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Re: Newbie: How to display query result in table?

Posted by IchBin on 07/07/06 00:21

Gulasch wrote:
> IchBin wrote:
>> flamer die.spam@hotmail.com wrote:
>>> <table>
>>> <tr>
>>> <?php
>>> $query = mysql_query(select blah..);
>>> echo '<td>'.$query.'</td>';
>>> ?>
>>> </tr>
>>> </table>
>>>
>>> now i assume your pulling multiple rows so you need a "while"
>>> statement, read up on them in the manual php.net they are simple to
>>> use, and excuse the psuedo code.
>>>
>>> flamer.
>>>
>>> Huevos wrote:
>>>
>>>> I scanned through hundreds of posts and tried several variations on
>>>> some
>>>> promising posts, but have not succeeded yet... All I want to do (to
>>>> start)
>>>> is have a web page display a table of the result of a query.
>>>>
>>>> I am using MySQL via geocities.yahoo.com. The MySQLAdmin auto
>>>> generated a
>>>> PHP version of my query as follows:
>>>>
>>>> $sql = 'SELECT * FROM `Varieties` WHERE 1'
>>>> . ' ORDER BY `nHeight` DESC, `nDiameter` DESC, `nMinTemp` ASC,
>>>> `tSun` DESC LIMIT 0, 30 ';
>>>>
>>>>
>>>> I tried to merge this into code from another suggestion in these
>>>> groups and
>>>> came up with:
>>>>
>>>> <HTML>
>>>> <HEAD>
>>>> </HEAD>
>>>> <BODY>
>>>> <?
>>>> // DATABASE QUERY
>>>> $bamboo='SELECT * FROM `Varieties` WHERE 1'
>>>> . ' ORDER BY `nHeight` DESC, `nDiameter` DESC, `nMinTemp` ASC,
>>>> `tSun` DESC LIMIT 0, 30 ';
>>>> $result= mysql_query($bamboo);
>>>> ?>
>>>>
>>>> <table width="100%" border="0" cellspacing="0" cellpadding="0">
>>>> <tr>
>>>> <td>LINK 1 DETAILS HERE<br>
>>>> DATABASE QUERY: <? $result; ?></td>
>>>> </tr>
>>>> </table>
>>>> <? } ?>
>>>> </BODY>
>>>> </HTML>
>>>>
>>>> Can anyone offer a suggestion?
>>>> Thanks in advance!
>>>>
>>>> Jeff Hawkins
>>>
>>
>> Not to add another question to this question but how would this be
>> done with a dropdown list?
>>
>> Thanks in Advance...
>> IchBin, Pocono Lake, Pa, USA http://weconsultants.phpnet.us
>> __________________________________________________________________________
>>
>>
>> 'If there is one, Knowledge is the "Fountain of Youth"'
>> -William E. Taylor, Regular Guy (1952-)
>
> If you have let's say
>
> --snip--
> $sql="SELECT field FROM table";
> $result=mysql_query($sql, $connection);
> --snap--
>
> you could do for example
>
> --snip--
> echo "<select size=1>";
> while($row=mysql_fetch_assoc($result))
> {
> echo "<option>". $row['field'] ."</option>";
> }
> echo "</select>";
> --snap--
>
> to get a Drop-Down-List.

I guess I should have tested before I responded before.. At first I
received the error: "Cannot use object of type stdClass as array in..."

Now I get the dropdown list box but I only see the following, in pairs,
for the # of rows returned"

Object id #1
Object id #2

Also can you specify two columns in the dropdown list box. One for a
Primary Key which is hidden and the other data for user to select. This
way I can get detail information for the selected item.

This is my code:

<?php
$detail = 'author_detail';
$quote = 'author_quotes';
//
// Open DB
include 'library/opendb.php';
//
// issue the DB CMD
$result = mysql_query("select lastname from $detail where lastname =
'Proverb'");

if (!$result) {
die('Database Error: ' . mysql_error());
exit;
}
//
// Handle the data coming back
echo "<select size=10>";
while ($row = mysql_fetch_object($result)) {
echo "<option>". $row ."</option>";
}
echo "</select>";
//
// Close DB
include 'library/closedb.php';
?>

Thanks in Advance...
IchBin, Pocono Lake, Pa, USA http://weconsultants.phpnet.us
__________________________________________________________________________

'If there is one, Knowledge is the "Fountain of Youth"'
-William E. Taylor, Regular Guy (1952-)

 

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