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Posted by Lloyd Harold on 11/14/07 14:23
Jerry Stuckle <jstucklex@attglobal.net> wrote:
> Hi, Lloyd,
>
> Sounds like a homework question?
>
> This isn't too bad, once you get the hang of it.
>
> First you need to execute the mysql query. Then, in a loop, retrieve
> each result and do what you want with it. For instance (error checking
> left out for clarity, but you should be checking the result of every
> MySQL call except the mysql_fetch_array()):
>
>
> $location = 4; // Assumed to be passed from somewhere & validated
>
> $link = mysql_connect('localhost', 'userid', 'password');
> mysql_select_db('mydb');
>
> $result = mysql_select("SELECT id, name, details FROM mytable WHERE
> location=$location"); // Sorry for the wrapping
> if (mysql_num_rows($result) == 0)
> echo "No results found<br>\n";
> else {
> while($data = mysql_fetch_array($result)) {
> // $data is an array with elements ['id'], ['name'] and ['details']
> echo
> "<p><a href='/details.php?id=$data['id']'>$data['name']</a></p>\n";
> // You could also display the details, here. Or, if you're not going to
> // display the details on this page, just leave them out of the query
> }
>
> Then on your details.php page, retrieve the data for the specific ID you
> want, and display them with code similar to the above (although you
> won't need a loop as you'll only have one result).
Thanks for your help and encouragement, Jerry.
I've tried the code and am seeing this error:
expecting `T_STRING' or `T_VARIABLE' or `T_NUM_STRING' in
"<p><a href='/details.php?id=$data['id']'>$data['name']</a></p>\n";
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