[SOLVED] Echoing Array???

[monkeypaw201]

so, i have this

 

if (!$pilot_id)

  {

  die('Could not connect: ' . mysql_error());

  }

  mysql_select_db("cerulean_site", $pilot_id);

  $result = mysql_query("SELECT * FROM pilots ORDER BY pilot_id DESC");

 

  $row = mysql_fetch_array($result);

$pilot_id=$row;

echo $-pilot_id;

 

and its displaying

 

Array

[kenrbnsn]

Please use

tags when inserting code in this forum, not

tags.

 

You want to get one field from the returned array:

<?php
$pilot_id = $row['pilot_id'];
echo $pilot_id;
?>

 

Ken

[monkeypaw201]

back to the original problem... it says:

 

Undefined Variable: row

[kenrbnsn]

Please post the code that's giving that error.

 

Ken

[rcorlew]

You have to put your results through a while loop like this:

 

<?php

if (!$pilot_id)
  {
  die('Could not connect: ' . mysql_error());
  }
  mysql_select_db("cerulean_site", $pilot_id);
  $result = mysql_query("SELECT * FROM pilots ORDER BY pilot_id DESC");

  while($row = mysql_fetch_array($result)) {
$pilot_id=$row;
echo $-pilot_id;
}

?>

[monkeypaw201]

rcorlew, i do not want a loop, i want only the 1st record displayed

 

Full Code:

 

<?php
$pilot_id = mysql_connect("localhost","username","password");
if (!$pilot_id)
  {
  die('Could not connect: ' . mysql_error());
  }
  mysql_select_db("cerulean_site", $pilot_id);
  $result = mysql_query("SELECT * FROM pilots ORDER BY pilot_id DESC");
  
$pilot_id = $row['pilot_id'];
echo $pilot_id;

[kenrbnsn]

You didn't fetch the row:

<?php
<?php
$pilot_id = mysql_connect("localhost","username","password");
if (!$pilot_id)
  {
  die('Could not connect: ' . mysql_error());
  }
  mysql_select_db("cerulean_site", $pilot_id);
  $result = mysql_query("SELECT * FROM pilots ORDER BY pilot_id DESC");
  $row = mysql_fetch_assoc($result); // add this line
$pilot_id = $row['pilot_id'];
echo $pilot_id;
?>

 

Ken

[monkeypaw201]

great! thanks!!