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sql query into textarea

Dec 25th, 2008

hallo there.

i am new in programming so i ask for your understanding

i want to put the results from a sql query into a textarea. then i want to be able to pick one of the records of textarea and delete it (textarea and database). (i dont have to use textarea i just thouhgt it would be a good idea)

something is not right. this is my code

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echo "<form method=post name=f2 action='dd-delete.php'>";
 
$qry="SELECT id, member, math, bathmos FROM bathmoi";
echo '<textarea name="show" rows="20" cols="70" readonly="readonly">';
 
while($result = mysql_fetch_array($qry)) {
echo "$result[id], $resuld[memeber], $result[mathima]";
}
echo "</textarea>";
 
echo "<input type=submit value=Submit>";
echo "</form>";
echo "<form method=post name=f2 action='dd-delete.php'>";

$qry="SELECT id, member, math, bathmos FROM bathmoi";
echo '<textarea name="show" rows="20" cols="70" readonly="readonly">';

while($result = mysql_fetch_array($qry)) {
echo "$result[id], $resuld[memeber], $result[mathima]";
}
echo "</textarea>";

echo "<input type=submit value=Submit>";
echo "</form>";

i receive:

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 PHP Syntax (Toggle Plain Text) 
<br />
<b>Warning</b>: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in <b>C:\xampp\htdocs\admin\drop_down\dd.php</b> on line <b>118</b><br />
<br />
<b>Warning</b>: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in <b>C:\xampp\htdocs\admin\drop_down\dd.php</b> on line <b>118</b><br />

can u help me? what should i do to see the records and be able to pick one of then in order to delete it (from database and textarea)?

Last edited by peter_budo; Dec 29th, 2008 at 5:50 am. Reason: Keep It Organized - For easy readability, always wrap programming code within posts in [code] (code blocks) and [icode] (inline code) tags.