cant update my table in the DB - Web Design & Web Development Forum: HTML, CSS, JavaScript, PHP

silverglade · #1 20 Oct 2009, 04:27 PM

hi, i have the following script which is supposed to update the table in the database, and output the results of the newly inserted data to the same page. it doesnt work. please if anyone can tell me why, id greatly appreciate it. here is the code.

PHP Code:


			
 <?php 
 
include("connect1.php");
 
 
////////////////////////////////////////
////////////////////////////////////////

/// query db and loop through rows example

$field2 = $_POST['username'];
$field3    = $_POST['password'];
$field4    = $_POST['Name'];
 
if(isset($_POST['Submit'])){ //if submit has been set, or clicked.
if($field2 && $field3 && $field4){

    mysql_query("INSERT INTO table2 (username,password,name)VALUES('$field2','$field3','$field4')");

}else{


echo "error";

}
}//isset
//////////////////////////////////////////////////////////output data into a table///////////////////////////////



$query     = mysql_query("SELECT * FROM table2");

    $table = "<table width=\"40%\" border=\"1\">
                  <tr>
                        <td>ID</td>
                        <td>username</td>
                        <td>password</td>
                        <td>Name</td>
                        
                  </tr>";

while($row = mysql_fetch_array($query)){

    $table .= "    <tr>
                    <td>".$row['record_id']."</td>
                    <td>".$row['username']."</td>
                    <td>".$row['password']."</td>
                    <td>".$row['Name']."</td>
                    
                </tr>";

}

$table .= "</table>";

echo $table;
?>


<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Derek's secret account updater</title>
</head>

<body>




<form id="form1" name="form1" method="post" action="secret_updater.php">
  <table width="31%" border="0">
    <tr>
      <td colspan="31%">Derek's secret Account updater</td>
    
    </tr>
    <tr>
      <td width="32%">username:</td>
      <td width="68%"><input name="username" type="text" id="username" /></td>
    </tr>
    <tr>
      <td>password:</td>
      <td><input name="password" type="text" id="password" /></td>
    </tr>
    <tr>
      <td>Name:</td>
      <td><input name="Name" type="text" id="Name" /></td>
    </tr>
     <tr>
       
      <td><input type="submit" value="Submit" /></td>
    </tr>
  </table>
</form>

</body>
</html>

Alan · #2 20 Oct 2009, 04:34 PM

PHP Code:


			
if(isset($_POST['Submit'])){ //if submit has been set, or clicked.

if($field2 && $field3 && $field4){

You have not set the "Submit" field in the form. You need to add name="Submit" to the submit input field.

As for the second line... what exactly are you testing for? It's very vague. Don't get me wrong, it should work... but you should be more specific, such as if(isset($field2) && !empty($field2)).

silverglade · #3 20 Oct 2009, 04:40 PM

AWESOME thank you very much!!! i spent over an hour staring at this problem. so thank you. derek

Alan · #4 20 Oct 2009, 05:21 PM

Quote:

Originally Posted by silverglade

AWESOME thank you very much!!! i spent over an hour staring at this problem. so thank you. derek

No problem. Your very welcome.

silverglade · #5 20 Oct 2009, 10:44 PM

my code isnt working again, it wont update the table, nor display the results, any help greatly appreciated. heres the code.

Code:

  <?php 
 
include("connect1.php");
 
 
////////////////////////////////////////
////////////////////////////////////////

/// query db and loop through rows example

$field2 = $_POST['username'];
$field3	= $_POST['password'];
$field4	= $_POST['Name'];
 

if($field2 && $field3 && $field4){

	mysql_query("INSERT INTO table2 (username,password,Name)VALUES('$field2','$field3','$field4')");

}else{


echo "error";

}

//////////////////////////////////////////////////////////output data into a table///////////////////////////////



$query 	= mysql_query("SELECT * FROM table1");

	$table = "<table width=\"40%\" border=\"1\">
				  <tr>
						<td>ID</td>
						<td>username</td>
						<td>password</td>
						<td>Name</td>
						
				  </tr>";

while($row = mysql_fetch_array($query)){

	$table .= "	<tr>
					<td>".$row['record_id']."</td>
					<td>".$row['username']."</td>
					<td>".$row['password']."</td>
					<td>".$row['Name']."</td>
					
				</tr>";

}

$table .= "</table>";

echo $table;
?>


<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Derek's secret account updater</title>
</head>

<body>




<form id="form1" name="form1" method="post" action="secret_updater.php">
  <table width="31%" border="0">
    <tr>
      <td colspan="31%">Derek's secret Account updater</td>
    
    </tr>
    <tr>
      <td width="32%">username:</td>
      <td width="68%"><input name="username" type="text" id="username" /></td>
    </tr>
    <tr>
      <td>password:</td>
      <td><input name="password" type="text" id="password" /></td>
    </tr>
    <tr>
      <td>Name:</td>
      <td><input name="Name" type="text" id="Name" /></td>
    </tr>
     <tr>
       
      <td><input name="Submit"  type="submit" value="Submit" /></td>
    </tr>
  </table>
</form>

</body>
</html>

silverglade · #6 20 Oct 2009, 11:14 PM

nevermind i copied my old posts code and redid the submit code. it works now. newbie lesson learned. IF IT AINT BROKE DONT FIX IT.