hi
thanks for the info. before i recieved your email i managed o cobble
somethign together from some tutorials. this is the code i am using
<?php
$host="localhost";
$username="gecko_gecko";
$password="3t7b8011";
$database="gecko_content";
mysql_connect($host,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");
$result = mysql_query("SELECT * FROM portfolio ORDER BY header ASC") or
die(mysql_error());
while($row = mysql_fetch_array( $result )) {
print ("<h1>{$row['header']}</h1><hr>");
print ("<p>{$row['body']}</p>");
}
?>
unfortunately, i have to have this code on each page and change the
table name for each one. i have looked at your code and i am not sure i
fully understand it. will this allow me to have one page with all the
php in, and depending on what link is clicked, will it then show that
information?
i have a database called "content" with 40 tables, each named after the
a section of the site. (probably not the best layout, but i am learning)
i have the input page all figured out, and now im working on the edit
and update pages.
thanks.
mark m...
Michael Sullivan wrote:
> On Tue, 2006-03-21 at 04:45 +0000, Mark Mckee wrote:
>
>>hi all
>>
>>I am definately stuck now. i have managed to insert data into my
>>table
>>using php and sql staements.
>>
>>I have also managed to display that data on a single page.
>>
>>the points where i am stuck
>>
>>1) how do i display all the data contained in the table and not just
>>the
>> one row?
>>
>
>
> To answer this question, here's an excerpt from my current project:
>
> if ($fullname == "Choir")
> $result = mysql_query("SELECT entry_name FROM ChoirEntry WHERE
> school_id='$schoolid' ORDER BY entry_name ASC;") or die("Could not find
> record.");
> else
> $result = mysql_query("SELECT entry_name FROM BandEntry WHERE
> school_id='$schoolid' ORDER BY entry_name ASC;") or die("Could not find
> record.");
>
>
> if (mysql_num_rows($result) > 0)
> {
> while($a_row = mysql_fetch_array($result, MYSQL_ASSOC))
> {
> foreach($a_row as $field)
> {
> print "<tr><td align='center'>$field</td></tr>\n";
> }
> }
> }
> else
> {
> print "<tr><td align='center'>None Entered Yet</td></tr>";
> }
>
>
> As you can see, I use a while loop to iterate through the results using
> mysql_fetch_array.
>
>
>>2) i have my input form up and ready. a place for the header, test and
>>a
>>drop down box containing the 40 pages i want to choose from to
>>update.
>>how do i send the correct header text and the chioce in the drop down
>>to
>> the correct table?
>
>
> <form method='post' action='nextpage.php'>
> <select name='choices'>
> <option>Option1</option>
> <option>Option2></option>
> <input type='submit' value='Send Choice">
>
> On nextpage.php have a line near the top of the script that says:
> $someVar = $_POST['choices'];
>
>>i think thats it for now lol. any help would be grately appreciated
>>
>>kind regards
>>
>>Mark...
>>
>>
>>
>>
>>Mark Mckee wrote:
>>
>>>thank you for the help.
>>>
>>>i wil keep trying to get it working.
>>>
>>>Michael Sullivan wrote:
>>>
>>>
>>>>On Tue, 2006-03-21 at 00:52 +0000, Mark Mckee wrote:
>>>>
>>>>
>>>>
>>>>>hi all
>>>>>
>>>>>i want to create an inut page for the content on my website. is it
>>>>>possible to have 1 data entry box and choose the specific what
>>
>>table
>>
>>>>>of
>>>>>a database to put it usign a drop down box?
>>>>>
>>>>>if so, can someone please point me in the right direction. i have
>>>>>been
>>>>>troubling over this for the last few days
>>>>>
>>>>>thanks in adavce
>>>>>
>>>>>mark m...
>>>>
>>>>
>>>>I don't see why this wouldn't be possible. You would just use the
>>
>>table
>>
>>>>name selected from the list box in constructing the query string you
>>
>>use
>>
>>>>in your mysql_query.
>>>>
>>>>Ex.
>>>>
>>>>I select 'SomeTable' from a list box called 'list' and hit the
>>
>>Submit
>>
>>>>button. Another page is called, in which is a line like this:
>>>>
>>>>$list = $_POST['list'];
>>>>
>>>>
>>>>
>>>>You code says:
>>>>
>>>>$query = "SELECT * FROM ".$list;
>>>>
>>>>Then you issue
>>>>
>>>>$result = mysql_query($query, $link2DB) or die ("Could not access
>>
>>table
>>
>>>>$list:".mysql_error());
>>>>
>>>>or whatever you want...
>>>>
>>>>
>>>>
>>>>
>>>>
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>>>>
>>>>
>>>>
>>>>
>>>
>>>
>>>
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