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Super newbie needs help

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Join Date: Jul 2006

Posts: 174

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tefflox

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Super newbie needs help

Oct 30th, 2007

I have written lots of php dealing with html strings, but it's my first time trying to get a very simple page to function. It has a textarea, and I guess that all I need to do is get the text into the database on the server. I really have no idea how to go about this, and I'm so frustrated that I can't find what I'm looking for. It's simple (again, guessing) as all there is, for now, is the text area. The client (user) types in some text, sends it, after previewing it through my nifty string work :-) and it goes to the database. It would be very nice of someone to walk me through this. I'm the type of learner who needs guidance over "solutions". Sorry if I seem dumb, but I just need help. I'm only a hobbyist programmer.

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Join Date: Jul 2006

Posts: 174

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tefflox

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Re: Super newbie needs help

Oct 30th, 2007

My html:

 Help with Code Tags 
 PHP Syntax (Toggle Plain Text) 
<form action="insert.php" method="post" name="form"> <br/><span style="position:relative: left: -20%; font-size: small;"></span><br/>
<textarea name="text" id="DynamicText" style="font-family: arial, helvetica, sans-serif; font-size: large;" cols="100" rows="5" wrap="soft" onclick="document.form.text.value='';">
Lorem ipsum . . .
  </textarea>
  <br/>
  <br/>
<input type="button" value="Preview" onclick="reloadTextDiv();" />
  </form>
<form action="insert.php" method="post" name="form"> <br/><span style="position:relative: left: -20%; font-size: small;"></span><br/>
<textarea name="text" id="DynamicText" style="font-family: arial, helvetica, sans-serif; font-size: large;" cols="100" rows="5" wrap="soft" onclick="document.form.text.value='';">
Lorem ipsum . . .
  </textarea>
  <br/>
  <br/>
<input type="button" value="Preview" onclick="reloadTextDiv();" />
  </form>

PHP:

 Help with Code Tags 
 PHP Syntax (Toggle Plain Text) 
  $username = "-------";
  $password = "-------";
  $database = "----";
 
  $text = $_POST['text']; 
 
  mysql_connect("mysql.gravityway.com", $username, $password); 
 
  @mysql_select_db($database) or die("Unable to select database"); 
 
  $query = "INSERT INTO ---- VALUES ('','$first')";
 
  mysql_query($query);
 
  mysql_close();
  $username = "-------";
  $password = "-------";
  $database = "----";
  
  $text = $_POST['text']; 
  
  mysql_connect("mysql.gravityway.com", $username, $password); 
  
  @mysql_select_db($database) or die("Unable to select database"); 
  
  $query = "INSERT INTO ---- VALUES ('','$first')";
  
  mysql_query($query);

  mysql_close();

Last edited by tefflox; Oct 30th, 2007 at 5:14 am.

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Join Date: Jul 2006

Posts: 174

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tefflox

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Re: Super newbie needs help

Oct 30th, 2007

Here is my progress. I have connected to the localhost database and inserted text.

Insertion code:

 Help with Code Tags 
 PHP Syntax (Toggle Plain Text) 
<?php
 
  $username = "root";
  $password = "-------";
  $database = "grav";
  $host     = "localhost";
 
  $text = $_POST['text']; 
 
  $cid  = mysql_connect($host, $username, $password); 
 
  if (!$cid) { echo("ERROR: " . mysql_error() . "\n"); }
 
  if ($_SERVER['REQUEST_METHOD'] == "POST") {
/*
  // the following 4 lines are needed if your server has register_globals set to Off
  $category = $_POST['category'];
  $sitename = $_POST['sitename'];
  $siteurl = $_POST['siteurl'];
  $description = $_POST['description'];
*/
  $SQL = " INSERT INTO posts VALUES ('', '$text')";
 
  $result = mysql_db_query($database, $SQL, $cid);
 
  if (!$result) {
      echo("ERROR: " . mysql_error() . "\n".$SQL."\n"); }
  else
      echo ("New Post Added\n");
 
  }
 
  mysql_close($cid);  
 
 
?>
<?php

  $username = "root";
  $password = "-------";
  $database = "grav";
  $host     = "localhost";
  
  $text = $_POST['text']; 
  
  $cid  = mysql_connect($host, $username, $password); 
  
  if (!$cid) { echo("ERROR: " . mysql_error() . "\n"); }
  
  if ($_SERVER['REQUEST_METHOD'] == "POST") {
/*
  // the following 4 lines are needed if your server has register_globals set to Off
  $category = $_POST['category'];
  $sitename = $_POST['sitename'];
  $siteurl = $_POST['siteurl'];
  $description = $_POST['description'];
*/
  $SQL = " INSERT INTO posts VALUES ('', '$text')";
  
  $result = mysql_db_query($database, $SQL, $cid);

  if (!$result) {
      echo("ERROR: " . mysql_error() . "\n".$SQL."\n"); }
  else
      echo ("New Post Added\n");

  }

  mysql_close($cid);  
  
     
?>

I need help with retrieving the data. This is mainly copynpaste code, but it's a start. I get an unknown category error. The table has two columns, the int key and a text field.

 Help with Code Tags 
 PHP Syntax (Toggle Plain Text) 
<?php
 
  $username = "root";
  $password = "-------";
  $database = "grav";
  $host     = "localhost";
 
    # connect to database
    $cid = mysql_connect($host,$username,$password);
    if (!$cid) { echo("ERROR: " . mysql_error() . "\n");    }
?>
<html>
 <head>
  <title>Display Posts</title>
 </head>
 <body>
 
<?php
    $category = "Posts";
 
    # setup SQL statement
    $SQL = " SELECT * FROM posts WHERE category = '$category' ";
 
    # execute SQL statement
    $retid = mysql_db_query($database, $SQL, $cid);
 
    # check for errors
    if (!$retid) { echo( mysql_error()); }
      else {
 
          # display results
          echo ("<h3><b>$category</b></h3>\n");
           while ($row = mysql_fetch_array($retid)) {
              $post = $row["post"];
              echo ($post."<br/>\n");
          }
    }
?>
 
</body>
</html>
<?php

  $username = "root";
  $password = "-------";
  $database = "grav";
  $host     = "localhost";

    # connect to database
    $cid = mysql_connect($host,$username,$password);
    if (!$cid) { echo("ERROR: " . mysql_error() . "\n");    }
?>
<html>
 <head>
  <title>Display Posts</title>
 </head>
 <body>
 
<?php
    $category = "Posts";

    # setup SQL statement
    $SQL = " SELECT * FROM posts WHERE category = '$category' ";

    # execute SQL statement
    $retid = mysql_db_query($database, $SQL, $cid);

    # check for errors
    if (!$retid) { echo( mysql_error()); }
      else {

          # display results
          echo ("<h3><b>$category</b></h3>\n");
           while ($row = mysql_fetch_array($retid)) {
              $post = $row["post"];
              echo ($post."<br/>\n");
          }
    }
?>

</body>
</html>

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Join Date: Aug 2007

Posts: 569

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Solved Threads: 71

ryan_vietnow ryan_vietnow is offline

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Re: Super newbie needs help

Oct 30th, 2007

•

I need help with retrieving the data. This is mainly copynpaste code, but it's a start. I get an unknown category error. The table has two columns, the int key and a text field.
Help with Code Tags

PHP Syntax (Toggle Plain Text)
<?php
 
  $username = "root";
  $password = "-------";
  $database = "grav";
  $host     = "localhost";
 
    # connect to database
    $cid = mysql_connect($host,$username,$password);
    if (!$cid) { echo("ERROR: " . mysql_error() . "\n");    }
?>
<html>
 <head>
  <title>Display Posts</title>
 </head>
 <body>
 
<?php
    $category = "Posts";
 
    # setup SQL statement
    $SQL = " SELECT * FROM posts WHERE category = '$category' ";
 
    # execute SQL statement
    $retid = mysql_db_query($database, $SQL, $cid);
 
    # check for errors
    if (!$retid) { echo( mysql_error()); }
      else {
 
          # display results
          echo ("<h3><b>$category</b></h3>\n");
           while ($row = mysql_fetch_array($retid)) {
              $post = $row["post"];
              echo ($post."<br/>\n");
          }
    }
?>
 
</body>
</html>
<?php $username = "root"; $password = "-------"; $database = "grav"; $host = "localhost"; # connect to database $cid = mysql_connect($host,$username,$password); if (!$cid) { echo("ERROR: " . mysql_error() . "\n"); } ?> <html> <head> <title>Display Posts</title> </head> <body> <?php $category = "Posts"; # setup SQL statement $SQL = " SELECT * FROM posts WHERE category = '$category' "; # execute SQL statement $retid = mysql_db_query($database, $SQL, $cid); # check for errors if (!$retid) { echo( mysql_error()); } else { # display results echo ("<h3><b>$category</b></h3>\n"); while ($row = mysql_fetch_array($retid)) { $post = $row["post"]; echo ($post."<br/>\n"); } } ?> </body> </html>

please replace this part of your code:

 Help with Code Tags 
 php Syntax (Toggle Plain Text) 
<?php
    $category = "Posts";
 
    # setup SQL statement
    $SQL = " SELECT * FROM posts WHERE category = '$category' ";
 
    # execute SQL statement
    $retid = mysql_db_query($database, $SQL, $cid);
 
    # check for errors
    if (!$retid) { echo( mysql_error()); }
      else {
 
          # display results
          echo ("<h3><b>$category</b></h3>\n");
           while ($row = mysql_fetch_array($retid)) {
              $post = $row["post"];
              echo ($post."<br/>\n");
          }
    }
?>
<?php
    $category = "Posts";

    # setup SQL statement
    $SQL = " SELECT * FROM posts WHERE category = '$category' ";

    # execute SQL statement
    $retid = mysql_db_query($database, $SQL, $cid);

    # check for errors
    if (!$retid) { echo( mysql_error()); }
      else {

          # display results
          echo ("<h3><b>$category</b></h3>\n");
           while ($row = mysql_fetch_array($retid)) {
              $post = $row["post"];
              echo ($post."<br/>\n");
          }
    }
?>

with this one:

 Help with Code Tags 
 php Syntax (Toggle Plain Text) 
$category = "Posts";
 
    # setup SQL statement
    $SQL = " SELECT * FROM posts WHERE category = '$category' ";
 
    # execute SQL statement
    $retid = mysql_query($query);
  $result=mysql_query($retid) or die (mysql_error());
   $num=mysql_num_rows($result);
 
# display results
while ($i<$num)
{
  $post=mysql_result($result,$i,"post");
echo "<h3><b>'.$category.'</b></h3>\n";
echo $post;          
$i++;
}
$category = "Posts";

    # setup SQL statement
    $SQL = " SELECT * FROM posts WHERE category = '$category' ";

    # execute SQL statement
    $retid = mysql_query($query);
  $result=mysql_query($retid) or die (mysql_error());
   $num=mysql_num_rows($result);
 
# display results
while ($i<$num)
{
  $post=mysql_result($result,$i,"post");
echo "<h3><b>'.$category.'</b></h3>\n";
echo $post;          
$i++;
}

I think that you are getting an unknown column because the field "post" is not the column name of what you are trying to display.
Look at your database, look what field do you want to display and replace "post" in

 Help with Code Tags 
 php Syntax (Toggle Plain Text) 
  $post=mysql_result($result,$i,"post");
  $post=mysql_result($result,$i,"post");

with that field.
Or better yet please display here all your table fields so I can guide you.

http://www.ryantetek.com/

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Join Date: Jul 2006

Posts: 174

Reputation: tefflox is an unknown quantity at this point

Solved Threads: 1

tefflox

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Junior Poster

Re: Super newbie needs help

Oct 31st, 2007

Hi, thanks. Your code works with the following change: $SQl = " SELECT post FROM posts ";

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Join Date: Jul 2006

Posts: 174

Reputation: tefflox is an unknown quantity at this point

Solved Threads: 1

tefflox

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Junior Poster

Re: Super newbie needs help

Oct 31st, 2007

Actually, there are problems with the code provided. I had mistakenly copied my own back in. This works ---

 Help with Code Tags 
 PHP Syntax (Toggle Plain Text) 
  # setup SQL statement
  $SQL = " SELECT post FROM posts ";
 
  # execute SQL statement
  $retid = mysql_db_query($database, $SQL, $cid);
 
  # check for errors
  if (!$retid) { echo( mysql_error()); }
 
  else {
 
    # display results
    echo ("<h3><b>Posts</b></h3>\n");
 
    while ($row = mysql_fetch_array($retid)) {
      $post = $row["post"];
      echo ($post."<br/>\n");
    }
  }
  # setup SQL statement
  $SQL = " SELECT post FROM posts ";

  # execute SQL statement
  $retid = mysql_db_query($database, $SQL, $cid);
 
  # check for errors
  if (!$retid) { echo( mysql_error()); }

  else {

    # display results
    echo ("<h3><b>Posts</b></h3>\n");

    while ($row = mysql_fetch_array($retid)) {
      $post = $row["post"];
      echo ($post."<br/>\n");
    }
  }

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Join Date: Aug 2007

Posts: 569

Reputation: ryan_vietnow is an unknown quantity at this point

Solved Threads: 71

ryan_vietnow ryan_vietnow is offline

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Posting Pro

Re: Super newbie needs help

Oct 31st, 2007

so,that's why I am wondering on where did you get that SQL statement.Actually mysql_fetch_array is another way of displaying data but for me my way of displaying data is what I have shown.But anyway,it's still the same!

http://www.ryantetek.com/