can a variable hold a link?

[conan318]

 

trying to post a link to the database

$amessage ="<a href="link.php">this is a link </a>";

 

then return that link later on another page.

[conan318]

$amessage ='<a href="link.php">this is a link </a>';

solved my own problem lol

[conan318]

$amessage ='<a href='freindacept?username=.$info['username']}.php'>Has sent you a freind request </a>';

 

same thing trying to post link to the database then return it later having trouble getting the syntax right.

 

Parse error: syntax error, unexpected T_STRING in /home/a7817347/public_html/freinds.php on line 6

[btherl]

You're using quotes to mark the start and end of the string, but you're also using quotes inside the string.  And the syntax for interpolating the variable is a bit off.  Try this:

 

$amessage ='<a href="freindacept?username={$info['username']}.php">Has sent you a freind request </a>';

 

BTW it is "friend", not "freind".  Also you might be wanting this:

 

$amessage ='<a href="freindacept.php?username={$info['username']}">Has sent you a freind request </a>';

[conan318]

here's the full code i have pretty much if someone clicks on a users profile and presses contact request link i wanting to send the link to confirm friends to alog which then shows up in the other users activity log if they press the link friends are confirmed then it takes both there users names and records them into the friends database.

 

still getting Parse error: syntax error, unexpected T_STRING

 

session_start();
require "database.php";
$myusername=$_SESSION['myusername'];
$getuname=$_POST['to_user'];
$amessage ='<a href="freindacept.php?username={$info['username']}">Has sent you a friend request </a>';
$message=$_POST['message'];




if (!isset($_GET['username'])) {
       $data = mysql_query("SELECT * FROM users LEFT JOIN status  ON status.username=users.username WHERE status.username = '".$_SESSION['myusername']."' ORDER BY id DESC LIMIT 1;") or die(mysql_error());
}  else  {
       $getuname = mysql_real_escape_string($_GET['username']);
       $data = mysql_query("SELECT * FROM users LEFT JOIN status  ON status.username=users.username WHERE status.username = '".$getuname."' ORDER BY id DESC LIMIT 1;") or die(mysql_error());



mysql_query("INSERT INTO alog (from_user, to_user, message ) VALUES ('$myusername', '$getuname','$amessage')") OR die("A log error <br>".mysql_error()); 
}



Echo 'freind request sent';

?>

still getting Parse error: syntax error, unexpected T_STRING

[Pikachu2000]

$amessage ="<a href=\"freindacept.php?username={$info['username']}\">Has sent you a friend request </a>\"";

[btherl]

Oops, thankyou for correcting that pikachu

[kenrbnsn]

Or

<?php
$amessage ="<a href='freindacept.php?username={$info['username']}'>Has sent you a friend request </a>'";
?>

 

Ken

[conan318]

it kinda working but when i click the link it takes me to http://datenight.netne.net/freindacept.php?username=

so its not passing the varaible

 

$data = mysql_query("SELECT * FROM alog, users WHERE from_user=users.username AND  alog.to_user='$myusername'  ORDER BY alog.id DESC LIMIT 10;") or die(mysql_error());
//Puts it into an array 
while($info = mysql_fetch_array( $data )) 
{

?>


<div class="chatroom">
  <?


Echo "<a href='viewprofile.php?username={$info['username']}'><img src='http://datenight.netne.net/images/".$info['img'] ."' width='30' height='30''></a>".$info['from_user']." ".$info['message'];


  }
?>
[code]

[kenrbnsn]

Is $info['username'] set to anything?

 

Put

<?php
echo '<pre>' . print_r($info,true) . '</pre>';
?>

at the start of your while loop. This will tell you what's being returned by the query.

 

Ken

[conan318]

it should because its working for the img link

[kenrbnsn]

The img link uses $info['img'] which is different than $info['username']

 

Just because it works for one, doesn't mean it works for the other.

 

Ken

[conan318]

the img link is using $info[username] and the location of the img is using $info.

 

 

[btherl]

Then it's likely that $info['img'] is set, but $info['username'] is not.  You need to find out why $info['username'] isn't set.