Please help with code

rcabrera

hi again guys !!! hope you had a great day....

I need help with this, I'm trying to obtain a set of results from a database put them on a table, and have an hyperlink to the idsocio field that, when clicked jump me to another page that could show me the details of that specific row (idsocio), being honest i'd read a few threads before but doesn't work at all. Here is part of the code

$result = mysql_query("SELECT * FROM coope where curriculum LIKE '%$TEXTO%'",$db);

if ($myrow = mysql_fetch_array($result)) {

echo "<table border=1>\n";

echo "<tr><td>id</td></tr>\n";

do {

echo "<tr><td><a href=\"curriculu.php?idsocio=".$myrow["idsocio"]."\"></a></td></tr>";

} while ($myrow = mysql_fetch_array($result));

echo "</table>\n";

} else {

echo "Sorry, no records were found!";

}

the only message I get it's that one

"; } while ($myrow = mysql_fetch_array($result)); echo "\n"; } else { echo "Sorry, no records were found!"; } ?>

by now i'm trying with just one field 'cause I tried to star from easier, in fact instead of the echo i wanted to use the printf funtion...

thanks for any help

The_Chancer

SLightly different style of coding but...

$result = mysql_query("SELECT * FROM coope where curriculum LIKE '%$TEXTO%'",$db); 
if (!($myrow))
{
print("No Results found from query SELECT * FROM coope where curriculum LIKE '%$TEXTO%'"); //lets you see what query was passed...
}
else
{
print("<table border=1>\n"); 

print("<tr><td>id</td></tr>\n"); 

while($myrow=mysql_fetch_array($result))
{
print("<tr><td><a href=\"curriculu.php?idsocio=$myrow['idsocio']\"></a></td></tr>");
}
print("</table>\n");

May be of some use... but I can't see any probs in the coding of that...

lunar

result = mysql_query("SELECT * FROM coope where curriculum LIKE \"%$TEXTO%\"",$db);

This way it should work 🙂

tomhath

Or try it this way:


$query = "SELECT * FROM coope where curriculum LIKE '%" . $TEXTO . "%' ";

echo $query;

$result = mysql_query ( $query, $db) or die ("Query failed: " . mysql_error() );