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PHP - Query Positioning Within A Page
I am trying to clean up some code in my pages and would like some advice on good practice for the structure.
Initially the code was generated by Dreamweaver and everything relating to the queries were place at the very top of the page, outside any of the HTML. A lot of the coding help I have had from this forum has put the queries etc in with the actual output php. Is there any right or wrong way to do this? I assume both ways have their pros and cons and both generate the same end result but I want to start getting into good habits before it is too late!! Thanks in advance Steve Similar TutorialsHey Guys, Ive been searching for a while, but havent really come up with any answers. At the minute i have a page where i can add a new topic to a database, but i want to be able to update the query on another page if that makes sense So as an example say i have this below thats just been inserted into the database with an ID of 2. <?php // Details $id = $_POST['id']; $title = $_POST['title']; $detail = $_POST['detail']; mysql_connect("localhost", "web148", "123") or die(mysql_error()); mysql_select_db("web148") or die(mysql_error()); mysql_query("SET names 'utf8'"); mysql_query("INSERT into sunnyargues(ID, title, body) VALUES ('2', ''超级马里奥意大利再遭种族歧视', '当被问及在友谊赛上的不友好一幕会不会重演时，曼奇尼说，“有这种可能性。” “我当然不希望这样的事情发生，但是谁也不能保证。”'"); echo "<h1>Article Updated!</h1>"; ?> On the sunny argues page i currently have, where it only selects ID 1: <?php mysql_connect("localhost", "web148", "123") or die(mysql_error()); mysql_select_db("web148") or die(mysql_error()); mysql_query("SET names 'utf8'"); $result = mysql_query("SELECT body FROM sunnyargues WHERE ID='1'"); $row = mysql_fetch_array( $result ); echo nl2br($row['body']); ?> Is there a way i can update this query with the new ID of the 2nd article once its been inserted into the database. I hope that makes sense friends whenever i include this coding to my page, everything goes blank. how could i fix this? Code: [Select] $time_count = 7*24*60*60; // count the 7 days Ads will be deleted $match_time = time()-$time_count; $query = "SELECT id FROM classifieds WHERE time<'".$match_time."' ORDER BY RAND() LIMIT 1"; $result = mysql_query($query); while($ad_id = mysql_fetch_array($result)) $delete =$ad_id[0]; mysql_query("DELETE FROM classifieds WHERE id='$delete'"); Hello all. I'm using this code to go through the database and output certain user-defined numbers Code: [Select] <?php mysql_connect ("pdb1.awardspace.com", "anastasov_db","moscow1945") or die (mysql_error()); mysql_select_db ("anastasov_db"); $term = $_POST['term']; $sql = mysql_query("select * FROM countries WHERE cocode = '$term'"); while ($row = mysql_fetch_array($sql)){ echo '<br/> Code: '.$row['cocode']; echo '<br/> Country: '.$row['coname']; echo '<br/><br/>'; } ?> Now how would I go about making a page that appears if the script can't find any results in the database? Thank you in advance hey @requinix, I know you and I have talked about this before but I just want to clarify this issue. Sadly, I have produced very few websites that have made use of dynamics in the form of database querying. But, the next project that I'm looking at will require it. The last thing I did with a query string involved the recording of downloads of file on a website. and on the page, I had this code: The archive is located he <a href="dl.php?f=archive.zip">archive.zip</a> and the ''dl'' page had this code:
if(!$_GET['f']) error('Missing parameter!');
$stmt = $conn->prepare("INSERT INTO tblDownloads (ip, file, date, time)
VALUES (?, ?, ?, ?)");
$stmt->bind_param("ssss", $ip, $file, $date, $time);
$stmt->execute();
$stmt->close();
$conn->close();
header("Location: archive.zip");
exit;
and all the variables in the above ''dl'' page are declared in a file that is required by it, called ""conn.php"". My question for anyone here is: I am going to be producing a website similar to the testing site called ""jsFiddle"": https://jsfiddle.net/ . And it is slated to have a huge amount of content on it. for example, a main page might have 100 links on it whereby, if a user clicks on any of them, it should take them to another page (or load async content) that illustrates and example of any given issue being asked about. So, I know we've talked about $_GET before Requinix, and how most websites use it to produce dynamic content. And we've also talked about the concept of markdown. But once again, can someone here remind me....what's the best way to go about doing this? Storing the actual ""answer to the issue"" content in a backend database and pulling/displaying it appropriately, kind of like what I am showing I did with the file in the above code? thanks. Adam Edited September 6, 2020 by ajetrumpetHi, I'm making an attempt to build a small cms based site. I have read in a lot of places it is best to use the id of a table row to fetch contents, but being as I am using url rewrite in frontend, it won't be showing query strings, so I thought I would ask if anyone saw any potential problems if I did the following. Page table looks like this: Pages - id - name - content In backend, the user has a form to name the page and use ckeditor to add rich content. I will use php to ensure that the name is alphanumeric and use strtolower & str_replace to ensure that My First HTML Page is sent to mysql as my-first-html-page and also check that a row with this name does not already exist. So on front end this page url will be mywebsite.com/my-first-html-page So to output the content I strip out the domain & / so I'm left with the name as entered in the db. $url = $_SERVER['REQUEST_URI']; $url = str_replace ('/','',$url); Then in my function to output content I will use WHERE name = $url hey guys im sure this is possible im wondering how a user can execute a mysql query by a click of a button but without the page reloading...thanks guys hi there alll newbie here lol just needed help with the code for the next page of results ? IE i have say 30 results returned from a sql query but only want to show ten at a time i no how to limit the result with the LIMIT 10 but how do i get it to put link show the other ten and so on. the code i have is here at the bottom of page http://www.phpfreaks.com/forums/index.php/topic,307971.0.html thanks in advance for any help kaine I have an MS Access Database that resides on the server that I will be hosting the pages I'm building. In my MS Access Database I have a query called "CheckedInEquipment" that I would like to have constantly display on a PHP page. It needs to be 'fluid', so the data is always up to date on the web page. (This is all in a sandbox XAMPP environment). Has anyone done this, can you lead me to examples or push me in the right direction? Carl OK, here it is. I have been trying to do this myself, but it has been driving me insane and I turn to professionals here for help.
I am a basic web developer for a company that I work for in other capacities. I have a reasonable understanding of HTML and that is about where my expertise ends. I am not typically a programmer, just a simple (extremely) part time designer that uses Muse and Dreamweaver when necessary.
However, recently my company has asked me to accomplish a task for their website. In plain English, they want a large database that exists currently as a CVS file made into a searchable web page. It is 21 columns by approximately 6,700 rows.
To explain what I need a little more technically, here are my ideas and where I have gotten to so far:
1. The company uses godaddy, into which I *believe* I have successfully imported the spreadsheet. I believe it is successful because through godaddy's SQL Control Panel (phpMyAdmin console), I can do the EXACT searches that the company needs, and it spits out the EXACT results that I need.
2. The end result needs to be a .php that I can upload to the website's root folder that can be then inserted into premade pages using:
<iframe src="SMQ.php" scrolling="yes" width="950" height="800"></iframe>3. On the .php page, I need to have a way to log in to the SQL server and a simple search box built in that will allow the user to input a very simple search string consisting of no more than 4 numbers and 3 letters at a time. No buttons, no check boxes, just a search box. 3. This query then needs to be output as a nice data table, similar to this: This in fact is a screenshot of a search I performed out of my SQL database, in phpMyAdmin using the column "Scott" for the search, and the number 226 as the search term. All column names are visible with the exception of the first column, entitled LINEID, made to be the key, and the output should not include the key but have everything else as above. 4. I can see what the simple line of php is that performed this task: SELECT * FROM `SMQSQL` WHERE `Scott` = '226' ORDER BY `LINEID` ASCbut I can't figure out how the hell to get this incorporated to a .php search. To sum it up, I need a .php page written that can connect to a SQL database, perform a data based search, and spit out a clean table when it is done. I had accomplished this in the past using an import into google docs and using it to perform a search and result display via the following code built into a php called SMQ.php:
<!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>Example of Google Spreadsheet Data Visualisation</title>
</head>
<body>
<form id="form1" method="post" action ="<?php echo $_SERVER['PHP_SELF']; ?>"> <label>
<input id="search" name="search" type="text" />
</label>
<label>
<input id="Scott #" name="Scott #" type="submit" value="Scott #" />
</label>
<img src="loading.gif" width="16" height="11" />
</form>
<p>
<?php
$search= $_REQUEST['search'];
if ($search > ''){ $search = $search;} else { $search = '';}
?>
<script type="text/javascript" src="http://www.google.com/jsapi"></script>
<script type="text/javascript">
google.load('visualization', '1', {packages: ['table']});
</script>
<script type="text/javascript">
var visualization;
function drawVisualization() {
var query = new google.visualization.Query(
'https://docs.google.com/spreadsheet/ccc?key=0AronCwm9QPefdGpIUllscGgtLUJod2pOazc0bjU0cUE&usp=sharing');
query.setQuery('SELECT A, B, C, D, E, F, G, H, I, J, K, L, M, N, O ,P ,Q ,R ,S ,T WHERE (A) LIKE ("<?php echo $search; ?>") order by A asc label A "Scott #", B "Den", C "Color", D "Cond", E "40", F "60", G "70", H "70J", I "75", J "75J", K "80", L "80J", M "85", N "85J", O "90", P "90J", Q "95", R "95J", S "98", T "98J"');
query.send(handleQueryResponse);
}
function handleQueryResponse(response) {
if (response.isError()) {
alert('Error in query: ' + response.getMessage() + '' + response.getDetailedMessage());
return;
}
var data = response.getDataTable();
visualization = new google.visualization.Table(document.getElementById('table'));
visualization.draw(data, { page: 'enable', page: 16, pageSize: 16, legend: 'bottom'});
}
google.setOnLoadCallback(drawVisualization);
</script>
<div id="table"></div>
</div>
</body>
</html>
But as you can see, this may not be the most secure thing in the world, plus we want to be able to expand it in the future and not be so simplistic, hence the need to switch to SQL.
Please let me know right away by contacting me at disead@gmail.com if this is something YOU might be able to help with. I'm sure for an experienced programmer, once you have the details from me that you need, it would take maybe 10 minutes to write. I don't have much, but I can pay a little bit for this one time job. If it ends up working out, I may be able to pay more down the line for more advanced options such as being able to do drop-down searches based on the column titled "ISSUE", as well as more things down the line as it grows.
Thank you so much, I hope to hear from someone soon!!!
What is the best way to display the next record on a web page, from a list of records, without executing the query over and over again each time you display the next record? For example, imagine my user has a list of messages displayed on a web page, and the user filters that list to display all messages with the title "Test", and let's say that returns 12 records. He then clicks on record 1 to read that message. Once he is done reading it, he then wants to read msg 2, and then msg 3 etc. He could hit the Back button in his browser and click on the next record from the list, but it would be more useful to just click on a link that says "Next" which displays the next message in the filtered list of messages (without having to go back to the list). A lot of web sites do this sort of thing, but what is the best way to do it? Do you have to run the original query again each time a new message is loaded, so you know which message is next, or can you somehow store the query to save the server from having to run the same query over and over again? If a user is viewing a list of pictures for example, they could be flicking through the pictures quite quickly, and I would not like to have the server running the same query over and over again every few seconds when it doesn't need to. Advice? (Thanks.) Good evening, I am currently doing a web application which requires pulling out of data from the database. I am still a novice in the programming industry, and is still seeking help from my colleagues and of course forum sites like phpdn. I have an existing code which my friend provided me. I have already done some modifications with the code. I have a problem though with the code, it executes database query upon loading of the page. I do understand how the code works, however, I am not able to modify the code to disallow the execution of query upon loading of page. Here is the code: <?PHP include("dbconnection.php"); $query = "SELECT * FROM records"; if(isset($_POST["btnSearch"])) { $query .= " WHERE last_name LIKE '%".$_POST["search"]."%' OR first_name LIKE '%".$_POST["search"]."%'OR territory LIKE '%".$_POST["search"]."%'OR job_title LIKE '%".$_POST["search"]."%'OR title LIKE '%".$_POST["search"]."%'OR employer LIKE '%".$_POST["search"]."%' " ; } $result = mysql_query($query, $connection) or die(mysql_error()); ?> I do know that this php code, the way it is written, is supposed to do that - to select data from my database (This code was provided by a friend). But my requirement for the project is actually to give it a search engine and display the information based from the search query. I have a search engine already together with the code, and it works pretty well. What I must do is to disallow the pulling of data from the first load, but just pull data if the search engine is used. Here's the whole code: <link href="add_client.css" rel="stylesheet" type="text/css"> <?PHP include("dbconnection.php"); $query = "SELECT * FROM records"; if(isset($_POST["btnSearch"])) { $query .= " WHERE last_name LIKE '%".$_POST["search"]."%' OR first_name LIKE '%".$_POST["search"]."%'OR territory LIKE '%".$_POST["search"]."%'OR job_title LIKE '%".$_POST["search"]."%'OR title LIKE '%".$_POST["search"]."%'OR employer LIKE '%".$_POST["search"]."%' " ; } $result = mysql_query($query, $connection) or die(mysql_error()); ?> <table width="760" border="0" align="center" cellpadding="0" cellspacing="0"> <tr> <td><table width="760" border="0" cellpadding="0" cellspacing="0"> <tr> <td width="199" align="center" valign="top"><a href="login.html"><img src="asia.gif" alt="" width="152" height="58" border="0" /></a> <script type="text/javascript" src="menu.js"></script></td> <td width="176" align="right" valign="bottom"><a href="main.php"><img src="Home.jpg" width="104" height="20" border="0"/></a></td> <td width="130" align="right" valign="bottom"><img src="View.jpg" width="104" height="20" border="0"/></td> <td width="146" align="right" valign="bottom"><a href="add_client.php"><img src="Add.jpg" width="104" height="20" border="0"/></a></td> <td width="109" align="right" valign="bottom"> </td> </tr> </table></td> </tr> <tr> <td><table width="760" border="0" cellpadding="0" cellspacing="0"> <tr> <td width="200" height="3" bgcolor="#1B1C78"><img src="images/topspacerblue.gif" alt="" width="1" height="3" /></td> <td width="560" bgcolor="#0076CC"><img src="images/topspacerlblue.gif" alt="" width="1" height="3" /></td> </tr> </table></td> </tr> <tr> <td height="553" align="center" valign="top" bgcolor="#F3FAFE"><br /> <form name="form" action="view_client.php" method="post"> <table width="351" border="0"> <tr> <td width="137" align="left" valign="middle">SEARCH RECORD:</td> <td width="144" align="center" valign="middle"><input type="text" name="search" /></td> <td width="56" align="left" valign="middle"><input type="submit" name="btnSearch" value="Search" /></td> </tr> </table> <br /> <table width="680" border="0" cellpadding="3" cellspacing="1" bordercolor="38619E" > <tr> <th width="100" align="center" bgcolor="#E0E8F3">Territory</th> <th width="110" align="center" bgcolor="#E0E8F3">Employer</th> <th width="110" align="center" bgcolor="#E0E8F3">Job Title</th> <th width="50" align="center" bgcolor="#E0E8F3">Title</th> <th width="110" align="center" bgcolor="#E0E8F3">First Name</th> <th width="110" align="center" bgcolor="#E0E8F3">Last Name</th> <th width="70" align="center" valign="middle" bgcolor="#E0E8F3"> </th> </tr> <?php if($result) { for($i=0; $i<mysql_num_rows($result); $i++) { $id = trim(mysql_result($result, $i, "id")); $territory = trim(mysql_result($result, $i, "territory")); $employer = trim(mysql_result($result, $i, "employer")); $job_title = trim(mysql_result($result, $i, "job_title")); $title = trim(mysql_result($result, $i, "title")); $first_name = trim(mysql_result($result, $i, "first_name")); $last_name = trim(mysql_result($result, $i, "last_name")); echo "<tr>"; echo "<td>".$territory."</td>"; echo "<td>".$employer."</td>"; echo "<td>".$job_title."</td>"; echo "<td>".$title."</td>"; echo "<td>".$first_name."</td>"; echo "<td>".$last_name."</td>"; echo "<td><a href='admin_edit.php?id=".$id."'>edit</a> | <a href='admin_delete.php?id=".$id."'>del</a> </td>"; echo "</tr>"; } } ?> </table> <br /> </form> <p> </p></td> </tr> <tr> <td height="38"><table width="760" border="0" cellpadding="0" cellspacing="0"> <tr> <td width="200" height="35" align="center" bgcolor="#1B1C78" class=white><img src="images/topspacerblue.gif" alt="" width="1" height="3" /> <a href="disclaimer.html"><font color="#FFFFFF">Legal Disclaimer</font></a> </td> <td width="560" align="center" bgcolor="#0076CC" class=white><img src="images/topspacerlblue.gif" alt="" width="1" height="3" /> Copyright © 2006 - 2010 Limited. All rights reserved. </td> </tr> </table></td> </tr> </table> Immediate response is well appreciated. Thank you very much! Code: [Select] The form below is meant to retrieve the named attribute of a form input called "limitedtextfield" , along side the current date and store those values in a database. The form contains some javascript which can be disregarded for this particular problem. [code] <form name="mymind" style= "position:relative;left:40px;" action="insert_status.php" method="post"> <input name="limitedtextfield" type="text" onKeyDown="limitText(this.form.limitedtextfield,this.form.countdown,55);" onKeyUp="limitText(this.form.limitedtextfield,this.form.countdown,55);" maxlength="55"><br> <font size="1">(Maximum characters: 55)<br> You have <input readonly type="text" name="countdown" size="3" value="55"> characters left.</font> <input type="hidden" name="submitted" value="TRUE"/> <input style="position:relative;left:0px;bottom:0px;" type="submit" name="submit" value="Submit!" /> </form> Here is the php script that processes the form. What it is meant to do is search the database and if no record is found for the authenticated member, input the relevant values into the database. If a record is found, it should update the record. Well every time I hit the submit button, all I get is a blank page. Any clue as to what is going wrong? PS: I checked the database and no values got inserted. Code: [Select] <?php //address error handling ini_set ('display_errors', 1); error_reporting (E_ALL & ~E_NOTICE); //authenticate user require('auth.php'); //Connect to database require ('config.php'); if (isset($_POST['submitted'])) { $query = "SELECT* FROM publications WHERE member_id ='".$_SESSION['id']."' AND cartegory='status'"; $result = @mysql_query($query); $num = @mysql_num_rows($result); if ($num> 0) { $update = mysql_query("UPDATE publications SET publication = '{$_POST['limitedtextfield']}' WHERE member_id = '".$_SESSION['id']."' AND cartegory = 'status'"); header("Location: member.php"); } else { $query = "INSERT INTO publications ( member_id, publication, cartegory, pub_date) VALUES ( '{$_SESSION['id']}','{$_POST['limitedtextfield']}', 'status', NOW() )"; header("Location: member.php"); } } ?> Good day and Merry Christmas to all, I just spent a good time of my christmas eve trying to figure out this problem. I hope one of you santas would be so kind as to help me with it. First off I have two tables; employee and employee_works both connected via employee_id key. Basically I have a parent window we'll call parent.php. inside the parent page is a search button that once clicked will open a child window we'll call child.php inside the child page is a list, lets say employees with name, employee_id, etc. My main concern is this: How do I populate parent.php based off the employee selection I made in the child window. Example: -Access parent.php -Click on search -Click on [ID: 004] [NAME: JOHN SMITH] [PHONE: 1233456] [DATE HIRED: JULY 16, 1992] <---format of a row in child.php -child.php automatically closes and parent.php now shows all data from employee_works with the employee_id = 004 Is this even possible? I know this is vary vague and would be willing to explain more if needed. My website is built mostly on javascript and php. Hi, I know this has to be possible but I have been unable to get this to work properly. I have a page with two forms. The first form is a list of sources associated with a subject, with checkboxes, so that it is possible to remove these associations, like so: Code: [Select] $qs = "SELECT s.source_id, s.source_name from source s, source_subject ss, subject sb where s.source_id = ss.source_id and sb.subject_id = ss.subject_id and sb.subject_id = $subject_id order by source_name"; $rs = mysqli_query($dbc, $qs) or die ('Error querying database'); while($row = mysqli_fetch_array($rs)) { echo '<input type="checkbox" value="' . $row['source_id'] . '" name="markdelete[]">' . $row['source_name'] . '<br />' ; } The second form is a list of sources that are not associated with this subject, with checkboxes enabling the addition of more sources to this subject, like so: Code: [Select] $r = "SELECT source_id, source_name FROM source WHERE source_id NOT IN (select s.source_id from source s, source_subject ss where s.source_id = ss.source_id and ss.subject_id = '$subject_id')"; $r1 = mysqli_query($dbc, $r) or die ('Update Error: '.mysqli_error($dbc)); while ($row = mysqli_fetch_array($r1)) { echo '<input type="checkbox" id="source_id" name="source_id[]" ' ; echo 'value="'. $row['source_id'] .'"'; echo '> ' . $row['source_name'] . '<br />'; } The first form works fine... when the "remove" submit button is clicked, it successfully removes any selected sources and then shows the remaining sources still associated, and then automatically populates this source in the below list (form 2) as one that is not selected. On the second form, when adding a new source to the subject, it successfully enters the data into the database, and removes this source from the list of unselected sources, but it does not seem to re-populate in the first form. The data is correct on the tables, but the page needs to reload the first query... how do I get it to do this? I'd like to be able to make this page editable, and re-editable (in case someone makes a mistake) without having to go back and reload the entire page. Note, both forms call the page itself; not sure if this is part of it. Using this: Code: [Select] <form method="POST" action="<?php echo $_SERVER['PHP_SELF'].'?subject_id='.$subject_id ; ?>"> Does anyone have any ideas? This topic has been moved to Third Party PHP Scripts. http://www.phpfreaks.com/forums/index.php?topic=355955.0 I wrote the page below to display a member's area of a website. The page is a mix of html and php code. the first part of is a block of php, which authenticates the user and displays the header. The middle part is a block of html which contains the bulk of the page. Embedded in this block of html are two sections of php, the first of which is supposed to display a user uploaded picture, enclosed in div tags, and the second of which, prints out the user's first name, from the database. The end of the script is a line of php that displays the footer. Now the page displays correctly when the section of php that contains the select query for displaying the picture(starting line 104) is omitted. But once I include that section, all i see is a blank page. Why is that section of php problematic? What can be done to fix it? Here is the data for the full page. Thanks for any help. <?php //address error handling ini_set ('display_errors', 1); error_reporting (E_ALL & ~E_NOTICE); //authenticate user require('auth.php'); //define title define('TITLE' , 'Members'); require ('header.html'); //need the header ?> <div id="main" style="background-color: #FFFFFF; height:71%; width:101%; border:0px none none; margin:auto; "> <!-- --> <div id="main_left" style="float:left; height:100%; width:20%; border:0px none none;"> <!--opens main left--> <div id="main_left_top" style="float:left; position:relative;bottom:5px;right:5px; height:31.25%; width:100%; background-color: #FFFFFF; border:1px solid #c0c0c0; margin:1px;"> <!--opens main left top--> </div> <!-- closes main left top--> <div id="main_left_center" style="float:left; background-color: #FFFFFF; height:33%; width:100%; border-color:#a0a0a0;border-style:outset;border-width:1px; margin:auto; "> <!--opens the white content area--> </div> <!-- closes main left center--> <div id="main_left_bottom" style="float:left; background-color: #FFFFFF; height:33%; width:100%; border-color:#a0a0a0;border-style:outset;border-width:1px; margin:auto; "> <!--opens the white content area--> </div> <!-- closes main left bottom--> </div> <!-- closes main left--> <div id="main_center" class="content_text" style="float:left; height:100%; width:58%; background-color: #FFFFFF; border:1px solid #c0c0c0;"> <!--opens main center--> <div id="image_box" style="float:left; background-color: #c0c0c0; height:150px; width:140px; border-color:#a0a0a0;border-style:outset;border-width:1px; margin:auto; "> <?php //address error handling ini_set ('display_errors', 1); error_reporting (E_ALL & ~E_NOTICE); $query = "SELECT* FROM images WHERE member_id ='{$_SESSION['id']}' AND cartegoty 'main' "; $result = mysql_query($query); $result_data = mysql_fetch_array($result); header("Content-type: image/jpeg") ; echo $result_data['image']; ?> </div> <a href="upload_image_page.php">click here to uplaod a picture</a> <h1>Welcome <?php echo ucfirst($_SESSION['firstname']);?></h1> <a href="member_profile.php">My Profile</a> | <a href="logout.php">Logout</a> <p>This is a password protected area only accessible to members. </p> <a href="blog_entries.php">Add to Hahap Tok Library</a> </div> <!-- closes main center--> <div id="main_right" style="float:left; background-color: #FFFFFF; height:100%; width:20%; border-color:#a0a0a0;border-style:outset;border-width:1px; margin:auto; "> <!--opens the white content area--> <div id="main_right_top" style="float:left; background-color: #FFFFFF; height:33%; width:100%; border-color:#a0a0a0;border-style:outset;border-width:1px; margin:auto; "> <!--opens the white content area--> </div> <!-- closes main left top--> <div id="main_right_center" style="float:left; background-color: #FFFFFF; height:33%; width:100%; border-color:#a0a0a0;border-style:outset;border-width:1px; margin:auto; "> <!--opens the white content area--> </div> <!-- closes main left center--> <div id="main_right_bottom" style="float:left; background-color: #FFFFFF; height:34%; width:100%; border-color:#a0a0a0;border-style:outset;border-width:1px; margin:auto; "> <!--opens the white content area--> </div> <!-- closes main left bottom--> </div> <!-- closes main right--> </div> <!-- closes main--> <?php require('footer.html'); ?> I'm trying to write a php page that displays data from a JOIN query for a specific ID table view brandinfo ID, brand, discounttype 1, antioni, no discount brandproducts brandID, producttype, price 1, Tshirt, 20.00 1, Pants, 30.00 1, Shoe, 40.00 the returned result is 1 antioni, no discount, Tshirt, 20.00, 2 antioni, no discount, Pants, 30.00 3 antioni, no discount, Shoe 40.00 The way I want the page to be displayed is ------------------ Antioni (at the top) Table 1. Tshirt 20.00 2. Pants 30.00 3. Shoe 40.00 no discount (at the bottom) ---------------------------- How should I construct the PHP page from the result since they're retrieved as rows? Hi Folks, This is my first post here and I was wondering if anyone could help me out at all please. I would basically like to change the logo on the home page based on whether a query string from a referring URL has a specific value. It's an affiliate URL for an online store, so essentially I need to get the referrer, check if the query string adnetwork = as and then replace the logo if it does, if not leave it as it is. Does that make sense? Any help would very much appreciated. I was thinking along the lines of: <?php if ($_GET('adnetwork') == 'as') { changeLogo(); } else { leaveLogo(); } ?> Am I along the right lines folks? Hi All, I have a form that enables the user to know events that occur in a specific city. Most of the time they get the results in a few seconds, despite the fact that many rows in various tables might be searched for detailed infos about the events in that city. However, this form partly relies on a table against which a MySQL event regularly does some operations. If the user uses the form while the event is being executed, they have to wait up to 30 seconds before getting the results, and sometimes only part of the html page is generated. Is there a way to avoid this lengthy waiting time ? I am wondering whether or not the best would be putting the site on maintenance while the event is being executed, with a "please come back in a few minutes" message. Thanks in advance for your help. Here is my code: // Start MySQL Query for Records $query = "SELECT codes_update_no_join_1b" . "SET orig_code_1 = new_code_1, orig_code_2 = new_code_2" . "WHERE concat(orig_code_1, orig_code_2) = concat(old_code_1, old_code_2)"; $results = mysql_query($query) or die(mysql_error()); // End MySQL Query for Records This query runs perfectly fine when run direct as SQL in phpMyAdmin, but throws this error when running in my script??? Why is this??? Code: [Select] You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '= new_code_1, orig_code_2 = new_code_2WHERE concat(orig_code_1, orig_c' at line 1 |
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