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20 Oct 2009, 03:27 PM
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Established Member
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Join Date: Sep 2009
Posts: 25
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cant update my table in the DB
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 $table;
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Derek's secret account updater</title>
</head>
<body>
<form id="form1" name="form1" method="post" action="secret_updater.php">
<table width="31%" border="0">
<tr>
<td colspan="31%">Derek's secret Account updater</td>
</tr>
<tr>
<td width="32%">username:</td>
<td width="68%"><input name="username" type="text" id="username" /></td>
</tr>
<tr>
<td>password:</td>
<td><input name="password" type="text" id="password" /></td>
</tr>
<tr>
<td>Name:</td>
<td><input name="Name" type="text" id="Name" /></td>
</tr>
<tr>
<td><input type="submit" value="Submit" /></td>
</tr>
</table>
</form>
</body>
</html>
Last edited by Alan; 20 Oct 2009 at 03:31 PM.
Reason: Changed bbcode
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20 Oct 2009, 03:34 PM
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Moderator
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Join Date: Feb 2007
Location: Ireland
Posts: 843
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PHP Code:
if(isset($_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)). 
__________________
“The best thing about a boolean is even if you are wrong, you are only off by a bit.”
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20 Oct 2009, 03:40 PM
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Established Member
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Join Date: Sep 2009
Posts: 25
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AWESOME thank you very much!!! i spent over an hour staring at this problem. so thank you. derek 
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20 Oct 2009, 04:21 PM
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Moderator
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Join Date: Feb 2007
Location: Ireland
Posts: 843
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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.
__________________
“The best thing about a boolean is even if you are wrong, you are only off by a bit.”
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20 Oct 2009, 09:44 PM
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Established Member
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Join Date: Sep 2009
Posts: 25
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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>
Last edited by silverglade; 20 Oct 2009 at 10:07 PM.
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20 Oct 2009, 10:14 PM
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Established Member
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Join Date: Sep 2009
Posts: 25
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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. 
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