Assign Hyperlinks to the Search result

[agarwalsrushti]

Hi,
Ive made a search page. It displays the name of the users which is the result of search. Now i want that if we click on the name it should open another page in which the profile of the user will be displayed. The code that i ve done till now, only works for the last user in the list of users displayed and it also doesnt opens in another window. I think the name of the users should be passed as an array to the account page where profile is displayed. But i was not able to do it. Plz help. Ive given the relevant code below plz let me know what is the problem.

search.php
[code=php]
if($resultnum>0 ) { // Echos out matches if anything was found
while($info=mys ql_fetch_array( $data)){
//Print it out to page
?><a href="searchacc ount.php">
<?php
$_SESSION['username'] = $info['UserName'];
echo "".$info['UserName'] . " <BR><BR>" ?></a>
<?php
}// end of while
}//end of if
else{
echo "There are no results to be displayed for your search";
}
[/code]

searchaccount.p hp
[code=php]
$uname = $_SESSION['username'];
$result1= " SELECT UserName, EmailAddress, FullName FROM usermaster WHERE UserName= '$uname'";
$data1 = mysql_query($re sult1) or die(mysql_error ());
[/code]

[Markus]

You should assign the username as a parameter to the url and loop through all user names.
[php]
$_select = "SELECT * FROM `tbl_name`";

while($_row = mysql_fetch_arr ay($_select))
{
echo $_row['username'] . " <a href=\"profile. php?id=$_row['username']\">profile</a><br />";
}
[/php]
Hope this helps.

[agarwalsrushti]

Thanks.
I tried uaing the above code but its giving error.

[code=php]
while($info=mys ql_fetch_array( $data)){
//Print it out to page
echo "".$info['UserName'] ."<a href=\"searchac count.php?id=$i nfo['UserName']\">profile</a><br/>";
}
[/code]
When i used this code it gave me the following error:
Parse error: syntax error, unexpected T_ENCAPSED_AND_ WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in C:\xampp\htdocs \skill_search\s earch.php on line 181

Then i removed the inverted commas at the start of echo.
[code=php]
while($info=mys ql_fetch_array( $data)){
//Print it out to page
echo $info['UserName'] ."<a href=\"searchac count.php?id=$i nfo['UserName']\">profile</a><br/>";
}
[/code]
When i used this code it gave me the following error:
Parse error: syntax error, unexpected '.' in C:\xampp\htdocs \skill_search\s earch.php on line 181

Plz let me know how to fix this problem.

[Markus]

Try
[code=php]
while($info=mys ql_fetch_array( $data)){
//Print it out to page
echo "$info['UserName'] <a href=\"searchac count.php?id=$i nfo['UserName']\">profile</a><br/>";
}
[/code]
When you use an echo with double quotes, variables will be parsed inside it automatically.

i.e.
[php]
echo "this will echo a $_variable";
echo 'this wont echo a $_variable';
[/php]

[agarwalsrushti]

I tried this now but its still giving me the error.
Parse error: syntax error, unexpected T_ENCAPSED_AND_ WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in C:\xampp\htdocs \skill_search\s earch.php on line 181

What could be the reason for this error.
Plz help

[Markus]

I tried this now but its still giving me the error.
Parse error: syntax error, unexpected T_ENCAPSED_AND_ WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in C:\xampp\htdocs \skill_search\s earch.php on line 181

What could be the reason for this error.
Plz help

what line is 181?

regards

[agarwalsrushti]

what line is 181?

regards

Thanks the problem got solved. The curly braces were missing which was giving error.
Thanks a lot.

[code=php]
echo "{$info['UserName']} <a href=\"searchac count.php?id={$ info['UserName']}\">profile</a><br/>";
[/code]

Can you just let me know how to make each profile open up in a different browser.

[Markus]

[code=php]
echo "{$info['UserName']} <a href=\"searchac count.php?id={$ info['UserName']}\" target=\"_blank \">profile</a><br/>";
[/code]


Try that

[agarwalsrushti]

Thanks. It worked out.
Thanks a lot.