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monkeypaw201

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Posts posted by monkeypaw201

    Help With UPDATE

    in PHP Coding Help

    Posted

    Put the query in a string, echo the query to make sure it is reading your id, and use mysql_error() after your query to see if there are any errors.

     

    Query in string outputs: UPDATE private_messages SET read = 'Yes'

    WHERE id = '1'

     

    mysql_error after query outputs:

     

    (nothing)

     

    Soooo Lost...

    Help With UPDATE

    in PHP Coding Help

    Posted

    Here is the code:

     

    <?php

$con = mysql_connect("localhost","user","pass");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }mysql_select_db("database", $con);

mysql_query("UPDATE private_messages SET read = 'Yes'
WHERE id = '$_GET[id]'");mysql_close($con);
?>

     

    Its not working... no errors, the page loads fine, it just doesnt update

    Double Query

    in PHP Coding Help

    Posted

    FIXED

     

    Ok, i kept messing with it and came out with this (Yes, It WORKS)

     

    <?php 
$con = mysql_connect("localhost","user","pass");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
  mysql_select_db("database", $con);
  $result = mysql_query("SELECT * FROM domiciles WHERE id = '$row_Pilot[hub]'");
  while($row_con = mysql_fetch_array($result))
  {
  $airport = $row_con['airport'];
  }
  mysql_close($con);
?>
<?php 
$con = mysql_connect("localhost","user","pass");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
  mysql_select_db("database", $con);
  $result = mysql_query("SELECT * FROM airports WHERE id = '$airport'");
  while($row_con = mysql_fetch_array($result))
  {
  $city = $row_con['city'];
  echo $city;
  }
  mysql_close($con);
?>

    Double Query

    in PHP Coding Help

    Posted

    rhodesa, my mistake, there should be an echo here:

    $airport = $row_con['airport']; echo $airport;

    i forgot to copy/paste it..

     

    ---------

     

    craygo , the $row_Pilot[hub] is coming from a previous query higher up on the page..

     

    ---------

     

    revraz, just because i'm more comfortable troubleshooting it.. i have had little experience merging them together, and more the more complex ones, i go back to the basics.. :P

     

    -------

    EDIT:: Would it help if i provide the whole page source?

    Double Query

    in PHP Coding Help

    Posted

    if it helps when i do this, i works (it echos the right ID)

     

    <?php 
$con = mysql_connect("localhost","user","pass");
if (!$con)
 {
 die('Could not connect: ' . mysql_error());
 }
 mysql_select_db("database", $con);
 $result = mysql_query("SELECT * FROM domiciles WHERE id = '$row_Pilot[hub]'");
 while($row = mysql_fetch_array($result))
 {
 $airport = $row_con['airport'];
 }
 /**
$air = mysql_connect("localhost","user","pass");
if (!$air)
 {
 die('Could not connect: ' . mysql_error());
 }mysql_select_db("database", $air);
 $result = mysql_query("SELECT * FROM airports WHERE id = '$airport'");
 while($row = mysql_fetch_array($result))
 {
 echo = $row_air['city'];
 }
mysql_close($air);  */
mysql_close($con);
 
		 ?>

     

    Turn on error reporting.

     

    :/ Never learned how...

    Double Query

    in PHP Coding Help

    Posted

    So, i have this code again, and i have a previous query that pulls soething from the database and provides an ID relative to another database, and then i need to query again from the database WHERE that ID matches the table. THEN i need to take THAT ID and re-query for a final match.

     

    <?php 
$con = mysql_connect("localhost","user","pass");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }mysql_select_db("database", $con);
  $result = mysql_query("SELECT * FROM domiciles WHERE id = '$row_Pilot[hub]'");
  while($row = mysql_fetch_array($result))
  {
  $airport = $row['airport'];
  }
  
  $air = mysql_connect("localhost","user","pass");
if (!$air)
  {
  die('Could not connect: ' . mysql_error());
  }mysql_select_db("database", $air);
  $result = mysql_query("SELECT * FROM airs WHERE id = '$airport'");
  while($row = mysql_fetch_array($result))
  {
  echo = $row_air['city'];
  }mysql_close($air);
mysql_close($con);
  
		 ?>

    Blank Page... No Errors??

    in PHP Coding Help

    Posted

    So, i have a page with the following code on it, and it just spits at me a blank page. No errors, no echos no nothing. When i view the source code, its empty... Im lost  ???

     

    <?php
if($_POST['password'] == $_POST['password2'])
{ //If Passwords Match

//Query for the latest Callsign
mysql_select_db($database_caliregional, $caliregional);
$query_PID = "SELECT * FROM users ORDER BY pilot_id DESC";
$PID = mysql_query($query_PID, $caliregional) or die(mysql_error());
$row_PID = mysql_fetch_assoc($PID);
$totalRows_PID = mysql_num_rows($PID);

//Create un-posted variables
$pilot_id = $row_PID['pilot_id']+1;
$date_hired = date("now");
$password = md5($_POST['password']);
$first_initial = substr($_POST['first_name'];
$forum_name = $first_initial . $_POST['last_name'];
$full_name = $_POST['first_name'] . $_POST['last_name'];
$IP = $_SERVER['REMOTE_ADDR'];
$password_salt = substr(str_shuffle("1234567890ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz"),0,4);
$fpassword = sha1($_POST['password'] . $password_salt);

//Insert Basic User Information
$user = mysql_connect("localhost","user","pass");
if (!$user)
  {
  die('Could not connect: ' . mysql_error());
  }mysql_select_db("database", $user);
  $sql="INSERT INTO users (pilot_id, first_name, last_name, email, password, date_hired, hub)
VALUES
('$pilot_id','$_POST[first_name]','$_POST[last_name]','$_POST[email]','$password','$date_hired','$_POST[hub]')";
if (!mysql_query($sql,$user))
  {
  die('Error: ' . mysql_error());
  }
  
if($_POST['forumacc'] == 'No')
{ //If No Forum Account
//Create Forum Account
$forum = mysql_connect("localhost","user","pass");
if (!$forum)
  {
  die('Could not forumnect: ' . mysql_error());
  }mysql_select_db("database", $forum);
  $sqll="INSERT INTO smf_members (memberName, dateRegistered, realName, passwd, emailAddress, memberIP, memberIP2, ID_POST_GROUP, passwordSalt)
VALUES
('$forum_name','$date_hired','$full_name','$fpassword','$_POST[email]','$IP','$IP','4','$password_salt')";
if (!mysql_query($sqll,$forum))
  {
  die('Error: ' . mysql_error());
  }
} //If No Forum Account

echo "Thank you for registering with California Regional. A confirmation e-mail has been sent to the address provided.";

} //If Passwords Match
else
{ //If Passwords DONT Match
echo "Oops! Looks like the passwords you entered don't match. Please go back and try again.";
} //If Passwords DONT Match
?>

    [SOLVED] MySQL / PHP Being Weird :: "Undefined T_VARIABLE"

    in PHP Coding Help

    Posted

    here it is:

     

    <?php
$pilot_id = mysql_connect("localhost","user","pass");
if (!$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']+1;
//--------------------------
$first_name=$_POST['first_name'];
$last_name=$_POST['last_name'];
$email=$_POST['email'];
$password=md5($_POST['password']);
$dob=$_POST['dob'];
$personal_location=$_POST['personal_location'];
$personal_country=$_POST['personal_country'];
$public_msn=$_POST['public_msn'];
$public_yim=$_POST['public_yim'];
$public_aim=$_POST['public_aim'];
$public_skype=$_POST['public_skype'];
$public_icq=$_POST['public_icq'];
$public_website=$_POST['public_website'];
$public_avatar=$_POST['public_avatar'];
$public_biography=$_POST['public_biography'];
$public_sagtv=$_POST['public_sagtv'];
$public_real_world_pilot=$_POST['public_real_world_pilot'];
$public_vatsim_controller=$_POST['public_vatsim_controller'];
$public_vatsim=$_POST['public_vatsim'];
$activation_code=substr(str_shuffle("ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789abcdefghijklmnopqrstuvwxyz"),0,35);
$hire_date=date("F j, Y");
//-------------------------
$transfer_airline=$_POST['transfer_airline'];
$transfer_proof_link=$_POST['transfer_proof_link'];
$transfer_hours_requested=$_POST['transfer_hours_requested'];

$pilot = mysql_connect("localhost","user","pass");
if (!$pilot)
 {
 die('Could not connect: ' . mysql_error());
 }mysql_select_db("cerulean_site", $pilot);
 $sql="INSERT INTO pilots (pilot_id, first_name, last_name, email, password, date_of_birth, personal_location, personal_country, activation_code, hire_date, public_msn, public_yim, public_aim, public_skype, public_icq, public_website, public_avatar, public_biography, public_sagtv, public_real_world_pilot, public_vatsim_controller, public_vatsim)
VALUES
('$pilot_id','$first_name','$last_name','$email','$password','$dob','$personal_location','$personal_country','$activation_code','$hire_date','$public_msn','$public_yim','$public_aim','$public_skype','$public_icq','$public_website','$public_avatar','$public_biography','$public_sagtv','$public_real_world_pilot','$public_vatsim_controller','$public_vatsim')";

if (!mysql_query($sql,$pilot))
 {
 die('Error: ' . mysql_error());
 }
mysql_close($pilot)

$hour = mysql_connect("localhost","user","pass");
if (!$hour)
 {
 die('Could not connect: ' . mysql_error());
 }mysql_select_db("cerulean_site", $hour);
 $sql="INSERT INTO pilot_hour_transfer (pilot_id, transfer_airline, transfer_proof_link, transfer_hours_requested)
VALUES
('$pilot_id','$transfer_airline','$transfer_proof_link','$transfer_hours_requested')";

if (!mysql_query($sql,$hour))
 {
 die('Error: ' . mysql_error());
 }
mysql_close($hour)

?>

     

    I get the following error:

     

    Parse error: parse error, unexpected T_VARIABLE on line 173

     

    Line 173 is:

     

    $hour = mysql_connect("localhost","user","pass");

     

    it works fine without the second mysql insert...

     

    Thanks in advance

    [SOLVED] Echoing Array???

    in PHP Coding Help

    Posted

    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;

    [SOLVED] Echoing Array???

    in PHP Coding Help

    Posted

    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

    [SOLVED] Undefined Varibale

    in PHP Coding Help

    Posted

    but this is giving me the same error:

     

    $pilot_id = mysql_connect("localhost","cerulean","bluesky1701");

    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['pilot_id'];

    [SOLVED] Undefined Varibale

    in PHP Coding Help

    Posted

    so, i am trying to run the following code:

     

    $pilot_id = mysql_connect("localhost","cerulean","bluesky1701");

    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'];

     

    and its throwing a weird error about an undefined variable

     

    Notice: Undefined variable: row
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