PHP - Seo Frendly Links

Hi,
     I need to get links from a web page and display them. User enters a url say abcxyz.com and all the links present on the page abcxyz.com will be displayed.  How can i do this?

Thanks.

Hey all, I'm currently coding a Few Blacklists for my Website, in which I want to code it so I'll have all blacklists on one page so the URL would be like:

localhost/blacklist.php?user
localhost/blacklist.php?staff

I've got a Blacklist coded which all works well, just I've attempted using the $_GET function to change the code to the blacklist that they clicked on...


<?php
session_start();

include ("includes/config.php");
include ("includes/functions.php");

logincheck();

$Username = $_SESSION['username'];

?>
<table width='50%' cellpadding='0' cellspacing='0' border='1' class='table' align='center'>
<tr>
     <td class='header' align='center' colspan='4'>Other Blacklists:</td>
    </tr>
    <tr>
     <td align='center'><a href='?user' target='mainFrame'>User Blacklist</a></td><td align='center'><a href='?money' target='mainFrame'>Most Money</a></td><td align='center'><a href='?racewins.php' target='mainFrame'>Race Wins</a></td><td align='center'><a href='?racelose' target='mainFrame'>Race Loses</a></td>
    </tr>
    <tr>
     <td align='center'><a href='?miles' target='mainFrame'>Most Miles</a></td><td align='center'><a href='?fosbl' target='mainFrame'>Faster Or Slower</a></td><td align='center'><a href='?police' target='mainFrame'>Police Chase</a></td>
    </tr>
</table>        
<br />
<table width='20%' cellpadding='0' cellspacing='0' border='1' class='table' align='center'>
     <tr>
         <td colspan='2' align='center' class='header'>User Blacklist:</td>
        </tr>
        <tr>
         <td align='center' class='omg' width='5%'>Num:</td><td align='center' class='omg' width='10%'>Username:</td>
        </tr>       

<?php
if ($_GET['user']){
$Query = mysql_query("SELECT * FROM users WHERE userlevel = '1' OR userlevel = '5' ORDER BY rep DESC LIMIT 25");
for ($Place < 25; $This = mysql_fetch_object($Query);){
$Place = $Place+1;
if ($This->username != $Username){
$Colour = "bgcolor=#424242";
}else{
$Colour = "bgcolor=#111111";
}
print ("<tr><td width='5%' align='center' $Colour>$Place</td><td width='20%' align='center' $Colour><a href='profile.php?viewuser=".$This->username."'>$This->username</a></td></tr>");
}
}


?>
    </table> 
    <br />

Soon as I added the $_GET['user]{ it stopped showing the Blacklist completely.

Sorry if it wasn't explained well, but hopefully you can catch what I'm trying to-do

Thanks.     

Hi, I have a string which contains a mix of Content and anchor links

Example:

//Example
$strContent = 'I am some <a href="/content.htm">Content</a> here, but I could have <a href="/link.php">a Link</a> <a href="http://www.example.com">Followed by another</a> Link, Mix external and not.';


I basically want to programatically pull out all of the Anchor links, and have an array of such

Array('/content.htm', '/link.php', 'http://www.example.com');


Help me please. Got a brain freeze as its Friday Afternoon

I have some links created dynamicaly with php. But i have the following problem. When i am in the root folder ieg. localhost/myPage/index.php the links work fine because they link in a file named entry.php. but when i am in a folder ieg. localhost/myPage/users/authors.php and i click the links they try to go to  localhost/myPage/users/entry.php and i want them to link not inside the folder but to the root folder localhost/myPage/entry.php.

Hi! I have an odd problem... I inserted a flash file in a php doc, and basically what happens is that on some computers, the link does not work without the www extension ... by that i mean if people go to http://mysite.com they will not be able to click the banner but if they put http://www.mysite.com it works...
Very odd as the link is part of the flash movie and it only happens on some computers not all ... anyone would have an idea as to what could cause this ?

Alternativly  what can I do with this piece of code that does not seem to do the trick either

Code: [Select]
<div class="subtitulo"><a href="http://hostelsuites.com/hostel-banner.php?Buenos+Aires+Hostel+Suites+Florida&id_hostel=30&idioma=3&id_banner=2"><?php echo ucfirst($lang["expanish"]) ?></a><br />
<br/></div>
Many thanks for any help !

I'm trying to make a link to an SQL database survive though page reloads, and it's not working. Connecting, I use this:

function link_to_converted_database()
{
 $link = mysqli_init();
 if (!$link)
  {
   die('mysqli_init failed');
  }
 if (!mysqli_options($link, MYSQLI_INIT_COMMAND, 'SET AUTOCOMMIT = 1'))
  {
   die('Setting MYSQLI_INIT_COMMAND failed');
  }
 if (!mysqli_options($link, MYSQLI_OPT_CONNECT_TIMEOUT, 50))
  {
   die('Setting MYSQLI_OPT_CONNECT_TIMEOUT failed');
  }
 if (!mysqli_real_connect($link, 'redacted', 'redacted', 'redacted'))
  {
   die('Connect Error (' . mysqli_connect_errno() . ') ' . mysqli_connect_error());
  }
 comment( ' Success... ' . mysqli_get_host_info($link) );

 if (mysqli_query($link,'USE conversion_database'))
  {
   comment("now using conversion_database.");
  }
 else
  {
   die('Conversion database unusable (' . mysqli_connect_errno() . ') ' . mysqli_connect_error());
  }
 $_SESSION['mysql_link']=$link;
 echo "<!--\n";
 var_dump($_SESSION['mysql_link']);
 echo "\n-->\n";
 return(true);
}

Subsequent SQL commands work fine using $_SESSION['mysql_link'], and the var_dump returns:

object(mysqli)#2 (17) {
  ["affected_rows"]=>
  int(0)
  ["client_info"]=>
  string(48) "mysqlnd 5.0.7-dev - 091210 - $Revision: 294543 $"
  ["client_version"]=>
  int(50007)
  ["connect_errno"]=>
  int(0)
  ["connect_error"]=>
  NULL
  ["errno"]=>
  int(0)
  ["error"]=>
  string(0) ""
  ["field_count"]=>
  int(0)
  ["host_info"]=>
  string(24) "192.168.1.101 via TCP/IP"
  ["info"]=>
  NULL
  ["insert_id"]=>
  int(0)
  ["server_info"]=>
  string(16) "5.1.48-community"
  ["server_version"]=>
  int(50148)
  ["sqlstate"]=>
  string(5) "00000"
  ["protocol_version"]=>
  int(10)
  ["thread_id"]=>
  int(194)
  ["warning_count"]=>
  int(0)
}

After a page changes, all my $_SESSION vars work fine except for  $_SESSION['mysql_link']. It's still there, but SQL queries bomb, and var_dump returns:

Warning: var_dump(): Property access is not allowed yet in C:\Program Files (x86)\Apache Software Foundation\Apache2.2\htdocs\voicereader\serverstate.php on line 595

Warning: var_dump(): Property access is not allowed yet in C:\Program Files (x86)\Apache Software Foundation\Apache2.2\htdocs\voicereader\serverstate.php on line 595

Warning: var_dump(): Couldn't fetch mysqli in C:\Program Files (x86)\Apache Software Foundation\Apache2.2\htdocs\voicereader\serverstate.php on line 595

Warning: var_dump(): Couldn't fetch mysqli in C:\Program Files (x86)\Apache Software Foundation\Apache2.2\htdocs\voicereader\serverstate.php on line 595

Warning: var_dump(): Couldn't fetch mysqli in C:\Program Files (x86)\Apache Software Foundation\Apache2.2\htdocs\voicereader\serverstate.php on line 595

Warning: var_dump(): Couldn't fetch mysqli in C:\Program Files (x86)\Apache Software Foundation\Apache2.2\htdocs\voicereader\serverstate.php on line 595

Warning: var_dump(): Couldn't fetch mysqli in C:\Program Files (x86)\Apache Software Foundation\Apache2.2\htdocs\voicereader\serverstate.php on line 595

Warning: var_dump(): Couldn't fetch mysqli in C:\Program Files (x86)\Apache Software Foundation\Apache2.2\htdocs\voicereader\serverstate.php on line 595

Warning: var_dump(): Couldn't fetch mysqli in C:\Program Files (x86)\Apache Software Foundation\Apache2.2\htdocs\voicereader\serverstate.php on line 595

Warning: var_dump(): Couldn't fetch mysqli in C:\Program Files (x86)\Apache Software Foundation\Apache2.2\htdocs\voicereader\serverstate.php on line 595

Warning: var_dump(): Couldn't fetch mysqli in C:\Program Files (x86)\Apache Software Foundation\Apache2.2\htdocs\voicereader\serverstate.php on line 595

Warning: var_dump(): Couldn't fetch mysqli in C:\Program Files (x86)\Apache Software Foundation\Apache2.2\htdocs\voicereader\serverstate.php on line 595

Warning: var_dump(): Couldn't fetch mysqli in C:\Program Files (x86)\Apache Software Foundation\Apache2.2\htdocs\voicereader\serverstate.php on line 595

Warning: var_dump(): Couldn't fetch mysqli in C:\Program Files (x86)\Apache Software Foundation\Apache2.2\htdocs\voicereader\serverstate.php on line 595
object(mysqli)#1 (17) {
  ["affected_rows"]=>
  NULL
  ["client_info"]=>
  NULL
  ["client_version"]=>
  int(50007)
  ["connect_errno"]=>
  int(0)
  ["connect_error"]=>
  NULL
  ["errno"]=>
  NULL
  ["error"]=>
  NULL
  ["field_count"]=>
  NULL
  ["host_info"]=>
  NULL
  ["info"]=>
  NULL
  ["insert_id"]=>
  NULL
  ["server_info"]=>
  NULL
  ["server_version"]=>
  NULL
  ["sqlstate"]=>
  NULL
  ["protocol_version"]=>
  NULL
  ["thread_id"]=>
  NULL
  ["warning_count"]=>
  NULL
}


There's got to be a way to save my SQL linkages so that I don't have to reconnect to the SQL database every time I want to see if new data has arrived, but this doesn't seem to be it. My PHP version is PHP 5.3.2.0.

Thanks for any help.

Hi all,

Im sorry if this is in the wrong section.


But nonetheless, I use php to display my news:
I.e - news.php?id=1 that displays all the news on that id.



What I want to know is using php how can I get the URL from news.php?id=1 too something like /news/titleofthestory ?

Hope I've given enough detail.

Thanks!!

I've been reading about trackbacks and pings but what I'm trying to do is a bit different. I'm trying to track a referral link. For example,

I post this referral link http://www.my-site.com/3ji89 on youtube or a blog..

and that referral links to..

http://www.another-site.com

in http://www.another-site.com is it possible to retrieve the referral link http://www.my-site.com/3ji89 ? Not youtube.com or the blog I posted the link on.

With my custom CMS I'm developing I have a menu structures type module where the admin can create a menu structures for their website. Inside the db is the links table and I'm trying to figure out what I can do if the link has an actual image. I created a field called link_image however I just thought of what if the link is apart of a sprite which would have most of the time 3 parts of the image (active, hover, normal) or whatever or has its own sprite. How should I handle this if I want to store this information in the database.

Hi,

I have a peice of code which publishes an image with a link from my database.  However I cant get it to use external links.  My code is:

echo "<a href=\"" .$link . "\">
<img src=\""   .$image ."\" />  </a><BR />";


I have tried all the options I can think of but I cant get it work.  Can anyone advise please?

hi guys i need a bit of help, im trying to build a page which lists all orders in the database, and next to each order put in some links. 
i could do this fine, but i've now changed certain things and each user has certain permissions.

my table is like this

_users
firstname | lastname | login | password | groupid

_usergroups
groupid | groupname

_usergroup_permissions
groupid | area | subarea

all should be self explanitory, but in the permissions table `area` refers to the page (orders, customers, system) and `subarea` refers to the actual permission (Add, Edit, Delete, Manage)

using the following script, im trying to:
1 - trying to get all the orders
2 - getting the groupid from the users table
3 - getting the permissions for the particular user's group


 $qry = mysql_query("SELECT o_id, sales_id, job_number, accounts_id, cust_id, status 
    FROM `_orders` ORDER BY o_id DESC LIMIT 50") or die(mysql_error());
    while ( $row = mysql_fetch_array($qry) ) {

$sql1 = sprintf("SELECT `group` FROM `_users` WHERE `login` = '%s'", $_SESSION['user']);

  

$qry1 =

mysql_query($sql1);

  

if(!$qry1){

die('could not query:' . mysql_error());}

  

$sql2 = sprintf("SELECT `subarea` FROM `_usergroups_permissions` WHERE `gid` = '%s' AND

 `area` =  'Order' ", mysql_result($qry1, 0));

$qry2 =

mysql_query($sql2);

if(!$qry2){

die('could not query:' . mysql_error());}


my group permissions are set to Add Edit and Delete, but there are additional privilages.

what i want to happen is for each user that logs in, next to each order are links and depending on if they have the permission, the link will be shown.

i tried using the following code in the while loop but it only showed upto the second if statement, and nothing else

$link = "<a href='#'>View</a> | ";    // show regardless of permissions (THIS WORKS

   if($racc == 0){$link .= "<a href='#'>Pick Up</a> | ";} // shown if the accounts id isn't set (THIS WORKS)

   if($qwerty == Edit ){
      $links .="<a href='#' onclick='return false'>Edit</a> | ";} //Shown if the permission level says the user can Edit (THIS WORKS)

      if($qwerty == Delete ){
      $links .="<a href='#' onclick='return false'>Delete</a> | ";} //Shown if the permission level says the user can Delete(DOESN'T WORK)
   
   
   if($qwerty == Approve ){
      $links .="<a href='#' onclick='return false'>Approve</a> | ";}  //Shown if the permission level says the user can Approve (DOESN'T WORKS)
         

   if($qwerty == Acceptance ){
      $links .="<a href='#' onclick='return false'>Acceptance</a> | ";}  //Shown if the permission level says the user can Acceptance (DOESN'T WORKS)
      
I asked about this on another forum and they said its because $qwerty cant be everything    and that i should look at using an array, unfortunately that has an undesired effect.

so to summaraize...for me the page should look like

order   -    salesman     -   customer   -  VIewLink | Addlink | Editlink | Deletelink | Approvelink

but for someone who (for example) doesn't have the ADD and Delete permissions
order   -    salesman     -   customer   -  VIewLink | Editlink | Approvelink .

if NONE of that makes sence, i can provide links and try to explain again. but what i want is different links to be shown depending one what permissions a user has


thanks for any help (sorry for waffling on a bit...its getting to that time of the day )
Jack

Hi,

I've been working on making my own FAQ system for my site with PHP & MySQL, I have the SQL displaying properly on the page but I cant seem to figure out how to make it so when the page loads just the questions show and when a question is clicked the answer shows below it. I've read a number of tutorials and when I do get it to work it will only show the answer to the first question when the link is clicked.

Here is the code I'm using to display the info from the database:

Code: [Select]
<div id="faqSQL">
   <?php foreach($result as $option) { ?>
      <div class="faq-question"><?php echo wordwrap($option['faqquestion'], 100, "\n", false);?></div>
      <div class="faq-answer"><?php echo wordwrap($option['faqanswer'], 100, "\n", false);?></div>
    <?php } ?>
</div>

If anyone can help it would be much appreciated.

Hello dear friends,

Let say we have

Code: [Select]
$to = "gigi";
then we have link like this

Code: [Select]
mysite.com/file.php?id=$1
We can change it to new form by add rule at .htaccess codes

Code: [Select]
RewriteRule -(.*)\.htm file.php?id=$1
So that the link be like this

Code: [Select]
mysite.com/$to-$id.htm

it will works , Now what if i want to make the link like this

Code: [Select]
mysite.com/$id/$to.htm
what will be the .htaccess code in this way ???

thanks

I am trying to get the output for a mysql query to be displayed in colums. I have it where it does this but it goes from left to right. I would rather it go from top to bottem, it would look better this way. I will also need it to brake into sevral pages when there are a certain amout of entrys shown. If you know of a  good how to on that let me know please.  The other problom that I am having is to make the output a link. I was able to do this before I attemped to devide everything but if I add in what I had before it comes back with no data.

Here's what I have now that put shows the data from left to right in 2 colums.

Code: [Select]
<?php
include"scripts/connect.php" ;

mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");
$query="SELECT * FROM movies WHERE type LIKE 'movie' ORDER BY title";

if(mysql_query($query))
{
$result=mysql_query($query);

$num=mysql_numrows($result);

}


for($i=0; $i<$num; $i++)
{
if($i%2==0){
echo "<tr><td>";
} 
else{
echo "<td>";
}

print mysql_result($result,$i,"title");

if ($i%2==0){
echo "</td>";
}
else{
echo "</td>";
}
}
echo "</tr>";
?>
Thanks in advance