Append to a URL

dcjones

Hi Guys & Girls.

If I have a field which contains a URL and I need to append a string to the end and the sring is made up of 2 or more words.

How do I code it. i.e.

URL: http://www.mysite.com?name=this is a test

when I more the pointer over the URL it is only picking up the first word in the string, in this example "it picks up "this" and not the rest . Why and how do I make it pick up the complete string.

Can anyone help please.

Thanks in advance

DCJ

netizen


$string = urlencode("This is a Test");

$html = "<a href=\"http://www.mywebsite.com/?string=$string\">My Link</a>";

This will produce:

http://www.mywebsite.com/?string=This+is+a+Test

dcjones

Hi and thanks for your reply, But

if the string to be appended to the URL can change depending on what record in the database the user selects i.e.

the records are like this. FIELD IS "NAME".

the user can see records like :

bridge hotel
river bank hotel
claverdon downs hotel

how do I make sure that the URL picks up the complete contents
of the field and appends it to its self. like.

http://www.mysite.com?bridge hotel
http://www.mysite.com?river bank hotel
http://www.mysite.com?claverdon downs hotel

Its the spaces in the part after the "?" how do I deal with them.

I hope I am making this clear, thanks for any help you can provide.

DCJ

I hope I am makes this clear.

dcjones

Hi again, this is the code I am trying to use:

any thoughts:

DCJ

James_McMurray2

Use this instead:

dcjones

Hi James, Thanks it sort of works, I am now picking up the correct append to the URL i.e

http://www.mysite.com/accomodation/php/accomchoice.php?idname=river+bank+hotel

That part is great and thank you, the problem I have now is when the script accomchoice.php is called using the link it as a wee wobbly,

The accomchoice.php shows this

Warning: Supplied argument is not a valid MySQL result resource in /homepages/41/d43000764/htdocs/TraveltoEngland/Accomodation/PHP/accomchoice.php on line 88

Bleow are lines 83 to 89. Can you see a problem there.

$connection = mysql_connect($server, $user, $password);
mysql_select_db("db66511019", $connection);
$result = mysql_query("SELECT * FROM accom WHERE name = $idname" , $connection);

while ($row = mysql_fetch_array($result))

Thanks for your time with this.

DCJ

dcjones

That did not look right Line 84 to 88

LINE 84 $connection = mysql_connect($server, $user, $password);

LINE 85 mysql_select_db("db66511019", $connection);

LINE 86 $result = mysql_query("SELECT * FROM accom WHERE name = $idname" , $connection);

LINE 87 BLANK

LINE 88 while ($row = mysql_fetch_array($result))

DCJ