Thread: link problem
hello sir, i have printed the contents of a database on a table using php. how to give link from each content of cell to another page or file(like pdf). table in php is been created using printf() statement and i am unable to insert code to give link for whatever i print. thanx in advance Arun
On Thu, 28 Feb 2002, arun kv wrote: > > > hello sir, > i have printed the contents of a database on a table using php. how > to give link from each content of cell to another page or file(like > pdf). table in php is been created using printf() statement and i am > unable to insert code to give link for whatever i print. > thanx in advance > Arun > >
At 07:02 28/02/02, Arun wrote: >hello sir, > i have printed the contents of a database on a table using php. how >to give link from each content of cell to another page or file(like >pdf). table in php is been created using printf() statement and i am >unable to insert code to give link for whatever i print. > thanx in advance > Arun How are you fetching the data from your query? If you use pg_fetch_array(), or one of the related functions, you can pick out the individual data elements from the array and put whatever formatting you like around them. best, Mo ===== "How can you expect a little child whose mother is an angel and whose father is a Cannibal King and who has spent her life sailing the seas to tell the truth always?" Astrid Lindgren (1907-2002)
On Thu, 2002-02-28 at 00:02, arun kv wrote: > > > hello sir, > i have printed the contents of a database on a table using php. how > to give link from each content of cell to another page or file(like > pdf). table in php is been created using printf() statement and i am > unable to insert code to give link for whatever i print. > thanx in advance > Arun Hi Arun, You should review the printf and sprintf documentation. It is a bit daunting at first, but makes sense after a bit of work. To answser your question as best I can without seeing source code, here is one way to make a link using a printf statement $link = "http://www.php.net/"; $text = "PHP Website"; printf ('<a href="%s">%s</a>', $link, $text); Notice that the first %s is replaced with the value of $link, while the second %s is replaced with the value of $text. It looks like you are just getting started with PHP - you can find a lot of helpful resources at http://www.php.net/support.php Good Luck and happy PHPing! -- Zak Greant MySQL Advocate PHP Quality Assurance Team