Fetching Multiple Rows

[La_Parka]

I'm trying to make an index page for my site with a navigation system that is dynamically generated using a MySQL database....but I can only get it to recall one row/one URL from the database, when I want it to recall every URL from the DB (each is in its own row, under the same field). How can this be done?

[rIkLuNd]

you could do like this:

<?php

// Connection stuff here

$get = mysql_query("SELECT * FROM table");
if(!$get)
die(mysql_error());

while($row = mysql_fetch_array($get))
{
print $row[url];
}

?>

[rIkLuNd]

Was this what you wanted?

[rIkLuNd]

I usually connect to mysql like this:

<?php

$link = mysql_connect($host,$user,$pass);
if(!$link)
die(mysql_error());

$sel = mysql_select_db($db);
if(!$sel)
die(mysql_error());

// ...

?>

[La_Parka]

Originally posted by rIkLuNd
Was this what you wanted?

Is that code for just displaying the fields like this: field1 field2

I wanted to put them in an HTML table e.g.:

FIELD
Row1
Row2
Row3
Row4
etc....

[La_Parka]

For reference, here's what I've tried:

$query = "SELECT * FROM mytable ORDER BY name";
$result = mysql_query($query) or die ("Query failed!");
$field = mysql_fetch_array($result);

echo "<TABLE>";
	echo "<TR>";
	echo "<TD>";
	echo "".$field['name']."";
	echo "</TD>";
	echo "</TR>";
echo "</TABLE>";

But it only prints one row from the field!

[rIkLuNd]

This is how you should do:

<?php

$query = "SELECT * FROM mytable ORDER BY name";
$result = mysql_query($query) or die ("Query failed!");

echo "<TABLE>";
while($field = mysql_fetch_array($result))
{
    echo "<TR>";
    echo "<TD>";
    echo "".$field['name']."";
    echo "</TD>";
    echo "</TR>";
}
echo "</TABLE>";

?>

[eledemar]

now lets say he wants to only output row 3. how would he do that without listing the others. (coded)

[rIkLuNd]

<?php

$query = "SELECT * FROM mytable WHERE id='3' ORDER BY name";
$result = mysql_query($query) or die ("Query failed!");

echo "<TABLE>";
while($field = mysql_fetch_array($result))
{
    echo "<TR>";
    echo "<TD>";
    echo "".$field['name']."";
    echo "</TD>";
    echo "</TR>";
}
echo "</TABLE>";

?>

[eledemar]

i was having same problem only i wanted just a certain field listed

my problem was:

where id="3"

fixed to:

where id='3'

thanks