examination</th>
<th>Results</th>
<th>WorkingDiagnosis</th>
<th>Investigation(s) Requested</th>
</tr>
<?php
$conn = mysqli_connect("localhost", "view", "", "referral");
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT patientdetails, hospitalnumber, DoB, referral, admission, specialty1, admissiondate, illness, awareness, question, history, medications, examination, results, workingdiagnosis, investigations FROM referralform";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while ($row = $result->fetch_assoc()) {
echo "<tr><td>" . $row["patientdetails"]. "</td><td>" . $row["hospitalnumber"] . "</td><td>" . $row["DoB"] . "</td><td>"
. $row["referral"] . "</td><td>" . $row["admission"] . "</td><td>" . $row["specialty1"] . "</td><td>" . $row["admissiondate"] . "</td><td>"
. $row["illness"] . "</td><td>" . $row["awareness"] . "</td><td>" . $row["question"] . "</td><td>" . $row["history"] . "</td><td>"
. $row["medications"] . "</td><td>" . $row["examination"] . "</td><td>" . $row["results"] . "</td><td>" . $row["workingdiagnosis"] . "</td><td>"
. $row["investigations"]. "</td></tr>";
}
echo "</table>";
} else { echo "0 results"; }
$conn->close();
?>
</table>
</body>
</html>
I guess what I am looking for is something like echo specialty1 IF it writes a particular specialty and only that specialty. Thank you. Edited April 2, 2020 by samanjI'm having a really hard time getting my carousel to display in a line when I call my images from the table:
<div id="user-gallery">
<h2>Recent Images</h2>
<button class="prev"><<</button>
<button class="next">>></button>
<?php
include("file/image/upload_file.php");
$select_query = "SELECT `images_path` FROM `images_tbl` ORDER by `images_id` DESC LIMIT 9";
$sql = mysql_query($select_query) or die(mysql_error());
while($row = mysql_fetch_array($sql,MYSQL_BOTH)){
?>
<div class="anyClass">
<ul>
<li><img src="<?php echo $row["images_path"]; ?>" alt="" width="125" height="70" ></li>
<li><img src="<?php echo $row["images_path"]; ?>" alt="" width="125" height="70" ></li>
<li><img src="<?php echo $row["images_path"]; ?>" alt="" width="125" height="70" ></li>
<li><img src="<?php echo $row["images_path"]; ?>" alt="" width="125" height="70"></li>
</ul>
</div>
<?php
}
?>
</div>
Without the carousel it will display perfectly, but as soon as I implement it, the images won't behave normal :/
Does anybody know how to show the below php results in a table format? <?php if(isset($_POST['submit'])){ if(isset($_GET['go'])){ $fname = $_POST['fname']; $lname = $_POST['lname']; $skill = $_POST['skill']; //connect to the database $db=mysql_connect ("127.0.0.1", "root", "") or die ('I cannot connect to the database because: ' . mysql_error()); //select the database to use $mydb=mysql_select_db("resource matrix"); //query the database table $sql="SELECT DISTINCT First_Name, Last_Name, l.Resource_ID FROM ((resource l inner join resource_skill ln on l.Resource_ID = ln.Resource_ID) inner join skill n on ln.Skill_ID = n.Skill_ID) WHERE First_Name LIKE '$fname' OR Last_Name LIKE '$lname' OR Skill_Name LIKE '$skill'"; //run the query against the mysql query function $result=mysql_query($sql); //create while loop and loop through result set while($row=mysql_fetch_array($result)){ $First_Name =$row['First_Name']; $Last_Name=$row['Last_Name']; $Resource_ID=$row['Resource_ID']; //display the result of the array echo "<ul>\n"; echo "<li>" . "<a href=\"a.php?id=$Resource_ID\">" .$First_Name . " " . $Last_Name . "</a></li>\n"; echo "</ul>"; } } else{ echo "<p>Please enter a search query</p>"; } } here is my code... please help <form method="get" action="shit3.php"> <input maxlength="50" size="50" type="text" name="q"> <input type="submit" value="SEARCH"> </form> <?php if (@$_GET['q']) { $var = $_GET['q'] ; $trimmed = trim($var); $limit=1; if ($trimmed == "") { echo "Please enter a search..."; exit; } if (!isset($var)) { echo "Container not found!"; exit; } mysql_connect("localhost", "xxx", "yyy") or die(mysql_error()); mysql_query("SET NAMES 'cp1251'"); mysql_select_db("agaexpor_container") or die("Unable to select database"); //select which database we're using $query = "select * from containertracking where vin like \"%$trimmed%\" order by vin"; $numresults=mysql_query($query); $numrows=mysql_num_rows($numresults); if (empty($s)) { $s=0; } $query .= " limit $s,$limit"; $result = mysql_query($query) or die("Couldn't execute query"); $count = 1 + $s ; while ($row= mysql_fetch_array($result)) { $vin = $row["vin"]; $container = $row["container"]; echo "$count.) VIN: $vin <br />"; echo "container is: $container <br />"; $count++ ; } } ?> Hi people. I am creating a carpet website for a good friend of mine as a favour. I am storing the carpets info in a MYSQL database and am currently trying to relay the info on a page. I have the array kind of done but it is not producing the results I want. Let me explain further: The columns in my database under the table "carpets" a id (auto incremented) colour type title price description imageloc ------ under the type I have various types of carpet (6-ish). They a Twist, Striped etc etc... Heres the php problem. I have all the images there and ready on the page and want to link them all with linking commands. i.e: the main page is at: http://www.ircdirect.co.uk/FTPServers/supremecarpets/index.php the carpets results page is at: http://www.ircdirect.co.uk/FTPServers/supremecarpets/carpetlist.php on the index.php page I want to link them so that for example: If he looks for a striped carpet he clicks "striped" and is shows carpetlist.php but with the striped carpets displayed. example link: <A HREF="carpetlist.php?type=striped">IMAGE HERE</A> I have tried various coding on the carpetlist.php and cant seem to get it to work. here is the snippets: <?php // Make a MySQL Connection $query = "SELECT * FROM carpets GROUP BY type"; $result = mysql_query($query) or die(mysql_error()); while($row = mysql_fetch_array($result)){ echo "<TABLE CELLPADDING=0 CELLSPACING=0 WIDTH=100% />"; echo "<TR />"; echo "<TD WIDTH=23 /><IMG SRC=images/main/search/topleft.png /></TD />"; echo "<TD BACKGROUND=images/main/search/top.png /> </TD />"; echo "<TD WIDTH=23 /><IMG SRC=images/main/search/topright.png /></TD />"; echo "</TR />"; echo "<TR />"; echo "<TD BACKGROUND=images/main/search/left.png />"; echo " "; echo "</TD />"; echo "<TD BACKGROUND=images/main/search/bg.png />"; echo "<FONT FACE=VERDANA SIZE=1 />"; echo $row['title']; echo "</TD />"; echo "<TD BACKGROUND=images/main/search/right.png />"; echo " "; echo "</TD />"; echo "</TR />"; echo "<TR />"; echo "<TD WIDTH=23 /><IMG SRC=images/main/search/bottomleft.png /></TD />"; echo "<TD BACKGROUND=images/main/search/top.png /> </TD />"; echo "<TD WIDTH=23 /><IMG SRC=images/main/search/bottomright.png /></TD />"; echo "</TR />"; echo "</TABLE />"; echo "<BR />"; } ?> It displays ALL the results which is not what I want. I want only striped carpets, blue carpets etc.... Help is needed and VERY much appreciated! Kind Regards, Ian I am writing a CRON job that will execute daily. First it will identify from a MySql table the date in a field 'FAPforSale_repost35' If the date is the today date it will then execute commands to delete photo images in a directory, delete the directory, and finally remove the record from the database.
I am on step one which is to build the array of records that match the days date. When I run the code, there are no errors but I am not getting results even though the records in the test table are set for today. Below is the select
<?php
define( "DIR", "../zabp_employee_benefits_processor_filesSm/", true );
require( '../zipconfig.php' );
require( DIR . 'lib/db.class.php' );
require_once( $_SERVER['DOCUMENT_ROOT'] . '/_ZABP_merchants/configRecognition.php' );
require_once( $_SERVER['DOCUMENT_ROOT'] . '/_ZABP_merchants/libRecognition/MailClass.inc' );
$todayRepost35 = date("Y-m-d");
echo $todayRepost35;
function repostEndSelect()
{
global $db;
$this->db = $db;
$data = $this->db->searchQuery( "SELECT `FAPforSale_IDnumber`, `FAPforSale_image1`, `FAPforSale_image2`, `FAPforSale_image3`, `FAPforSale_repost35` FROM `FAP_forSaleTest` Where `FAPforSale_repost35` = '$todayRepost35' ");
$this->FAPforSale_IDnumber = $data[0]['FAPforSale_IDnumber'];
$this->FAPforSale_image1 = $data[0]['FAPforSale_image1'];
$this->FAPforSale_image2 = $data[0]['FAPforSale_image2'];
$this->FAPforSale_image3 = $data[0]['FAPforSale_image3'];
$this->FAPforSale_repost35 = $data[0]['FAPforSale_repost35'];
echo $this->FAPforSale_IDnumber;
echo $this->FAPforSale_image1;
echo $this->FAPforSale_image2;
echo $this->FAPforSale_image3;
echo $this->FAPforSale_repost35;
} // ends function...
echo( ' Finished...' );
?>
Thanks in advance for any suggestions or direction. Chapter two will be when I start testing the commands to delete.
Hi
I have 2 tables that I am searching infor for. One holds the phone info and the other the repairs that are available.
I have the code below and it displays on the webpage in a table for the phone information and then another table the repairs but they seem to be separate tables and headers. inbetween them is the phone and the details which separates the repairs but its not in a table.
Hopefully the code below will explain better.
What I would like it the phone details in one table and then a table full of all the repairs (I am not bothered about a new heading for each repair).
$result = mysql_query("SELECT * FROM phone, phonerepairs WHERE phone.model_no='".$_POST['model']."' AND phone.phone = phonerepairs.phone");
while($row = mysql_fetch_array($result))
{
echo '<td width="70" align="center"><img src="'.$row['icon'].'" width="66"></td>';
echo '<td width="100" align="center">'.$row['model_no'].'</td>';
echo '<td width="130" align="center">'.$row['model'].'</td>';
echo '<td width="50" align="center">'.$row['year'].'</td>';
echo '<td width="150" align="center">'.$row['capacity'].'</td>';
echo '</tr>';
echo '</table><br>';
echo '</p>';
echo '<p>';
echo '<div id="pageheading"><h3><div id="title">Repairs we carry out for '.$row['model'].'</div></h3></div>';
echo '<table width="99%" cellpadding="5" cellspacing="0" border="1">';
echo '</p>';
echo '<p>';
echo '<tr>';
echo '<td align="center"><strong>Image</strong></td>';
echo '<td align="center"><strong>Fault</strong></td>';
echo '<td align="center"><strong>Repair</strong></td>';
echo '<td align="center"><strong>Cost</strong></td>';
echo '</tr>';
echo '<tr>';
echo '<td width="70" align="center"><img src="'.$row['icon'].'" width="66"></td>';
echo '<td width="100" align="center">'.$row['fault'].'</td>';
echo '<td width="130" align="center">'.$row['description'].'</td>';
echo '<td width="50" align="center">£'.$row['cost'].'</td>';
echo '</tr>';
echo '</table><br>';
Thanks in advance
Martyn
So I have 2 queries, that has the potential to return alot of data, foreach loops running. The first shows each group heading, and the foreach nested is calling another query specific to the group heading. The result currently is a lengthy delay in results showing. (Now this may/may not be the most ideal code practise in this instance however... short term solution discussion please). For Each State Read Each States Details from DB For Each State Listing Display Details Next Next Is there a command or other where I can say, Display HTML page as it stands while it continues to process the FOREACH loops? EG: For Each State Read Each States Details from DB For Each State Listing Display Details REFRESH HTML DISPLAYED Next Next I was wondering how does one go about showing results from SELECT query in columns in a html table. I have a list of products in a table, and would like to show them on the page in 4 columns. I have done many searches on google to try and find the sulution, but the majority of what im finding instead is about displaying a table from phpmyadmin as a table in html. If its a large operation to do this, I would be very happy if someone could poiint me in the direction of a tutorial maybe. Here is the code I have so far to display the products, but for some reason, it only show 1 row instead of all the rows from my table. Code: [Select] <?php $dbhost = "localhost"; $dbuser = "user"; $dbpass = "pass"; $dbname = "dbname"; mysql_connect ($dbhost, $dbuser, $dbpass)or die("Could not connect: ".mysql_error()); mysql_select_db($dbname) or die(mysql_error()); $result = mysql_query("SELECT * FROM mcproducts"); while($row = mysql_fetch_array($result)) { $products_local_id = $row['products_local_id']; $productname = $row['product_name']; $thumburl = $row['image_from_url']; $productlink = $row['product_local_url']; $thumbnail = $row['product_image_small']; $currencysymbol = $row['product_currency']; $price = $row['product_price']; $flagicon = $row['product_country_from']; } ?> <html> <head> <link href="style/stylesheet.css" rel="stylesheet" type="text/css" /> </head> <body> <div class="displaybox"> <div class="productimage"> <a href="<?php echo $productlink; ?><?php echo $products_local_id; ?>"><img src="<?php echo $thumburl; ?><?php echo $thumbnail ?>" width="150" height="150"></a> </div> <div class="productdescription"> <div class="pro_name"> <a href="<?php echo $productlink ?><?php echo $products_local_id; ?>"><?php echo $productname; ?></a> </div> <div class="pro_description"> </div> <?php if ($flagicon=="Ireland") { $flagicon = "<img src=\"flags/ireland.jpg\">"; } elseif ($flagicon=="UK") { $flagicon = "<img src=\"flags/uk.jpg\">"; } else echo ""; ?> <div class="pro_description"><?php echo $flagicon; ?><?php echo $currencysymbol ?> <?php echo $price ?></div> </div> </div> </body> </html> Many thanks, DB This topic has been moved to Ajax Help. http://www.phpfreaks.com/forums/index.php?topic=330251.0 Our product is running on LAMP(PHP) architecture. For MySQL operation PDO library has been used. We are in the process of moving the product to a new Linux server. On the new server we are facing MySQL query output related issue. At the end of the post I have listed the server configuration details of both the servers and the sample PHP program with the output. It works fine in server 1 but when the same code moved to Server 2, the output is not correct. Would appreciate any points to solve this problem The following piece of code is tested and output is also given. <?php /*** mysql hostname ***/ $hostname = 'localhost'; /*** mysql username ***/ $username = 'ram'; /*** mysql password ***/ $password = 'tellmehow'; try { $dbh = new PDO("mysql:host=$hostname;dbname=inhouse", $username, $password); $sql = "SELECT first_name,last_name,user_id,email FROM users LIMIT 1"; $stmt = $dbh->prepare($sql,array(PDO::MYSQL_ATTR_USE_BUFFERED_QUERY => TRUE)); $stmt->execute(); $result = $stmt->fetch(PDO::FETCH_OBJ); $stmt->closeCursor(); /*foreach($result as $key=>$val) { echo $key.' - '.$val.'<br />'; }*/ print_r($result); $dbh = null; } catch(PDOException $e) { echo $e->getMessage(); } ?> Output for server 1: stdClass Object ( [first_name] => super [last_name] => admin [user_id]=>1 [email] => ram@greynium.com ) output for server 2: stdClass Object ( [first_name] => super [last_name] => admin [users] => ram@greynium.com ) On server 2 I am not getting the output for field 'user_id' from the table 'users'. For the field 'email' from 'users' table, it is displaying the name of the table itself (users). Not sure what is wrong. The configuration of the servers are given below Server 1 -------- OS Details: Linux 2.6.9-5.ELsmp PHP Details: PHP 5.2.3 (cli) (built: Aug 28 2007 11:48:30) Copyright (c) 1997-2007 The PHP Group Zend Engine v2.2.0, Copyright (c) 1998-2007 Zend Technologies with Suhosin v0.9.22, Copyright (c) 2007, by SektionEins GmbH PDO details: PDO PDO support => enabled PDO drivers => sqlite2, sqlite, mysql PDO Driver for MySQL, client library version => 5.0.37 PDO Driver for SQLite 3.x => enabled Server 2 --------- OS Details: Linux 2.6.18-8.el5 PHP Details: eAccelerator requires Zend Engine API version 220060519. The Zend Engine API version 220090626 which is installed, is newer. Contact eAccelerator at http://eaccelerator.net for a later version of eAccelerator. PHP 5.3.2 (cli) (built: Jul 6 2010 17:57:31) Copyright (c) 1997-2010 The PHP Group Zend Engine v2.3.0, Copyright (c) 1998-2010 Zend Technologies PDO Details: PDO PDO support => enabled PDO drivers => mysql, sqlite, sqlite2 PDO Driver for MySQL => enabled PDO Driver for SQLite 3.x => enabled I am using html5 and all the pages have h2 headings as titles. They all show up in google search results fine. The only thing that doesn't show up fine are some of the descriptions under each heading. Some meta descriptions show up fine under the correct heading title, while rest shows ALL the heading tags AS description under each search result.
For eg.
Cars
www.mywebsite.com/category?id=5&name=cars Hi , I have one question .. Can I split showing of content of dynamic list in 2 parts , when I echo list in code .. Code: [Select] <?php // Run a select query to get my letest 8 items // Connect to the MySQL database include "../connect_to_mysql.php"; $dynamicList = ""; $sql = mysql_query("SELECT * FROM products ORDER BY date_added DESC LIMIT 8"); $productCount = mysql_num_rows($sql); // count the output amount if ($productCount > 0) { while($row = mysql_fetch_array($sql)){ $id = $row["id"]; $product_name = $row["product_name"]; $price = $row["price"]; $date_added = strftime("%b %d, %Y", strtotime($row["date_added"])); $dynamicList .= '<table width="100%" border="2" cellspacing="2" cellpadding="2"> <tr> <td width="17%" valign="top"><a href="product.php?id=' . $id . '"><img style="border:#666 1px solid;" src="inventory_images/' . $id . '.jpg" alt="' . $product_name . '" width="77" height="102" border="2" /></a></td> <td width="83%" valign="top">' . $product_name . '<br /> $' . $price . '<br /> <a href="product.php?id=' . $id . '">View Product Details</a></td> </tr> </table>'; } } else { $dynamicList = "We have no products listed in our store yet"; } mysql_close(); ?> Code: [Select] <p><?php echo $dynamicList; ?><br /> </p> It works ok, and putting my files, everything works, but when I put 8 pictures with price and other details, it just show one image with details and another image below with details, and the third image below and so on .. Can I split dynamic list to show 4 images with details on the left side and 4 on the right side? Thank you in advance for help , if is possible |
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