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PHP - Linking Php Web Page With Asp3 Classic
Dear all,
Is it possible to link a php web page into a ASP3 classic page? Suppose there's a link on a ASP page into a PHP page, also there a link in the PHP page to the ASP page. I was thinking about use session variable to go back and forth between the page, but I don't know how to use the variable in the PHP or ASP code. Is this a correct way? Or is there any other approach? regards, ethereal1m Similar TutorialsI have only been working with PHP for about a week, so please be patient. I need to convert the following ASP Classic code to PHP. I have the SQL partially working.
<label>Select a bulletin:</label>
<select>
<option value="">Select One</option>
<?php
$servername = "localhost";
$username = "username";
$password = "password";
$conn = new mysqli($servername, $username, $password);
if (!$conn) {
die('Could not connect: ' . mysql_error());
}
$sql = "SELECT bulletinid, bulletindate, bulletinlink, bulletintext FROM tblbulletins";
mysql_select_db(MyDB);
$retrieved = mysql_query( $sql, $conn );
if(! $retrieved ) {
die('Could not get data: ' . mysql_error());
}
while($row = mysql_fetch_array($retrieved, MYSQL_ASSOC)) {
$bullentinid = $row['bulletinid'];
$bullentindate = $row['bulletindate'];
$bullentinlink = $row['bulletinlink'];
echo $link;
}
mysql_close($conn);
?>
</select>
The records that are retrieved a | 01 | 05/01/19 - 05/07/19 | link 1 | <embed src="bul/05-01.pdf#"> | | 02 | 05/08/19 - 05/14/19 | link 2 | <embed src="bul/05-08.pdf#"> | | 03 | 05/15/19 - 05/21/19 | link 3 | <embed src="bul/05-15.pdf#"> | The problem I'm having is converting line 10 in ASP Classic to PHP:
01. <% Function bulletins %>
02. <% sSQL = "SELECT bulletinid, bulletindate, bulletinlink, bulletintext FROM tblbulletins ORDER BY bulletinid" %>
03. <% Set oRecordSet = adoCon.Execute(sSQL) %>
04. <% If Not oRecordSet.EOF Then %>
05. <% arrayResult1 = oRecordSet.GetRows() %>
06. <% End If %>
07. <% Set oRecordSet = Nothing %>
08. <% iRowNumber1 = ubound(arrayResult1,2) %>
09. <% For iCounter1= 0 to iRowNumber1 %>
10. <% Response.Write ("<option value="pg01.php?mode=view&ID=<%=arrayResult1(0,iCounter1)%>"><% = arrayResult1(1,iCounter1) %></option>")
11. <% Next %>
12. <% End Function %>
13. <% 'arrayResult1(0,iCounter1) = bulletinid %>
14. <% 'arrayResult1(1,iCounter1) = bulletindate %>
15. <% 'arrayResult1(2,iCounter1) = bulletinlink %>
16. <% 'arrayResult1(3,iCounter1) = bulletintext %>
17. <label>Select bulletin:</label>
18. <select>
19. <% Call bulletins %>
20. </select>
Once the link was clicked, it would open pg01.php to the specific saint that was selected. Here is the ASP Classic code on pg01.asp that needs to be converted to PHP:
<% IF Request.querystring("mode") = "view" THEN %>
<% Session("bulletinid") = Request.querystring("ID") %>
<% bulletinID = Clng(Session("bulletinID")) %>
<% sSQL = "SELECT * FROM tblbulletins WHERE bulletinid = " & bulletinID %>
<% Set oRecordSet = adoCon.Execute(sSQL) %>
<% IF Not oRecordSet.EOF THEN %>
<% arrayResult = oRecordSet.GetRows() %>
<% END IF %>
<% END IF %>
<% Set oRecordSet = Nothing %>
<div class="main">
<h4><% =arrayResult(1,0) %></h4>
<p><%=arrResultSet(4,0)%></p>
</div>
I'm working on a site that provides database results. To get to the final report, I'm going through a few pages where the user selects specific options. This builds up arrays that need to be passed from one page to another. As a result the final page has a pretty lengthy variable list. Since I want to provide a link to certain customers where they don't have to go through all the option select stuff, I'm using the GET method on the final page. Due to all the arrays I need to pass, I'm hitting hyperlink address character limitations, and I fear I will eventually hit browser URL length limits. Is there any way to provide links to customers without using the GET method so I could keep the URL length down? Ok. Thought I'd be able to do this myself but ran into a couple of snags. This is the template page for the image thumbnails. WHen a thumbnail is clicked it is dragged into this template. I put the code for the comments on the page but it's just not working properly. When I post comment the comment posts but the picture turns into my BAD PHOTO ID error message. Have some ideas on whats going but it's getting kind of messy. If you need to see the pages from the comment system I will post. Thanks phpeople. Where's fugix? Code: [Select] <!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>What do you know about.......</title> <style type="text/css"> #What { text-align: center; } </style> </head> <body> <div id="What"><img src="images/what.png" width="477" height="36" alt="do you know something i don't" /></div> <?php $hostname='fd.db.fdffd.d.com'; $username='d'; $password='d!'; $dbname='d'; $usertable='d'; mysql_connect('yd.com', 'yd9', 'Ad!') or die(mysql_error()); mysql_select_db("yd9") or die(mysql_error()); $id = (int)$_GET['id']; if( $id > 0 ) { $result = mysql_query("SELECT `name`, `photopath`, `state` FROM `photo` WHERE `id` = $id") or die(mysql_error()); $row = mysql_fetch_array( $result ); echo " <img src=\"/{$row['photopath']}\" height=\"400px\" width=\"400px\" />". "<br /> "; echo $row['name']. "\n". "<br/>"; echo $row['state']. "\n"; } else { echo 'BAD PHOTO ID'; } ?> <?php $hostname='durce.com'; $username='yod9'; $password='Ad1!'; $dbname='yd'; $usertable='d'; mysql_connect('yd, 'yd9', 'Ad!') or die(mysql_error()); mysql_select_db("yd") or die(mysql_error()); $name=$_POST['name']; $comment=$_POST['comment']; $submit=$_POST['submit']; if($submit) { if($name&&$comment) { $query=mysql_query("INSERT INTO comment (id,name,comment) VALUES ('','$name','$comment')"); } else { echo "Please fill out all the fields."; } } ?> <body> <div id="comments"> <form action="image_show.php" method="POST"> <label>Name: </label><br /><input type="text" name="name" input id="name"value="<?php echo "$name" ?>" /><br /><br /> <label>Comment: </label><br /><input type="text" input id="comment"/><textarea name="comment" "cols="25" rows="7"></textarea><br /><br /><br /> <input type="submit" name="submit" value="Comment" /><br /> </form></div> <hr width="1100px" size="5px" /> </body> <?php $query=mysql_query("SELECT * FROM comment ORDER BY id DESC"); while($rows=mysql_fetch_assoc($query)) { $id=$rows['id']; $name=$rows['name']; $comment=$rows['comment']; $linkdel="<a href=\"delete.php?id=" . $rows['id'] . "\">Delete User</a>"; echo '<font color="red">Name:</font> ' . $name . '<br />' . '<br />' . '<font color="red">Comments:</font> ' . '<br />' . $comment . ' ' . ' ' . ' ' . ' ' . $linkdel . '<br />' . '<br />' . '<hr size="5px" width="500px" color="blue" />' ; } ?> </html> </body> </html> Okay so im not that new with PHP but im still learning. So im creating a php website with dreamweaver CS3 and phpmyadmin, which is based on a Music Library. I have a page where you can browse by the artists name. When you click on that it shows you all the artists with that name. But its the next part im having difficulty with. Keeping this as a one page system, i want the artists albums to be displayed when you click on a particular artist. I tried some methods but it doesnt seem to work [Refer to the images from below] Image1 shows the alphabetical list..clicking on any of these takes you to Image 2 which displays all the artists. By clicking on one of these artists i want it to take me to a seperate page which shows all their albums. And by clicking on the albums it shows song names from those particular albums. These song names will then be linked to actual lyrics of the song. There are three tables i have on phpmyadmin which i want the data to be from. These are albums -album_id -artist -album_name -tracklisting -genre -release_date -price -album_image -album_description artists -artist_ID -artist -artist_info -image media -media_id -song_url -video_url -artist -album -song_title -lyrics The ones in bold are the field names id like to be included in the php script. The edition of mhpmyadmin i have doesnt enable linked tables. any help would be greatly appreciatted. Thanks. I found this code which makes a BMI Calculator (Form) for me however when I click on submit it takes the user back to the index page ie. domain.com/index.php. How do I change it to go to, say, domain.com/calculator.php ? The code is below: <? /** * @package Module Body Mass Index Calculator for Joomla! 1.5 * @version $Id: mod_bodymassindexcalculator.php 599 2010-03-20 23:26:33Z you $ **/ defined( '_JEXEC' ) or die( 'Restricted access' ); $heightcm=$_POST["heightcm"]; $weightkg=$_POST["weightkg"]; if ($heightcm!="" && $weightkg!="") { $heightm = $heightcm / 100; $bmi=round($weightkg / ($heightm*$heightm),1); echo "Heigth, m: ".$heightm."<br />"; echo "Weigth, kg: ".$weightkg."<br />"; echo "Body Mass Index (BMI): ".$bmi."<br />"; echo "<strong>"; if ($bmi<16.5) {echo "Severely Underweight</strong><br />";} if ($bmi>=16.5 && $bmi<=18.4) {echo "Underweight</strong><br />";} if ($bmi>=18.5 && $bmi<=24.9) {echo "Normal</strong><br />";} if ($bmi>=25 && $bmi<=29.9) {echo "Overweight</strong><br />";} if ($bmi>=30 && $bmi<=34.9) {echo "Obese Class I</strong><br />";} if ($bmi>=35 && $bmi<=39.9) {echo "Obese Class II</strong><br />";} if ($bmi>=40) {echo "Obese Class III</strong><br />";} echo "<br />"; } $domain = $_SERVER['HTTP_HOST']; $path = $_SERVER['SCRIPT_NAME']; $queryString = $_SERVER['QUERY_STRING']; $url = "http://" . $domain . $path; $url3 = "http://" . $domain . $_SERVER['REQUEST_URI']; $mystring1="?"; $s1=strpos($url3,$mystring1); if($s1==0) {$url2=$url3;} if($s1!=0) {$url2=substr($url3,0,$s1);} $path = $url2; //1 foot = 0.3048 meters //1 inch = 2.54 centimeters //1 pound = 0.45359237 kilograms $n1=230; echo "<table style=\"width: 100%\" cellspacing=\"0\" cellpadding=\"0\" align=\"center\"><tr><td valign=\"top\">"; //echo "<h3>BMI Calculator</h3>"; echo "<form action=\"".$path."\" method=\"post\" >"; echo "<strong>Height</strong><br />"; echo "<select name=\"heightcm\" >"; for ($i=30; $i<=$n1; $i++){ echo "<option value=\"$i\">".$i." cm / ".floor($i / 30.48)." ft ".round(($i-(floor($i / 30.48)*30.48)) / 2.54, 1)." in </option>";} echo "</select>"; echo "<br />"; echo "<strong>Weight</strong><br />"; echo "<select name=\"weightkg\" >"; for ($i=30; $i<=$n1; $i++){ echo "<option value=\"$i\">".$i." kg / ".round($i / 0.45359237,2)." pounds </option>";} echo "</select>"; echo "<br />"; //echo "<input name=\"searchterm\" type=text size=\"27\" class=\"ns1\">"; echo "<br />"; echo "<input type=\"submit\" value=\"Calculate\" name=\"B1\">"; echo "</form><br />"; //DON'T REMOVE THIS LINK - DO NOT VIOLATE GNU/GPL LICENSE!!! echo "<a href=\"http://nutritioncaloriecounter.com\">Nutrition Calorie Counter</a>"; //DON'T REMOVE THIS LINK - DO NOT VIOLATE GNU/GPL LICENSE!!! echo "</td></tr></table>"; ?> is there a way in php to link from the root dir ? like in html you just use the '/' at the start for the link " <a herf="/link.php" ></a> " but i noticed this does not work when using php like include or require. so is there anywey to tell a link to start from root dir? without using the ../../link.php I have this code : <img class= "image_frame1" src="images/groupproductimages/<?php echo $offer['cimage'];?>" /> It pulls pics I need the pics to link to urls using this code <a href="index.php?option=com_grouppurchase&view=todaysdeal&id=<?php echo $offer['cid'];?>"> I know I am missing something stupid This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=321572.0 Hi I got three tables (employers , company , and Jobs) Employer table holds info about employer. Company hold info about the company and jobs table holds info about jobs. I was just wandering what would be a good way to link these tables in the database ? Here is the coding for each table. Employer Table Code: [Select] CREATE TABLE IF NOT EXISTS `employers` ( `id` int(11) NOT NULL, `username` varchar(50) NOT NULL, `password` varchar(50) NOT NULL, `email` varchar(50) NOT NULL, PRIMARY KEY (`id`) ) ENGINE=InnoDB DEFAULT CHARSET=utf8; Jobs Table Code: [Select] CREATE TABLE IF NOT EXISTS `Jobs` ( `id` int(11) NOT NULL, `JobTitle` varchar(200) default NULL, `Company` varchar(200) default NULL, `Salary` varchar(30) default NULL, `Description` varchar(2000) default NULL, `CompanyURL` varchar(200) default NULL, `PhoneNumber` varchar(30) default NULL, `Requirements` varchar(2000) default NULL, `JobCategory` varchar(100) default NULL, `JobType` varchar(100) default NULL, `Apply_To` varchar(1000) NOT NULL, `Email` varchar(200) NOT NULL, `modified_at` datetime NOT NULL, `PostedOnDate` timestamp NOT NULL default CURRENT_TIMESTAMP on update CURRENT_TIMESTAMP, `Address` varchar(250) NOT NULL, `State` varchar(200) NOT NULL, `City` varchar(200) NOT NULL, `Country` varchar(100) NOT NULL, `Zipcode` varchar(100) NOT NULL, `JobID` varchar(100) NOT NULL, `WorkExperience` varchar(2000) NOT NULL, `EducationRequirement` varchar(2000) NOT NULL, `WebsitePostedFrom` varchar(200) NOT NULL, PRIMARY KEY (`id`) ) ENGINE=InnoDB DEFAULT CHARSET=utf8; Company Table Code: [Select] CREATE TABLE IF NOT EXISTS `Company` ( `id` int(11) unsigned NOT NULL auto_increment, `CompanyName` varchar(100) NOT NULL, `Address` varchar(100) NOT NULL, `Logo` varchar(100) NOT NULL, `PhoneNumber` varchar(25) NOT NULL, `ContactPerson` varchar(25) NOT NULL, PRIMARY KEY (`id`) ) ENGINE=InnoDB DEFAULT CHARSET=utf8 AUTO_INCREMENT=2 ; Hi All, I googled this and there is endless results. I went through a lot of them but couldn't get this working properly. How do I link from within my web site root to files outside the root? It works for me using relative links i.e. ../../phpfiles/includes but that is going to get messy and I can't get a way of doing absolute links to work. If someone could lay that out so a newbie can get it clearly I would really appreciate it! Also - I understand why I should put all of my php files outside the web root but is this a guaranteed way to secure these files other than someone hacking my ftp access? I've looked at a few site hierarchy examples - Am I right that the only pages within the web site root should be template pages with calls to required files (outside the root), session checks, and content includes and all other includes that have php executable code should be outside the root? I really appreciate the advice and insight. Thank you! Hi: I have a login file where a user goes to a db based on the dbtype selected. Now $dbtype1 links to a db on Server 1 ( that is the server on which this script is running and $dbtype2 links to a db on Server 2 I created a connection-link file connectlink.php as under but while $dbtype1 works without a problem , $dbtype2 gives me an error 'no access to db' and user dbun1@localhost not found on db1.db . What am I missing in the connectlink.php file please ? Thanks login.php =========== <? ...... if($_POST['submit']){ $dbtype = $_POST['dbtype']; if ($dbtype == 'type1') { $section = 'type1'; require("../x/type1/type1.php"); } if ($dbtype == 'type2') { $section = 'type2'; require("../x/type2/connectlink.php"); //the dir 'x' is a common name on both the servers'' } //then it processes $userfile and give this Click <a href="'.$section.'/'.$userFile.'?Userid='.$userid.'"> here ?> connectlink.php ============== <? //this file contains db info, log and checks if user is authorised to access the db - ist check. error_reporting(E_ERROR | E_PARSE | E_CORE_ERROR); $dbusername2='a'; $dbpassword2='b'; $dbname2='c'; $servername = 'ip address of server or localhost ?'; $link2=mysql_connect ("$servername","$dbusername2","$dbpassword2", true); if(!$link2){ die("Could not connect to MySQL");} mysql_select_db("$dbname2",$link2) or die ("could not open db".mysql_error()); $dbusername1='d'; $dbpassword1='e'; $dbname1='f'; $servername = 'localhost'; $link1=mysql_connect ("$servername","$dbusername1","$dbpassword1", true); if(!$link1){ die("Could not connect to MySQL");} mysql_select_db("$dbname1",$link1) or die ("could not open db".mysql_error()); $sql = "SELECT * FROM Users WHERE Userid='$userid'",$link2; $result = mysql_query($sql); if ($myrow = mysql_fetch_array($result)){ $login_success = 'Yes'; $status = "On"; .... $sql2= "insert into Log (....) values(.....)",$link2;; $result2 = mysql_query($sql2) or die ('no access to database: ' . mysql_error()); // echo mysql_error(); } } else { $failureMessage = '<p class="data"><center><font face="Verdana" size="2" color="red">Login Failure. You are not authorised to access this database .<br /></font></center></p>'; print $failureMessage; $logoutMessage = 'Click <a href="../NEWDBS/mainlogout.php"> here </a>to logout </p>'; print $logoutMessage; exit; } ?> So I have 3 tables: student, course_student and course. The student table is a student's registration info and course is the different courses the student can register for. The course_student table should contain the id fields of the other 2 which I should then use to link the tables and for example display all student's registered for current courses. Student table has an id field: sno and course table has : cid. Course_student contains both. How do I go about linking the tables so the sno id field of the student table and the cid from course updates with those in the course_table? I got two lists on a website. One list includes bunch of items and simple info. Another list includes pictures of those items and detailed information. I would like to set it up that way when users clicks on item from list #1 it would automatically jump to an item with the picture and detailed info on a second list. Hi, I have an issue with a banner I integrated in a page (http://hostelsuites.com) if you go to the page you'll see a banner on the left side , called "Combo Andes". The problem I am encountering is that this banner is linking to inside pages, and it works all fine for me but does not on some other computers, although they are using the same explorer or firefox versions.... I have no idea as to what could trigger this type of error, any help would be muche welcome. Some details that might help : The link is winthin the flash movie The code that brings the banner up depending on the chosen language is <div class="subtitulo"><?= ucfirst($lang["combomendoza"]) ?></div> Thank you very much in advance This topic has been moved to Miscellaneous. http://www.phpfreaks.com/forums/index.php?topic=326928.0 I have been trying to figure out how this is done? One php file and all that changes is the name of the brand and the logo. This has to be clickable from a menu and also if a user changes the brand name it changes to the appropriate name and logo. Any ideas as to how this is done? If you can guide me in the right direction or give me an example of how this is done would be greatly appreciated. Thanks. http://www.drivermanager.com/en/download-confirmation.php?brand=compaq&logo= <?php $con = database_connect(); $sql = "SELECT * FROM anime1, episode1 WHERE animeid='$animeid'"; $result = mysql_query($sql); while ($row = mysql_fetch_assoc($result)) { $title = $row['title']; $ep = $row['ep']; } ?> keep giving me back error Warning: mysql_fetch_assoc() expects parameter 1 to be resource, boolean given in C:\wamp\www\studying\take 2\addin12.php on line 45 In a relational database, you "link" tables together. For example, Customer (parent) ---> Product (child) How do you do that with Objects? I have a "User" class and an "AddressBook" class. A User can have zero or more Addresses (in their AddressBook). If I create a User object and an AddressBook object, I need a way to link them up similar to how they are linked up in the database. Make sense?! TomTees hi all, I am trying to link 2 mysql tables and display some information from each of them. I have a list of all the possible items for sale in table1 and I am trying to count the number of rows in the other table2 where the items exist. E.g. 'table1' manufacturer model man1 item1 man1 item2 man1 item3 'table2' id model 1 item3 2 item3 3 item2 And the result would show: item1(0) item2(1) item3(2) It would list all the items from table1 and show next to it how many rows are related to that item from table2. I have inserted a quote where I have tried many times to enter something similar to that show in the note below - but I can not get it to work - it just shows the total number of models in table1 for a given manufacturer. The php I have made so far is: <?php case 'manufacturer': $query = " SELECT * FROM table1"; $query .= " WHERE manufacturer = '".$data."' "; $query .= " ORDER BY model "; $result = mysqli_query($cxn,$query); $returnData[''] = "Select a Model..."; while($row = mysqli_fetch_assoc($result)){ // I THINK I NEED TO INSERT SOMETHING LIKE $query2 = "SELECT * FROM table2 WHERE model = table1.model"; $k=$row['model']; $k2=$row2['model']; $counter[$k]+=1; $returnData[$k]=$k; } foreach($counter as $k => $row) { $returnData[$k] .= " ($row)"; } break; ?> Hi, I would like some guidance. I have discussed with a few people and they have told me its best to store an image only in either one of these Mysql BLOB or in a folder on the website. I understand how to link the image, stored in a blob with the users SESSION but I dont understand how will I link the image stored on a folder with the user who is logged in at that time. Basically in more clear words what I am after is if the user logged in uploads an avatar and I store the avatar on a folder how will I tell php that this is the avatar for the user logged in. Another question is there a way I can create a folder with PHP for each individual user registered on my website example with their ID? |
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