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PHP - <a Href> Link To Fill Page With Different Data
Hi people,
I've been having trouble with this for a while now. Basically, i have this website: http://el07jpg.info . You can log in with username:doctor1 password:password1 Once you've logged on, u can click on "Patients". Then, if you click on a patient, it shows their info. Then at the bottom, I have links to glucose readings and heart rate readings... but when I click on either of those links, it goes back to the main page, showing all the patients.. The weird thing is that I got this to work for the first step, u click on the <a href> link of a patient, and it brings you to their page. Then I've done the EXACT same thing to link to the glucose or heart rate readings of that specific patient, but it doesn't work. I really can't see what i've done wrong. Although I noticed in the address bar, when I click on glucose readings for example, the url is http://el07jpg.info/index.php?op=patients&id=&reading=glucose (id should be equal to a number, like id=4, but there's no number... Any help would be much appreciated! Cheers! Code for op_patients.php page: <?php //Page if a set of records is selected from a single patient's page if ($_REQUEST[reading]+0) { $GLOBALS['html']['css'][]="default.css"; $reading_source=$_REQUEST[reading]+0; $sql="select * from {$reading_source}_readings where patientRFID='{$patient[patientRFID]}' "; $results=mysql_select_assoc($sql); if (!$results){ $GLOBALS[err][]="No records exist!"; exit; } $GLOBALS['html']['title']="Patient: {$patient[first_name]} {$patient[last_name]}'s {$reading_source} readings "; echo $reading_source; } //Page if a single patient is selected if ($_REQUEST[id]+0) { $GLOBALS['html']['css'][]="default.css"; $id=$_REQUEST[id]+0; $sql="select * from patients where doctorID='{$_SESSION[user][id]}' and patientID='$id' "; $patient=mysql_select_assoc($sql); if (!$patient){ $GLOBALS[err][]="The patient can't be found!"; exit; } $GLOBALS['html']['title']="Patient: {$patient[last_name]}, {$patient[first_name]} "; $output.=" <br> <table border='1' cellpadding='5' cellspacing='5' width='600'> <tr><td rowspan=6><img src=\"$patient[photo]\"> </td> <th>Name: </th><td>{$patient[first_name]} {$patient[last_name]} </td></tr> <tr><th>D.O.B: </th><td>{$patient[dob]} </td></tr> <tr><th>Email: </th><td>{$patient[email]} </td></tr> <tr><th>Phone Number: </th><td>{$patient[phone_no]}</td></tr> <tr><th>Address: </th><td>{$patient[address]}</td></tr> <tr><th><br> Comments:<br> </th> <td>{$patient[comments]} </td></tr> </table> <table border='1' cellpadding='5' cellspacing='5' width='600'> <tr><th colspan=3><br>Medical Readings<br><br></th></tr> <tr> <td align='center' width='300'> <a href='index.php?op=patients&id={$v[patientID]}&reading=glucose'>Glucose Readings</a> </td> <td colspan=2 align='center' width='300'> <a href='index.php?op=patients&id={$v[patientID]}&reading=heart'>Heart Rate Readings</a> </td> </tr> </table> "; echo $output; exit; } //Main Page showing all patients $GLOBALS['html']['title']="Patients"; $GLOBALS['html']['css'][]="default.css"; if(!$_SESSION[user][id]) { $er="Access Denied! Please Log in."; echo "<div style='color:red;'>$er</div>"; } else { $sql="select * from patients where doctorID='{$_SESSION[user][id]}' order by last_name"; $d=mysql_getarray_assoc($sql); $output=''; foreach ($d as $v){ $output.=" <tr><td> <a href='index.php?op=patients&id={$v[patientID]}'>{$v[last_name]}, {$v[first_name]}</a> </td></tr> "; } echo " <table width=80% align=left border='1' cellpadding='5' cellspacing='5' width='100%'> <tr><th class=popupheader align='left'> Patient Name<br><br> </th></tr> $output </table> "; } ?> Similar TutorialsHi all, I find myself in need again on PHP coding and the support last time was outstanding (no creeping here) so I am trying my luck again! I am trying to use a HREF link (lots of individual ones) to search a Mysql database and return the results in a PHP page. Example of HREF links on the page are he https://flighteducation.co.uk/Create.html and in theory you click a career and it 'gets' and 'shows' the data from the database he https://flighteducation.co.uk/Careers.php so, you click the 'Animator' link ->it goes to my database and drags back all the data to the Careers.php page
So far this is what I have managed to not get right: HTML (screen shot added/attached Links.png)
<a href="Careers Results.php?jobTitle=Animator"> PHP
$id = $_GET['jobTitle'];
if( (int)$id == $id && (int)$id > 0 ) {
$link = mysqli_connect('localhost','MYUSERNAME','MYPASSWORD','MYDATABASE'); // Connect to Database
if (!$link) {
die('Could not connect: ' . mysqli_connect_error());
}
$sql='SELECT * FROM careers WHERE jobTitle=' .$id;
$result = mysqli_query($link,$sql);
$row = mysqli_fetch_array($result);
echo $row['jobTitle'];
echo $row['jobDescription'];
}
else {
echo "Record NOT FOUND";
}
Up to now the code returns "Record NOT FOUND" so it is passing through the php to the end. I am new to PHP and trying to get my head around it all, and in theory this is the kind php code that I should be looking at, but I am still very new to it!! Any help or advice very much appreciated again!!!
Thanks
Hi Support, I have a form, where it collects user description input. I can collect the inputs and store it with newline. The issue is - how to collect the http link to actual hyperlink ref during display. The following is my code: <textarea name="description" cols="50" rows="10" id="description"><?php echo str_replace("<br>", "\n", $description);?></textarea></td> For example, User input: Hi, Check it out - http://www.google.com/ I would like to display google link as href so that Viewers can click the link and go to the page. Right now, it is not href and user need to copy the link to new tabs or pages and then it can come. Thanks for your help. Regards, Ahsan Is there anyway of saving data to phpmyadmin, and linking it to another page with one button. I can do it separately but I can't seem to do it using one button. Does anyone has any ideas! Im fairly new to PHP and have a quick question. I am creating a mail system from scratch and I have two main SQL rows, ID and read. I need to put a link around the ID so that it only pops up in a pop up window that message. This is the coding I have so far, feel free to point out where I have gone wrong: <? include('../config/config.php'); $Query = sprintf("SELECT * FROM qa LEFT JOIN EmployeeInfo ON(qa.Agentsname = EmployeeInfo.Name) WHERE EmployeeInfo.Name LIKE '%s'", mysql_real_escape_string($_SESSION['SESS_NAME'])); $result = mysql_query($Query); $num_rows = mysql_num_rows($result); echo "<table border='0' cellspacing='10' cellpadding='10'>"; echo "<tr>"; echo "<td><strong>Message ID</strong></td><td><strong>Team Lead</strong></td><td><strong>Unread/Read</strong></td><td><strong>Score</strong></td>"; while($row = mysql_fetch_array($result)) { echo "<tr><td><a href='http://10.0.11.5/form/employees/attendance/qaview.php?id='". $row['ID'] . ">" . $row['ID'] . "</a></td><td>" . $row['Name'] . "</td><td>" . $row['read'] . "</td><td>" . $row['Score'] . "</td></tr>"; } echo "</table>"; ?> and this is the qaview.php page <? include('../config/config.php'); $query = doquery("SELECT * FROM qa WHERE ID='$ID'"); $update_query=doquery("UPDATE qa SET read='1' WHERE ID='$ID'"); $row = mysql_fetch_array($query); echo "<td><strong>" . $row['Name'] . "</strong></td></tr>"; ?> Hi, I have a directory page the displays all names beginning with B across 3 columns, that part works fine but when I try to add a link on each of the results the link doesn't work. It's not picking up the $id or $name variables for some reason, can anyone tell me why this is? Any help would be very much appreciated. Code: [Select] <?php $data = mysql_query("SELECT id, name FROM new WHERE name LIKE 'B%' ORDER by name ASC LIMIT 30"); $prev_row =''; echo '<table width="690px">'; while($info = mysql_fetch_array( $data )) { $letter = strtoupper(substr($info['name'],0,1)); if ($letter != $prev_row) { if($count % 3) { for ($i = ($count % 3); $i < 3; $i++) echo '<td> </td>'; echo '</tr>'; $count =0; } $prev_row = $letter; echo '<tr><td class="directory-letter" colspan="3">',$letter,'</td></tr>'; } $id = $row['id']; $name = $row['name']; if ($count % 3 == 0) echo '<ul><tr>'; $count++; echo '<td width="230px"><li class="directory-store-name"><a href="directory/name.php?id=$id" title="$name" />', $info['name'],"</li></td>\n"; if ($count % 3 == 0) echo '</tr></ul>'; } if($count % 3) { for ($i = ($count % 3); $i < 3; $i++) echo '<td> </td>'; echo '</tr>'; } echo '</table>'; ?> 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 Im rebuilding my website and have decided to make the dynamic images for my main photography gallery be layed out using div tags and css. I can get the images to load in and layout correctly but for some reason the links on the images only works on the last loaded in image. I check the code on the page and all the links are there but only the last one is active. The code worked when it was all wrapped in table tags. Is this a PHP or css issue stopping all the links expect for the last to be active. This is the css code I am using. #maincontent{width:1000px;margin:0 auto;padding:0 20px} #maingallery{height:382px;position:relative;width:1000px;text-align:center;margin-top:15px} #maingallery #text{position:absolute;top:50px} .gallery_item {float:left; margin: 20px 0 0 0; width: 200px; height: 67; padding: 0 10px 0 10px} .gallery_item .p {padding: 10px; font-size:12px} .gallery_item a:hover, a:active { display : block; border : none; } Any help would be greatly appreciated. Not sure if there would be any thing conflicting the links being active in either php or css. Thanks Matt Hello Everyone, I'm using PHP with MySQL. One of the fields in the database is website. The code I'm using is as follows: variable: $web=mysql_result($result,$i,"web"); html/php code: <td><font face="Arial, Helvetica, sans-serif"><a href="<? echo "$web"; ?>">Website</a></font></td> What it should do is have the clickable word Website with the correct url from the database. Instead it's repeating the website the page is on. For example it's doing this: http://mywebsite/www.enteredwebsite.com Any help is appreciated. After several attempts, I now have data displayed on my index.php page, but the data is from all rows. Luckily I have two rows. The page also has a menu - with two links. What I would like help with is: How to display index.php with just the data for it - i.e. home page data. How to display data if either the ‘home’ or ‘copyright’ links are clicked. I understand you can use $_GET[‘id’], and isset(), but I don’t know how to do that. I include the full html page code:
<?php
// database connection
require_once('admin/databasecon.php');
?>
<!DOCTYPE html>
<html>
<?php include 'includes/headsection.php'; ?>
<body>
<?php
// displaying data
$table = "pages"; // table
$sqli = "SELECT * FROM $table ORDER BY id ASC";
$result = $conn->query($sqli);
?>
<!-- topMenu -->
<table id="topMenu">
<tr>
<td><h1 class="siteName">Scarab Beetle</h1></td>
<?php
if ($result = mysqli_query($conn, $sqli)) {
while($row = $result->fetch_assoc()) {
echo "<td class='navItem'>" . "<a href=index.php>" . $row["menuheader"] . "</a>" . "</td>";
}
}
?>
</tr>
</table>
<!-- topMenu end -->
<!-- timeline menu -->
Menu goes here
<!-- timeline menu end -->
<!-- page title -->
<?php
if ($result = mysqli_query($conn, $sqli)) {
while($row = $result->fetch_assoc()) {
echo "<h1 class='centered'>" . $row["pageheader"] . "</h1>";
}
}
?>
<hr>
<!-- page content -->
<?php
if ($result = mysqli_query($conn, $sqli)) {
while($row = $result->fetch_assoc()) {
echo $row["pagetext"];
}
}
?>
<hr>
<div class="clear"></div>
<!-- footer content -->
<?php include 'includes/footersection.php'; ?>
<!-- footer content end -->
</body>
</html>
Any help to achieve what I want will be appreciated. Thank you. I have a some value that are being generated from a database then thrown into <li><href> to create a list that user can click and fetch data through ajax right now its in a form select/menu and works fine however I need to convert to a list and use and onKeyDown event Code: [Select] <form> <select name="users" size="<?php echo $num_rows;?>" onchange="showUser(this.value)" > <?php do { ?> <option value="<?php echo $row_Recordset1['item_id']?>"><?php echo $row_Recordset1['item_id'].' '. $row_Recordset1['item_name']?></option> <?php } while ($row_Recordset1 = mysql_fetch_assoc($Recordset1)); $rows = mysql_num_rows($Recordset1); if($rows > 0) { mysql_data_seek($Recordset1, 0); $row_Recordset1 = mysql_fetch_assoc($Recordset1); } ?> </select> </form> I need to correct this Code: [Select] <ol> <?php do { ?> <li onKeyDown="showUser(this.value)"><a href="getmenu.php?item_id="<?php echo $row_Recordset1['item_id']?>"> <?php echo $row_Recordset1['item_name']?></a></li> <?php } while ($row_Recordset1 = mysql_fetch_assoc($Recordset1)); $rows = mysql_num_rows($Recordset1); if($rows > 0) { mysql_data_seek($Recordset1, 0); $row_Recordset1 = mysql_fetch_assoc($Recordset1); } ?> </ol> Hi guys; I've managed to find my way into something of a maze. <td> <a href="'Details/21/index.php'"?id= . $id .'>" . $row['id'] . "</a> </td> I'm looking at this line in the code. I realise it's currently in the wrong syntax but I'm just trying multiple different variations. At the moment I'm actually just trying to make it go to a static link but my real goal is to - get it to pick up the id number, of the id row I click, and concatenate that with the rest of my address string. Then use that as the href. Something like this: - "'Details/' + $id + '/index.php'" It's really confusing me though inside this loop and for some reason the id number is being picked up as an int, by the looks of things. It's getting above my level of understanding. Any chance one of you masters would through a n00b a lifeline?
This is my code below
<!DOCTYPE html>
<html>
<head>
<title>LifeSaver DB</title>
<h1> LifeSaver Database </h1>
</head>
<body>
<table>
<tr>
<th>Id</th>
<th>Location</th>
<th>Initials</th>
<th>TimeStamp</th>
<th>Notes</th>
</tr>
<?php
$conn = mysqli_connect("localhost", "meh", "pas", "DB");
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM LifeSaver1 ORDER BY id DESC";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
//for href row
$id = $row['id'];
$Footage = ['Footage'];
echo
"<tr>
<td> <a href="'Details/21/index.php'"?id= . $id .'>" . $row['id'] . "</a> </td>
<td>" . $row["Location"] . "</td>
<td>" . $row["Initials"]. "</td>
<td>" . $row["TimeStamp"]. "</td>
<td>" . $row["Notes"] . "</td>
</tr>";}
//show table
echo "</table>";
} else { echo "0 results"; }
$conn->close();
?>
</table>
</body>
<style>
table, td, th {
border: 1px solid black;
margin: auto;
}
table {
border-collapse: collapse;
color: #000; <!--font colour -->
font-family: monospace;
font-size: 18px;
text-align: center;}
th {
background-color: #337AFF;
color: white;
font-weight: bold; }
tr:nth-child(odd) {background-color: #add8e6}
</style>
</html>
Edited February 8, 2020 by JonnyDriller Hello.
I have a bit of a problem. When I fetch the link field from the database.i don't see an actual link on the page.
One more thing, what type of field should I use to store the link in the database? Probably there is where I went wrong.
All help is
Hi, Is it possible to build a PHP Template page that selects and publishes a row of data from a MySQL Database when a linked is clicked? I would design: Template.php Text links (perhaps on homepage of navigation bar): ProductA, ProductB and ProductC If you click link ProductA Template.php would display data for ProductA and likewise for ProductB and ProductC. I would also like search engines for find ProductA, ProductB and ProductC PHP pages. (Not just my single Template.php) Any ideas as to how this could be done without designing individual PHP pages for each product? Kind regards, Matthew. I have a page that reads from a database to list employees from different states. Under each name is a link that pulls up a new window with a contact form. My goal is to get the name from that link to appear on that new window. (Which I will then use to ultimately add the name into the email that was generated so the recipient can know who the email was intended for.) I've tried using the GET and POST methods but nothing seems to work. Any help would be appreciated greatly. I will name my firstborn after anything you desire. I've included my code below. (The last line is what I've been primarily working with) Code: [Select] //Open Database Connection $db = open_db_connection(); $sql = "SELECT DISTINCT 1 from users where active = 'Y' and show_on_contactus = 'Y'"; $results = mysql_query($sql, $db); $emaillink = "<a target='main' onclick=window.open('contactform/contactCentral.php?name=$state_row[state]','popup','width=380,height=400,scrollbars=no,resizable=no,toolbar=yes,directories=no,location=no,menubar=yes,status=no,left=0,top=0'); return false>"; if(!mysql_num_rows($results) == FALSE) while($row = mysql_fetch_array($results, MYSQL_BOTH)) { echo "<ul id=\"directors\">\n"; $state_sql = "SELECT * from users where active = 'Y' and show_on_contactus = 'Y' and state_full = 'indiana' order by last_name"; $state_results = mysql_query($state_sql, $db); if(!mysql_num_rows($results) == FALSE) { while($state_row = mysql_fetch_array($state_results, MYSQL_BOTH)) { echo "<li>\n"; echo " <div class=\"fl dirimg\">\n"; if($state_row[photo] == "") { echo "<img border=\"0\" width=\"61\" height=\"85\" src=\"images/exe_placeholder.jpg\" alt=\"$state_row[first_name] $state_row[last_name]\" />\n"; } else { echo "<img border=\"0\" width=\"61\" height=\"85\" src=\"n2team/pictures/$state_row[photo]\" alt=\"$state_row[firstname] $state_row[last_name]\" />\n"; } echo " </div>\n"; echo " <span class=\"fr\">$state_row[first_name] $state_row[last_name]<br />\n"; echo " $state_row[title]<br />\n"; echo " $state_row[city], $state_row[state]<br />\n"; echo " $emaillink Contact $state_row[first_name] </a></span></li>\n"; } ok so now all my logo link goes to the same page..i dont know how to fix it.i want all of them connect to a link(each of the company has their own company_link id) which i named company_link which is already in the database..so i need to call it.there are about 20 total logos.. except for the $company id=33..just remain that one.. Code: [Select] <? if(isset($_GET[ttid])) { $ttid = $_GET[ttid]; } $connection=mysql_connect("$server", "$username", "$password") or die("Could not establish connection"); mysql_select_db($database_name, $connection) or die ("Could not select database"); $query = "select master_event.* , (DATE_FORMAT(event_datefrom, '%d %M %Y')) as datefrom, (DATE_FORMAT(event_dateto, '%d %M %Y')) as dateto, ucase(event_name) as eventname from master_event where master_event.event_id = '$ttid '"; $result=mysql_query($query); while($row = mysql_fetch_array($result)){ $eventname = $row['eventname']; $eventdesc = $row['event_desc']; $companydescription = $row['company_description']; $eventvenue = $row['event_venue']; $eventfee = $row['event_fee']; $datefrom = $row['datefrom']; $dateto = $row['dateto']; echo "<font color='#000000' face='arial' ><b> $eventname </b> </font> <br>"; echo "<font color='#000000' face='arial' ><i>Date</i> : $datefrom - $dateto <br>"; echo "<i>Venue</i> : $eventvenue <br>"; $querypstype = "SELECT DISTINCT master_pstype.pstype_id, pstype_desc FROM master_pstype INNER JOIN master_psevent ON master_psevent.pstype_id= master_pstype.pstype_id where master_psevent.event_id = '$ttid' ORDER BY pstype_id"; $resultpstype=mysql_query($querypstype); while($rowpstype = mysql_fetch_array($resultpstype)){ $pstypeid = $rowpstype['pstype_id']; $pstypedesc = $rowpstype['pstype_desc']; echo "<br><font color='#8B3A3A' size='1' face='georgia' ><b><i>$pstypedesc</i></b></font> <br>"; $queryps = "SELECT master_psevent.*, pstype_desc, company_name, company_link, company_description, logo_filename FROM master_psevent LEFT JOIN master_pstype ON master_psevent.pstype_id = master_pstype.pstype_id LEFT JOIN master_ps ON master_ps.ps_id = master_psevent.ps_id WHERE master_psevent.event_id = '$ttid' and master_pstype.pstype_id= '$pstypeid' ORDER BY pstype_desc,company_name "; $resultps=mysql_query($queryps); while($rowps = mysql_fetch_array($resultps)){ $companyname = $rowps['company_name']; $companyid = $rowps['ps_id']; $companylinkori = $rowps['company_link']; $companydescription = $rowps['company_description']; if ($companyid==33){ $companylink = "http://(confidential)/index.php?view=describe"; } else { if ($companylinkori <> "") { //$companylink = "http://".$rowps['company_link']; $companylink = "http://(confidential)/index.php?view=".$companyid; } else { $companylink = ""; } } $logopath = $rowps['logo_filename']; if ($companylink <> ""){ echo "<font color='#4A708B' face='arial' ><a href=$companylink style=\"color:white;text-decoration: none; \" target='new'><img src=".$logo_dir."/".$logopath." width=\"15%\"></img></a>  </font>"; } else { echo "<font color='#4A708B' face='arial' ><img src=".$logo_dir."/".$logopath." width=\"15%\"></img>  </font>"; } ?> <br> <? } echo "<br>"; } } ?> Can someone help me debug this code...this is a search page where it searches the keyword typed in the mysql db and it limits 10 results per page...and at the bottom if there are more than 10 results it says next 10 and its a link but when it is clicked it goes to the code and says please type a search, here is the code can someone please help me with this I would really appreciate it. Code: [Select] <?php $page = $PHP_SELF; // Get the search variable from URL $var = @$_GET['search'] ; $searchtype = @$_GET['searchtype']; $isbn = "ISBN"; $school = "School"; $title = "Title"; $subject = "Subject"; $trimmed = trim($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 $connect = mysql_connect("","]",]") or die("Not connected"); //specify database mysql_select_db("collegebooxboox") or die("could not log in"); if ($searchtype == ($title)) { // Build SQL Query $query = "select * from boox where name like \"%$trimmed%\" order by name"; // EDIT HERE and specify your table and field names for the SQL query } elseif ($searchtype == ($school)) { // Build SQL Query $query = "select * from boox where school like \"%$trimmed%\" order by school"; // EDIT HERE and specify your table and field names for the SQL query } elseif ($searchtype == ($isbn)) { // Build SQL Query $query = "select * from boox where isbn like \"%$trimmed%\" order by isbn"; // EDIT HERE and specify your table and field names for the SQL query echo "isbn"; } elseif ($searchtype == ($subject)) { // Build SQL Query $query = "select * from boox where subject like \"%$trimmed%\" order by subject"; // EDIT HERE and specify your table and field names for the SQL query } $numresults=mysql_query($query); $numrows=mysql_num_rows($numresults); // 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 . "" in "" . $searchtype ."" </p>"; // begin to show results set echo "Results "; $count = 1 + $s ; if ($searchtype = $title) { // now you can display the results returned while ($row= mysql_fetch_array($result)) { $title = $row["name"]; echo "$count. $title   </br><table width='297' border='1' align='center'> <tr> <td width='152'>Book Title:</td> <td width='129'>$row[name]</td> </tr> <tr> <td>Author:</td> <td>$row[author]</td> </tr> <tr> <td>ISBN#</td> <td>$row[isbn]</td> </tr> <tr> <td>Date Posted:</td> <td>$row[date]</td> </tr> <tr> <td>Posted By:</td> <td><a href='backpack.php' onclick='document['packback'].submit()'>$row[username]</a></td> </tr> <tr> <td>School:</td> <td>$row[school]</td> </tr> </table></br>"; $count++ ; } } elseif ($searchtype = $school) { // now you can display the results returned while ($row= mysql_fetch_array($result)) { $title = $row["school"]; echo "$count. $title" ; $count++ ; } } elseif ($searchtype = $subject) { // now you can display the results returned while ($row= mysql_fetch_array($result)) { $title = $row["subject"]; echo "$count. $title" ; $count++ ; } } elseif ($searchtype = $isbn) { // now you can display the results returned while ($row= mysql_fetch_array($result)) { $title = $row["isbn"]; echo "$count. $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>"; ?> Hello. Say i want to view another user profile. I click on user name and i see the profile page. So how can i link user name to his profile? I have tried linking it to $id (this is where user id is stored in the database), but no luck. Thank you. Hello all, I am working on a webpage which needs to be finished Friday so I am a bit stressing out, hence this post. I have made the following script: // Uitkomst van POST waardes echo"<pre>"; print_r($_POST); echo"</pre>"; // LET OP: De tabel "test_optie" is niet compleet gevuld en daarom werkt het script alleen als je start bij categorie 1. // Het is dus GEEN fout in de script, alleen de database is niet helemaal gevuld kostte me teveel tijd! echo" <html> <head> <style> select {float:left;width:250px;margin:0 5px 0 0;} </style> </head> <body> <form name='form' action='chainselect.php' method='post' /> <select name='veld1' onChange='document.form.submit()'> "; // Data opvragen $sql_categorie = mysql_query("SELECT * FROM plaatsnaam ORDER BY id") or die (mysql_error()); // Counter $a = 0; // Standaard geselecteerd echo"<option value='0' selected>Selecteer een plaats</option>"; while($row_categorie = mysql_fetch_array($sql_categorie)){ $a++; echo"<option value='$row_categorie[id]' ";if($_POST[veld1] == "$a"){echo"selected";}echo">$row_categorie[naam]</option>"; } echo" </select> <select name='veld2' onChange='document.form.submit()'> "; // Data opvragen $sql_rubriek = mysql_query("SELECT * FROM branche WHERE catid = '$_POST[veld1]'") or die (mysql_error()); $ant_rubriek = mysql_num_rows($sql_rubriek); // Huidige id veld2 om zodoende huidig veld te selecteren indien gewijzigd if(isset($_POST[veld2])){ $b = "$_POST[veld2]"; } // Als er geen data gevonden is dan de onderstaande option laten tonen. if($ant_rubriek <= 0){ echo"<option value='0'>-</option>"; } else{ // Standaard geselecteerd echo"<option value='0' selected>Selecteer een branche</option>"; while($row_rubriek = mysql_fetch_array($sql_rubriek)){ echo"<option value='$row_rubriek[id]' ";if($_POST[veld2] == "$row_rubriek[id]"){echo"selected";}echo">$row_rubriek[naam]</option>"; } } echo" </select> <select name='veld3' onChange='document.form.submit()'> "; // Data opvragen $sql_optie = mysql_query("SELECT * FROM filiaal WHERE catid = '$_POST[veld1]' AND rubid = '$_POST[veld2]'") or die (mysql_error()); $ant_optie = mysql_num_rows($sql_optie); // Huidige id veld3 om zodoende huidig veld te selecteren indien gewijzigd if(isset($_POST[veld3])){ $c = "$_POST[veld3]"; } // Als er geen data gevonden is dan de onderstaande option laten tonen. if($ant_optie <= 0){ echo"<option value='0'>-</option>"; } else{ // Standaard geselecteerd echo"<option value='0' selected>Selecteer een filiaal</option>"; while($row_optie = mysql_fetch_array($sql_optie)){ echo"<option value='$row_optie[id]' ";if($_POST[veld3] == "$row_optie[id]"){echo"selected";}echo">$row_optie[naam]</option>"; } } echo" </select> </form> </body> </html> "; ?> What I need now is the results shown linked to a page. U can see the dropdowns in action he www.inventar.nl/chainselect.php For example: In the first dropdown I select Groningen, in the 2nd drop I select Groningen, Restaurant and in the 3rd Groningen, Restaurant, Ni Hao Wok. This gives me the following: Quote Array ( [veld1] => 1 [veld2] => 2 [veld3] => 3 ) Now I have a page named nihao.php, this page needs to be linked to this array result. So if I do the above dropdowns it needs to open nihao.php Can anyone help me with this problem? I am really confused! Thanks in advance! Wheelie |
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