Duplicate entry....

[oops73]

Hi,

i have trouble with this code:

	$query = "SELECT $IDcode FROM users WHERE IDcode = '$IDcode'";

$num_rows = mysql_num_rows($result)>0;
if ($num_rows) {
print ("Sorry - this IDcode is already in registerd!");
} else {
$query = "INSERT INTO .... ";
$result = mysql_query($query, $connection);
}

I want to check if IDcode is already in database. if so, go back to form.

I keep getting this errors:

Notice: Undefined variable: result in c:\installerade program\easyphp1-8\www\page_3.php on line 56

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in c:\installerade program\easyphp1-8\www\page_3.php on line 56

Notice: Undefined variable: connection in c:\installerade program\easyphp1-8\www\page_3.php on line 61

Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in c:\installerade program\easyphp1-8\www\page_3.php on line 61
Error : Duplicate entry '731216' for key 2

Please help me out.

/Joel

[bodzan]

 $query = "SELECT $IDcode FROM users WHERE IDcode = '$IDcode'";
$num_rows = mysql_num_rows($result)>0;
if ($num_rows) {
print ("Sorry - this IDcode is already in registerd!");
} else {
$query = "INSERT INTO .... ";
$result = mysql_query($query, $connection);
} 

It should be:

 $query = "SELECT $IDcode FROM users WHERE IDcode = '$IDcode'";
$result = mysql_query($query,$connection);
$num_rows = mysql_num_rows($result);
if ($num_rows>0) {
print ("Sorry - this IDcode is already in registerd!");
} else {
$query = "INSERT INTO .... ";
$result = mysql_query($query, $connection);
} 

[oops73]

still the same errors =(

Notice: Undefined variable: connection in c:\

Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in c:\

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in c:\

Notice: Undefined variable: connection in c:\

Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in c:\
Error : Duplicate entry '731216' for key 2

it looks like mysql_num_rows & mysql_query commands is the trouble.... why????

[leatherback]

To me it look slike you ar enot connected to the database properly.

Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in c:\

J.

[oops73]

Originally posted by leatherback
To me it look slike you ar enot connected to the database properly.

Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in c:\

J.

Any ideas how to check that the connection is ok.
I have IDcode as UNIQUE in mysql.

/joel

[bodzan]

...since it is reported missing I assume that there is something wrong with it...

Well, if not the connection show us the lines where you connect to the MySQL...

[oops73]

Here is my entire code for the page.

<?php


session_start();

// connection to database // 
$host = 'localhost';		// host
$login = 'root';	// login
$password = '';		// password 				
$base = 'testbase';	// name of database

if (isset($_COOKIE[session_name()])) {
	$url = $_SERVER['PHP_SELF'];
} else if (defined('SID')) {
	$url = $_SERVER['PHP_SELF'].SID;
}


// condition ? if yes : if no //
isset($_SESSION['Fname']) ? 	$Fname = $_SESSION['Fname'] : 	$Fname = NULL;
isset($_SESSION['Lname']) ? 	$Lname = $_SESSION['Lname'] :	$Lname = NULL;
isset($_POST['IDcode']) ? 		$IDcode = $_POST['IDcode'] : 	$IDcode = NULL;
isset($_SESSION['Region']) ? 	$Region = $_SESSION['Region'] :	$Region = NULL;
isset($_SESSION['Depart']) ? 	$Depart = $_SESSION['Depart'] :	$Depart = NULL;

// WORK now //
if (isset($_POST['action']) && ('add' == $_POST['action'])) {
	////////// ACTION /////////
	// condition ? if yes : if no //
	isset($_POST['Fname']) ? 	$Fname = $_POST['Fname'] : 	$Fname = NULL;
	isset($_POST['Lname']) ? 	$Lname = $_POST['Lname'] :	$Lname = NULL;
	isset($_POST['IDcode']) ? 	$IDcode = $_POST['IDcode'] : $IDcode = NULL;
    isset($_POST['Region']) ? 	$Region = $_POST['Region'] : $Region = NULL;
	isset($_POST['Depart']) ? 	$Depart = $_POST['Depart'] : $Depart = NULL;
	// check if empty // 
	if (empty($Fname)) {
		$error['emptyFname'] = 'Empty Fname.';
	} else if (empty($Lname)){
		$error['emptyLname'] = 'Empty Lname.';
	} else if (empty($IDcode)){
	    $error['emptyIDcode'] = 'Empty IDcode.';
    } else if (empty($Region)){	
	    $error['emptyRegion'] = 'Empty Region.';
	} else if (empty($Depart)){	
		$error['emptyDepart'] = 'Empty Depart.';
	} else {	


	// connexion to database and table //
	$db = mysql_connect($host, $login, $password) or die ('Error : '.mysql_error());
	mysql_select_db($base, $db) or die ('Error : '.mysql_error());

    // Check IDcode already in database //
	$query = "SELECT $IDcode FROM users WHERE IDcode = '$IDcode'"; 
    $result = mysql_query($query, $connection); 
    $num_rows = mysql_num_rows($result); 
    if ($num_rows>0) { 
    print ("Sorry - this IDcode is already in registerd!"); 
    } else { 
    $query = "INSERT INTO .... "; 
    $result = mysql_query($query, $connection); 
}


    // End of Check IDcode already in database //


	// Requête SQL //
	$sql = 'INSERT INTO users VALUES("","'.$Fname.'","'.$Lname.'","'.$IDcode.'","'.$Region.'","'.$Depart.'")';
	$result = mysql_query($sql) or die ('Error : '.mysql_error() );

	// close database and table //
	mysql_close();

	// Result //
	if (FALSE == $result){ 
		$error['sql'] = "There has been an error adding in the database! Please try again.";
	} else {
		$error['ok'] = "Thank you ".$_SESSION['Fname']. ", you are now registerd in our database.";
	}
}
}


<p align="center">SIMPLE PREVIEW &amp; SUBMIT PAGE</p>
<FORM method="post" action="<?=$url;?>">
  <table width="50%" border="1" align="center">
    <tr> 
      <td>First Name:</td>
      <td><input type="text" name="Fname" value="<?=$Fname;?>">
        <font color="#FF0000">*</font> </td>
    </tr>
    <tr> 
      <td>Last Name:</td>
      <td><input type="text" name="Lname" value="<?=$Lname;?>">
        <font color="#FF0000">*</font> </td>
    </tr>
    <tr> 
      <td>IDcode:</td>
      <td><input type="text" name="IDcode" value="<?=$IDcode;?>">
        <font color="#FF0000">*</font> </td>
    </tr>
    <tr> 
      <td>Region &amp; Departement:</td>
      <td><input name="Region" value="<?=$Region;?>"></select>
        <font color="#FF0000">*</font> 
        <input  name="Depart" value="<?=$Depart;?>"></select>
        <font color="#FF0000">*</font></td>
    </tr>
    <tr> 
      <td>&nbsp;</td>
      <td><input name="action" type="hidden" value="add" /> <input type="submit" name="Submit" value="SEND TO DATABASE">

  </td>
</tr>
  </table>
  </FORM>
<p>&nbsp;</p>
<p><a href="page_4.php" target="_blank">SEE database!</a></p>
<?php
// message modification ou erreurs
if (isset($_POST['action']) && ('add' == $_POST['action']) && isset($error)) {
	$error = implode("<br />", $error);
	echo '<div style="color:#999;">'.$error.'</div>';
}
unset($error); ?>

[bodzan]

I bet this line:

mysql_select_db($base, $db) or die ('Error : '.mysql_error()); 

Should be:

$connection = mysql_select_db($base, $db) or die ('Error : '.mysql_error()); 

[oops73]

the error is pointing to these two lines:

$result = mysql_query($query,$connection); 

$num_rows = mysql_num_rows($result); 

im starting to get nuts over here =)

[oops73]

Is it possible that i have put the code in the wrong place??

[leatherback]

The error is pointing to the fact that $connection is not valid. THat is why bodzan changed the section where the connection is made.

J.

[bodzan]

...since you didn't define your $connection nowhere...
Correct the thing I told you to and the error should be solved...

[oops73]

i did that and still the same errors..

[bodzan]

..since this is beginning to make not much sense...

$result = mysql_query($query);

[oops73]

Originally posted by bodzan
..since this is beginning to make not much sense...

$result = mysql_query($query);

[/B]

now we are getting somewhere...

I only got this error now:

Sorry - this IDcode is already in registerd!Error : Duplicate entry '731216' for key 2

its working to the point that it print the message that i want ,but then i need it to go back to the form itself not just show the error.

Thanks so far!!!!

[bodzan]

...your getting an error in this line:

$sql = 'INSERT INTO users VALUES("","'.$Fname.'","'.$Lname.'","'.$IDcode.'","'.$Region.'","'.$Depart.'")';
        $result = mysql_query($sql) or die ('Error : '.mysql_error() ); 

I don't know why it is there in the first place because the if/else is working to avoin the particular error you're getting i.e. duplicate entry and since it has an error the code never finishes...

I'm guessing that this particular line is a leftover so try deleting it and then try your code once more...

[bodzan]

...I made a stupid suggestion back there...

$db = mysql_connect($host, $login, $password) or die ('Error : '.mysql_error());
$connection =        mysql_select_db($base, $db) or die ('Error : '.mysql_error());

    // Check IDcode already in database //
    $query = "SELECT $IDcode FROM users WHERE IDcode = '$IDcode'"; 
    $result = mysql_query($query, $connection); 

Should be:

$db = mysql_connect($host, $login, $password) or die ('Error : '.mysql_error());
        mysql_select_db($base, $db) or die ('Error : '.mysql_error());

    // Check IDcode already in database //
    $query = "SELECT $IDcode FROM users WHERE IDcode = '$IDcode'"; 
    $result = mysql_query($query, $db); 

i.e. Where you use $connection should be the $db since the second and optional parameter that mysql_query() accepts is a link to the connection that you made where $db is...

Sorry, about that...

[oops73]

Originally posted by bodzan
...I made a stupid suggestion back there...

$db = mysql_connect($host, $login, $password) or die ('Error : '.mysql_error());
$connection =        mysql_select_db($base, $db) or die ('Error : '.mysql_error());

    // Check IDcode already in database //
    $query = "SELECT $IDcode FROM users WHERE IDcode = '$IDcode'"; 
    $result = mysql_query($query, $connection); 

Should be:

$db = mysql_connect($host, $login, $password) or die ('Error : '.mysql_error());
        mysql_select_db($base, $db) or die ('Error : '.mysql_error());

    // Check IDcode already in database //
    $query = "SELECT $IDcode FROM users WHERE IDcode = '$IDcode'"; 
    $result = mysql_query($query, $db); 

i.e. Where you use $connection should be the $db since the second and optional parameter that mysql_query() accepts is a link to the connection that you made where $db is...

Sorry, about that... [/B]

Im with you, there was some errors in your post =)
Its working little better now.

however here is the "new" code. Im not sure about the part where the form shall return back upon error.

<?php


session_start();

// connection to database // 
$host = 'localhost';		// host
$login = 'root';	// login
$password = '';		// password 				
$base = 'testbase';	// name of database

if (isset($_COOKIE[session_name()])) {
	$url = $_SERVER['PHP_SELF'];
} else if (defined('SID')) {
	$url = $_SERVER['PHP_SELF'].SID;
}


// condition ? if yes : if no //
isset($_SESSION['Fname']) ? 	$Fname = $_SESSION['Fname'] : 	$Fname = NULL;
isset($_SESSION['Lname']) ? 	$Lname = $_SESSION['Lname'] :	$Lname = NULL;
isset($_POST['IDcode']) ? 		$IDcode = $_POST['IDcode'] : 	$IDcode = NULL;
isset($_SESSION['Region']) ? 	$Region = $_SESSION['Region'] :	$Region = NULL;
isset($_SESSION['Depart']) ? 	$Depart = $_SESSION['Depart'] :	$Depart = NULL;

// WORK now //
if (isset($_POST['action']) && ('add' == $_POST['action'])) {
	////////// ACTION /////////
	// condition ? if yes : if no //
	isset($_POST['Fname']) ? 	$Fname = $_POST['Fname'] : 	$Fname = NULL;
	isset($_POST['Lname']) ? 	$Lname = $_POST['Lname'] :	$Lname = NULL;
	isset($_POST['IDcode']) ? 	$IDcode = $_POST['IDcode'] : $IDcode = NULL;
    isset($_POST['Region']) ? 	$Region = $_POST['Region'] : $Region = NULL;
	isset($_POST['Depart']) ? 	$Depart = $_POST['Depart'] : $Depart = NULL;
	// check if empty // 
	if (empty($Fname)) {
		$error['emptyFname'] = 'Empty Fname.';
	} else if (empty($Lname)){
		$error['emptyLname'] = 'Empty Lname.';
	} else if (empty($IDcode)){
	    $error['emptyIDcode'] = 'Empty IDcode.';
    } else if (empty($Region)){	
	    $error['emptyRegion'] = 'Empty Region.';
	} else if (empty($Depart)){	
		$error['emptyDepart'] = 'Empty Depart.';
	} else {	


	// connexion to database and table //
	$db = mysql_connect($host, $login, $password) or die ('Error : '.mysql_error());
	 mysql_select_db($base, $db) or die ('Error : '.mysql_error());
	// OLD LINE mysql_select_db($base, $db) or die ('Error : '.mysql_error()); //

    // Check IDcode already in database //
	$query = "SELECT $IDcode FROM users WHERE IDcode = '$IDcode'"; 
    $result = mysql_query($query ,$db); 
    $num_rows = mysql_num_rows($result); 
    if ($num_rows>0) { 
    print ("Sorry - this IDcode is already in registerd!"); 
    } else { 
    $query = "page_3.php"; 
    $result = mysql_query($query); 
}

    // End of Check IDcode already in database //


	// Requête SQL //
	$sql = 'INSERT INTO users VALUES("","'.$Fname.'","'.$Lname.'","'.$IDcode.'","'.$Region.'","'.$Depart.'")';
	$result = mysql_query($sql) or die ('Error : '.mysql_error() );

	// close database and table //
	mysql_close();

	// Result //
	if (FALSE == $result){ 
		$error['sql'] = "There has been an error adding in the database! Please try again.";
	} else {
		$error['ok'] = "Thank you ".$_SESSION['Fname']. ", you are now registerd in our database.";
	}
}
}
?>

<p align="center">SIMPLE PREVIEW &amp; SUBMIT PAGE</p>
<FORM method="post" action="<?=$url;?>">
  <table width="50%" border="1" align="center">
    <tr> 
      <td>First Name:</td>
      <td><input type="text" name="Fname" value="<?=$Fname;?>">
        <font color="#FF0000">*</font> </td>
    </tr>
    <tr> 
      <td>Last Name:</td>
      <td><input type="text" name="Lname" value="<?=$Lname;?>">
        <font color="#FF0000">*</font> </td>
    </tr>
    <tr> 
      <td>IDcode:</td>
      <td><input type="text" name="IDcode" value="<?=$IDcode;?>">
        <font color="#FF0000">*</font> </td>
    </tr>
    <tr> 
      <td>Region &amp; Departement:</td>
      <td><input name="Region" value="<?=$Region;?>"></select>
        <font color="#FF0000">*</font> 
        <input  name="Depart" value="<?=$Depart;?>"></select>
        <font color="#FF0000">*</font></td>
    </tr>
    <tr> 
      <td>&nbsp;</td>
      <td><input name="action" type="hidden" value="add" /> <input type="submit" name="Submit" value="SEND TO DATABASE">

  </td>
</tr>
  </table>
  </FORM>
<p>&nbsp;</p>
<p><a href="page_4.php" target="_blank">SEE database!</a></p>
<?php
// message modification ou erreurs
if (isset($_POST['action']) && ('add' == $_POST['action']) && isset($error)) {
	$error = implode("<br />", $error);
	echo '<div style="color:#999;">'.$error.'</div>';
}
unset($error); ?>

[oops73]

so how do i get the "error" to go back to the form??

[bodzan]

$sql = 'INSERT INTO users VALUES("","'.$Fname.'","'.$Lname.'","'.$IDcode.'","'.$Region.'","'.$Depart.'")';
        $result = mysql_query($sql) or die ('Error : '.mysql_error() ); 

And what is it supposed to do? If your getting an error reading that there is a duplicate entry for some of the keys there, it is because of this line. From what I understood from your code you're checking for a duplicate entry and then display an error if there is one and that is O.K. BUT this particular line goes on and enteres the data that has duplicate key entry in it no matter what happened with the check...

I would suggest you loose it... And go from there...