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PHP - Display Record With Image Link
Greetings!
I have a website www.lanceronlinejobs.com/our_franchises.php I have a franchise images. I want to display each franchise record from database whenever a user click on franchise link. Here is my code. ourfranchises.php code: Code: [Select] <?php include('fconnection.php'); $sql = mysql_query("SELECT * FROM tbl_franchise ORDER BY id DESC") or die(mysql_error()); $row3 = mysql_fetch_array($sql); ?> <a href="franchiseDetails.php?city=<?php echo $row3['city'];?>"><img src="images/lbatkhela.jpg" width="150" height="150" alt="Batkhela" /></a>---------------------------------------------------------------------- franchiseDetails.php code: Code: [Select] <?php $id = $_GET['id']; $query = "SELECT * FROM tbl_franchise WHERE id = '$id' ORDER BY id DESC LIMIT 1"; $result = mysql_query($query); if (!$result) { echo "NO RECORD FOUND"; } else { while($row3 = mysql_fetch_array($result)): ?> Manager Name: <?php echo $row3['manager_name'];?> Please help me. Any help would be appreciated. Thanks. Similar TutorialsHey guys I'm not amazing at PHP, but know just enough to be dangerous. Basically I'm trying to get these records from my database to display in rows of 4. For some reason every time, I'm missing a record (the first record returned in my query). It's probably something simple is my code. Would be grateful if anyone could lend a suggestion. Code: [Select] <?php $a=0; $b=3; while ($ArrayData = mysql_fetch_array($QueryData)) { //if a=0, start new row if ($a==0) { echo '<tr>'; } //create cell //if thumbnail exists, use it, otherwise throw in the unavailable image if ($ArrayData["PhotoThumb"]!="") { $ImageToUse = $ArrayData["PhotoThumb"]; } else { $ImageToUse = "image_unavailable_thumb.jpg"; } echo ' <td width="33%" align="center" valign="top"><p><a href="item.php?ItemID='.$ArrayData["ItemID"].'"><img src="images/'.$ImageToUse.'" border="0" /></a></p> <p><a href="item.php?ItemID='.$ArrayData["ItemID"].'" class="cart_body_text">'.$ArrayData["ItemName"].'</a><br /> <span class="style1"> <strong>$'.sprintf("%01.2f", $ArrayData["UnitPrice"]).'</strong></span><br /> </p> </td>'; //increment by 1 $a++; //if a = 3 then close the row using </tr>, ready to start a new one if ($a==$b) { echo ' </tr>'; //reset counter, ready to start new row $a=0; } } if ($a > 0 ) { echo '</tr>'; } ?> Thanks guys! can i use a link to drop a record, or do u need to do a form and use a button to post to a page? This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=353620.0 Hello, First time posting here, my MYSQL version is 5.1.52 and I have a created a search function to find a user record and if the name is typed correctly the name and details are displayed on the search page, but if the name is typed incorrectly nothing is appearing. Basically I would like to be able to say if no record found and redirect back to the previous page. Code: Code: [Select] <?php add = $_POST['add']; // Database details $host=""; // Host name $username=""; // Mysql username $password=""; // Mysql password $db_name=""; // Database name $tbl_name="Rewards"; // Table name // Connect to server and select databse. mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot SELECT DB"); //Add query $sql = mysql_query("UPDATE Rewards SET Rewards = (Rewards + 10) WHERE Name '%$add%'"); // Database details $host=""; // Host name $username=""; // Mysql username $password=""; // Mysql password $db_name=""; // Database name $tbl_name="UserList"; // Table name // Connect to server and select databse. mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot SELECT DB"); //Update Total Points query $sql = mysql_query("UPDATE UserList SET TotalPoints = (TotalPoints + 10) WHERE Name '%$add%'"); // Database details $host=""; // Host name $username=""; // Mysql username $password=""; // Mysql password $db_name=""; // Database name $tbl_name="UserList"; // Table name // Connect to server and select databse. mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot SELECT DB"); //Search Query $sql = mysql_query("SELECT * FROM UserList WHERE Name '%$add%'"); while ($row = mysql_fetch_array($sql)){ echo '<br/><img src="images/line1.jpg" alt="" width="200" height="2" />'; echo '<br/> Username: '.$row['Username']; echo '<br/>'; echo '<br/><img src="images/line2.jpg" alt="" width="200" height="2" />'; echo '<br/> Name: '.$row['Name']; echo '<br/> Year Group: '.$row['YearGroup']; echo '<br/> Form Class: '.$row['FormClass']; echo '<br/>'; echo '<br/><img src="images/line3.jpg" alt="" width="200" height="2" />'; } // Database details $host=""; // Host name $username=""; // Mysql username $password=""; // Mysql password $db_name=""; // Database name $tbl_name="Rewards"; // Table name // Connect to server and select databse. mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot SELECT DB"); $sql = mysql_query("SELECT * FROM Rewards WHERE Name '%$add%'"); while ($row = mysql_fetch_array($sql)){ echo '<br/> Rewards: '.$row['Rewards']; } // Database details $host=""; // Host name $username=""; // Mysql username $password=""; // Mysql password $db_name=""; // Database name $tbl_name="Sanctions"; // Table name // Connect to server and select databse. mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot SELECT DB"); $sql = mysql_query("SELECT * FROM Sanctions WHERE Name '%$add%'"); while ($row = mysql_fetch_array($sql)){ echo '<br/> Sanctions: '.$row['Sanctions']; } // Database details $host=""; // Host name $username=""; // Mysql username $password=""; // Mysql password $db_name=""; // Database name $tbl_name="UserList"; // Table name // Connect to server and select databse. mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot SELECT DB"); $sql = mysql_query("SELECT * FROM UserList WHERE Name '%$add%'"); while ($row = mysql_fetch_array($sql)){ echo '<br/> Total Points: '.$row['TotalPoints']; echo '<br/>'; echo '<br/><img src="images/line4.jpg" alt="" width="200" height="2" />'; } Any help would be greatly appreciated. Many Thanks P Quinn MOD EDIT: [code] . . . [/code] tags added. Hello all. I am using an API from jambase.com to display events on a php page. Everything works great, the xml file is read and the data is displayed on the page as it should. My only problem is that when the data is displayed there is always a blank area where the first record of data should go. Any ideas how to get rid of this? http://www.wddclients.com/mag33/events_search.php (artist search field not enabled yet) Hopefully this is something pretty minor but if anyone wants to see specific code I can supply that. Thanks so much for all help! I have finaly got my removing recods page working, but i do have one problem... i have a delete button which goes to a new php page which contains the DELETE FROM, this deletes the record automaticaly. There are two paths which i could choose from the problem is i am stuck on both of them. 1. a link that will activate the delete code. 2. once the record has been deleted then it will automaticaly go back the the first page. can any one help me here please? I have the following code: Code: [Select] $result = mysql_query("SELECT * FROM ESSAY_QUESTIONS WHERE SUBJECT = 'ENGLISH'") or die(mysql_error()); while($essay_data=mysql_fetch_array($result)){ $question = $essay_data['QUESTION']; $id = $essay_data['ID']; echo "<a href=\"englishessays.php?id=$id>"; echo "$question </a>"; echo "<br><br>"; } It seems to be almost working but it just displays one link and the address is all the rest of the code including </a>"; echo "<br><br>"; and other questions. I presume there's an error with how I've written the echo statements so can anyone see it? Or can anyone suggest a better way to do this? On the next page, the php will read the ID from the address and display the information form that record. I am looking to accomplish the following but have been hitting a brick wall: * User enters a single keyword into a text field, and conducts a search keyword search. * The results are pulled from the campusbooks.com API(API docs attached) * result are then output in <div> class and includes all the book details and its corresponding url./img etc I'm trying to simplify this process but I continue to receive syntax errors. A step by step practical explanation would do me justice! I know with a submit button I can have a new record created when clicking it, but is it possible to do with a text link? Hey guys, I can't wrap my head around how to make this work right... I have three tables: Code: [Select] CREATE TABLE `games` ( `g_id` int(11) NOT NULL AUTO_INCREMENT, `name` varchar(150) DEFAULT NULL, PRIMARY KEY (`g_id`)); CREATE TABLE IF NOT EXISTS `game_player` ( `r_id` int(11) NOT NULL AUTO_INCREMENT, `p_id` int(11) DEFAULT NULL, `g_id` int(11) DEFAULT NULL, `bool` int(1) NOT NULL DEFAULT '0', PRIMARY KEY (`r_id`)); CREATE TABLE IF NOT EXISTS `players` ( `p_id` int(11) NOT NULL AUTO_INCREMENT, `playerid` varchar(150) NOT NULL, PRIMARY KEY (`p_id`), UNIQUE KEY `playerid` (`playerid`)); The players table is my list of users, and they're tied to the list of games via the game_player table. So here's my issue... I'm trying to show the full list of games, and then check mark each record where the player does play it. This is what I have so far - it shows all the games, but it's not checking the boxes. Code: [Select] $result = mysql_query("SELECT * FROM games") or die(mysql_error()); while($row = mysql_fetch_array($result)) { $newquery = "SELECT * FROM game_player, players WHERE game_player.p_id = players.p_id AND game_player.g_id = ".$row['g_id']. " AND players.playerid = {$userid}"; $query = mysql_query($newquery) or die(mysql_error()); if($query['bool'] == 1) { $set_checked = " CHECKED"; } else{ $set_checked = ""; } echo "<input type=\"checkbox\" name=\"box1\" value=\"".$query['g_id']."\"" . $set_checked . "/>".$row['name']."<br />\n"; } Hi, n0obie here. I'm trying to identify where my customers are coming from via emails I've sent/received. However, many email providers (namely Gmail) have circumvented this by disallowing IP address geolocation/image caching. Hi I've got this database I created with fields ProductId ProductName Image I've managed to get it to list the ID,productname, and Image urls in a list. My next step is to have the image field actually display an image and make it clickable: heres what I've done so far: Code: [Select] <?php $con = mysql_connect("localhost","root",""); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("productfeed", $con); $result = mysql_query("SELECT * FROM productfeeds"); echo "<table border='0'> <tr> <th>Firstname</th> <th>Lastname</th> <th>Image</th> </tr>"; while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['ProductID'] . "</td>"; echo "<td>" . $row['ProductName'] . "</td>"; echo "<td>" . $row['ImageURL'] . "</td>"; echo "</tr>"; } echo "</table>"; mysql_close($con); ?> Heres what I want to do: Code: [Select] while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['ProductID'] . "</td>"; echo "<td>" . $row['ProductName'] . "</td>"; // my changes beneath echo "<td>" . <a href="<?php echo $row['ImageURL'];?>"> <img src="<?php echo $row['LinkURL']; ?>"> </a>. "</td>"; echo "</tr>"; } echo "</table>"; mysql_close($con); ?> Can you guys point me in the right direction? Many thanks My code: Code: [Select] <?php $username = $_SESSION['loggedin']; $extract_user_rank = mysql_query("SELECT `rank` FROM users WHERE username='$username'"); if($extract_user_rank['rank'] > 0) { echo "<a href='control_panel.php'>Admin Control Panel</a>"; } else { //return nothing } ?> I have my rank in the database set to 1. Why doesn't it output the ACP link? I wanted to store a link in the db and display the same link on the front end which will take me to the site. i tried with href but i get some string error. Also i wanted to add a youtube embed tag to display it again on query to the db. in short is possible to store the url in the db and display the same url. thanks I have a line like this it prints text link but I prefer image link how should i edit it I would appreciate some feedback Code: [Select] $templates['etiket'] = array('name' => t('ETİKET'), 'module' => 'uc_invoice_pdf', 'path' => $templates_uc_invoice_pdf_path, 'pdf_settings' => $pdf_settings); Hello guys, I want to make an xml feed into my webpage, but the xml is generated from an .asp file on another website, so it's not actually an xml file, but xml output. Now, I want to retrieve this xml output on my webpage, but I'm having difficulties while doing this. What I've tried so far: - With PHP: Code: (php) [Select] <?php echo file_get_contents("http://link-on-other-website.asp?XML=1"); ?> - With javascript: Code: (javascript) [Select] <script type="text/javascript"> if (window.XMLHttpRequest) {// code for IE7+, Firefox, Chrome, Opera, Safari xmlhttp=new XMLHttpRequest(); } else {// code for IE6, IE5 xmlhttp=new ActiveXObject("Microsoft.XMLHTTP"); } xmlhttp.open("GET","http://link-on-other-website.asp?XML=1", false); xmlhttp.send(); xmlDoc=xmlhttp.responseXML; document.write("<table border='1'>"); var x=xmlDoc.getElementsByTagName("RECORD"); for (i=0;i<x.length;i++) { document.write("<tr><td>"); document.write(x[i].getElementsByTagName("NAME")[0].childNodes[0].nodeValue); document.write("</td><td>"); document.write(x[i].getElementsByTagName("DATE")[0].childNodes[0].nodeValue); document.write("</td></tr>"); } document.write("</table>"); </script> When using the PHP code, I can retrieve it obviously, but I get an <?xml ... > declaration within my body, which is not allowed. When using the Javascript code, I can't even retrieve anything. And the goal is to style this xml output to match my website. That's the whole purpose of this xml link, because I couldn't style the iframe I had earlier. Can someone help me on this, on how to properly embed this asp generated xml output, within my HTML5 document? ps: I've asked for the xml-feed link from that other website, so using the above PHP code is legit in my case. I have a "Members" page that displays my organizations members info via My SQL. Currently, the database displays "State" quick links at the top and has the members organized by State down the page. If you click on one of the State links at the top, it will navigate to the section of the page with that state and associated members. I want the members associated with a specific state to be displayed only once I click the associated state link -- instead of all of the information showing at once like it is now. The page I am referring to can be seen at this link: http://homesforhorse...rs.com/members/
<?php
update_option('image_default_link_type','none');
include("/home/cingen/config_admin.php");
function listMembers() {
$sql = mysql_query("SELECT c.*, s.*
FROM (".TABLE_MEMBERS." c LEFT JOIN ".TABLE_STATE." s on c.state = s.state_abbr)
WHERE c. status = '1'
ORDER BY c.country, c.state, c.organization ASC");
while ($row = mysql_fetch_array($sql)) {
$display_members = false;
$organization = stripslashes($row['organization']);
$website = stripslashes($row['website']);
if ($website) {
$link = "<a href='http://".$website."' target='_blank'>";
$endlink = "</a>";
} else {
$link = "";
$endlink = "";
}
$display_members .= $link.$organization.$endlink."<br />";
if ($row['address']) $display_members .= stripslashes($row['address'])." ".stripslashes($row['address2'])."<br />";
if ($row['city']) $display_members .= stripslashes($row['city']).", ";
if ($row['state']) $display_members .= stripslashes($row['state'])."";
if ($row['zip']) $display_members .= " ".$row['zip'];
$display_members .= "<br />";
if ($row['contact_name']) $display_members .= "Contact: ".stripslashes($row['contact_name']);
if ($row['contact_title']) $display_members .= ", ".stripslashes($row['contact_title']);
if ($row['phone']) $display_members .= "<br />Tel: ".stripslashes($row['phone']);
if ($row['email']) $display_members .= "<br />".$row['email'];
if ($row['website']) $display_members .= "<br /><a href='http://".$row['website']."' target='_blank'>".$row['website']."</a><br/>";
if ($row['year_est']) $display_members .= "Founded in ".$row['year_est'].".";
if ($row['org501c3'] == "1") $display_members .= " A 501(c)3 non-profit.";
if ($row['gfas'] == "1") $display_members .= "<br />GFAS: Accredited Sanctuary.";
if ($row['gfas'] == "2") $display_members .= "<br />GFAS: Verified Sanctuary.";
if ($row['member_category']) $display_members .= "<br />".$row['member_category'];
$display_members .= "<br /><br />";
$entries[$row['country']][$row['state_name']][] = $display_members;
}
$countrylinks = false;
$statelinks = false;
$display = false;
if(is_array($entries)){
$display .= '
<div class="memberlist">';
foreach($entries as $country=>$state_members){
$countrylinks .= '<a href="#'.$country.'">'.$country.'</a> ';
$display .= '
<h2 id="'.$country.'">'.strtoupper($country).'</h2>
<div class="country">';
if(($state_members)){
foreach($state_members as $state=>$members){
$statelinks .= '<a href="#'.$state.'">'.$state.'</a> ';
$display .= '
<h3 id="'.$state.'">'.strtoupper($state).'</h3>
<div class="state">';
if(is_array($members)){
foreach($members as $key=>$member){
$display .= '
<div class="member">
'.$member.'
</div>';
}
}
$display .= '
</div>';
}
}
$display .= '
</div>';
}
$display .= '
</div>';
}
$statelinks1 = '
<h2>Members List</h2>
<strong>Quick Links</strong><br /><br />
'.$statelinks.'<br /><br />'
.$display;
return $statelinks1;
}
add_shortcode('memberlist', 'listMembers');
function listRescueStandards() {
$display_members = '';
$sql = mysql_query("SELECT vc.*, s.*, m.*
FROM ".TABLE_COMPLIANCE." vc, ".TABLE_STATE." s, ".TABLE_MEMBERS." m
WHERE vc.member_id = m.cid
AND m.status = '1'
AND m.state = s.state_abbr
ORDER BY m.state, m.organization ASC");
while ($row = mysql_fetch_array($sql)) {
$organization = stripslashes($row['organization']);
if ($row['website']) {
$link = "<a href='http://".$row['website']."' target='_blank'>";
$endlink = "</a>";
} else {
$link = "";
$endlink = "";
}
if($x!=$row['state_name']){
$display_members .= "<br /><strong>".strtoupper($row['state_name'])."</strong><br />";
$x = $row['state_name'];
}
$display_members .= $link.$organization.$endlink."<br />
".stripslashes($row['address'])." ".stripslashes($row['address2'])."<br />
".stripslashes($row['city']).", ".stripslashes($row['state'])." ".$row['zip']."<br />";
if ($row['contact_name']) $display_members .= "Contact: ".stripslashes($row['contact_name']);
if ($row['contact_title']) $display_members .= ", ".stripslashes($row['contact_title']);
if ($row['phone']) $display_members .= "<br />Tel: ".stripslashes($row['phone']);
if ($row['fax']) $display_members .= "<br />Fax: ".stripslashes($row['fax']);
if ($row['email']) $display_members .= "<br />".$row['email'];
if ($row['website']) $display_members .= "<br /><a href='http://".$row['website']."' target='_blank'>".$row['website']."</a>";
if ($row['year_est']) $display_members .= "<br />Founded in ".$row['year_est'].".";
if ($row['org501c3'] == "1") $display_members .= "<br />This organization IS registered with the IRS as a 501(c)3.";
if ($row['org501c3'] != "1") $display_members .= "<br />This organization is NOT registered with the IRS as a 501(c)3.";
$display_members .= "<br /><br />";
}
return "<div class='memberlist'>" . $display_members . "</div>";
}
add_shortcode('standardslist', 'listRescueStandards');
Hi, I've read a lot of places that it's not recommended to store binary files in my db. So instead I'm supposed to upload the image to a directory, and store the link to that directory in database. First, how would I make a form that uploads the picture to the directory (And what kinda directories are we talking?). Secondly, how would I retrieve that link? And I guess I should rename the picture.. I'd appreciate any help, or a good tutorial (Haven't found any myself). Hello, Five images will be displayed inside a division. There will be a previous and next button/link. If someone click the next button the next image will be added in that div and the first image will be gone from that div. The previous button/link will do the same thing. Is it possible with php? I am confused if it's a javascript or ajax question. Thanks. Would like to be able to click on a radio button that represents an image. Once selected and submitted, have that image display on another page. I have an idea, but need some guidance. BTW, is using php only doable? Is there a simpler or more elegant way to do this? Thanks all! |
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