PHP - Page Link

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.

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: &quot;" . $trimmed . "&quot; returned zero results</p>";

// google
 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: &quot;" . $var . "&quot; in &quot;" . $searchtype ."&quot; </p>";

// begin to show results set
echo "Results&nbsp;&nbsp;";
$count = 1 + $s ;

if ($searchtype = $title)

{
// now you can display the results returned
  while ($row= mysql_fetch_array($result)) {
  $title = $row["name"];
  
  
   echo "$count.&nbsp;$title&nbsp;&nbsp
 </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.&nbsp;$title" ;
   $count++ ;
   }
 
}
elseif ($searchtype = $subject)
{
// now you can display the results returned
while ($row= mysql_fetch_array($result)) {
  $title = $row["subject"];
 echo "$count.&nbsp;$title" ;
   $count++ ;
   }

}
elseif ($searchtype = $isbn)
{
// now you can display the results returned
while ($row= mysql_fetch_array($result)) {
  $title = $row["isbn"];
   echo "$count.&nbsp;$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-$limit);
  print "&nbsp;<a href=\"$PHP_SELF?s=$prevs&q=$var\">&lt;&lt; 
  Prev 10</a>&nbsp&nbsp;";
  }

// 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 "&nbsp;<a href=\"$PHP_SELF?s=$news&q=$var\">Next 10 &gt;&gt;</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