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plodos

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    combobox selection problem

    in PHP Coding Help

    Posted 

    <form>
<select name="contact">

<option value="List">List</option>
<option value="Plen">Plen</option>

</select>


<!-- if the variable "List" is sent then display the field -->

<?php if(isset($_GET['List'])){?>

<td>Code</td>

<td>
<input name="add list" type="text" id="add list" value="<?php echo $_GET['add list'];?>" size="28"/></td> 

<?php ;}?>

</form>

     

    If the user select the "List" from the combobox, From will add "add list" input field, this must be visible/appeared in the form...

     

    If user select "Plen", there is no need to show "add list" field...

     

    You understood what I mean :)

     

    But this code is not working, I didnt do it....How can do it?, Who knows the true code..Pls help me

    verificationimage is not working ?

    in PHP Coding Help

    Posted

    verificationimage.php

    <?php
header('Content-type: image/jpeg');

$width = 50;
$height = 24;

$my_image = imagecreatetruecolor($width, $height);

imagefill($my_image, 0, 0, 0xFFFFFF);

// add noise
for ($c = 0; $c < 40; $c++){
$x = rand(0,$width-1);
$y = rand(0,$height-1);
imagesetpixel($my_image, $x, $y, 0x000000);
}

$x = rand(1,10);
$y = rand(1,10);

$rand_string = rand(1000,9999);
imagestring($my_image, 5, $x, $y, $rand_string, 0x000000);

setcookie('tntcon',(md5($rand_string).'a4xn'));

imagejpeg($my_image);
imagedestroy($my_image);
?>

     

    Im using this link for show the image....

    <img src="verificationimage.php?<?php echo rand(0,9999);?>" width="50" height="24" align="absbottom" />

     

    At the before I didnt see any problem..when I changed my hosting company, this script is not show anything(JPEG files)

    What can be the reason?

    how to? ... #top #bottom #center such and such

    in HTML Help

    Posted

    index.html

    <a href=”country_list.html#bulgaria”>Click for Bulgaria </a>
<a href=”country_list.html#greece”>Click for Greece </a>

     

    country_list.html

    <html>
greece
xxxxxx
aaaaaa
bbbbbb
tttttttt 
.........
bulgaria
.........
.........
.........
.........
.........
</html>

     

    Sometimes, I see the some codes like #top #bottom #center

     

    bulgaria information is center of the coutry_list.html...

     

    If the user click the bulgaria in the index page.....user will directly go to bulgaria which is center in the country_list.html

    but I dont know how to do it :S

     

    who can give me the example, pls...

    how to? ... #top #bottom #center such and such

    in HTML Help

    Posted

    index.html

    <a href=”htmllink.html#bulgaria”>Click for Bulgaria </a>
<a href=”htmllink.html#greece”>Click for Greece </a>

     

    country_list.html

    <html>
greece
xxxxxx
aaaaaa
bbbbbb
tttttttt 
.........
bulgaria
.........
.........
.........
</html>

    Sometimes, I see the some codes like #top #bottom #center

     

    bulgaria information is center of the coutry_list.html...

    If the user click the bulgaria in of the index page.....user will directly go to bulgaria center in the country_list.html

    but I dont know how to do it :S

     

    who can give me the example, pls...

    onSubmit, how to combobox validate?

    in Javascript Help

    Posted 

    <script LANGUAGE="JavaScript">
<!--
function ValidateForm(form){
ErrorText= "";
if ( form.age.selectedIndex == 0 ) 
{ 
alert ( "Please select your Age." ); return false; 
}
if (ErrorText= "") 
{ 
form.submit() }
}
-->
</script>

<form name="feedback" action="mailto:aaa@aaa.com" method=post   onSubmit="ValidateForm(this.form)">
Your Age: <select name="age"> 
<option value="">Please Select an Option:</option> 
<option value="0-18 years">0-18 years</option> 
<option value="18-30 years">18-30 years</option> 
<option value="60+ years">60+ years</option> 
</select>
<input type="button" name="SubmitButton" value="Submit">
<input type="reset" value="Reset">
</form>

    I want to validate the combobox is selected or not...but onSubmit is not working :S

    I have already onclick evet I dont know how to combine two onclick event in one function:S

     

    What can be the errors..I couldnt find it :s

    php check empty string ??

    in PHP Coding Help

    Posted

    of course I used trim and empty...but problem is, I didnt write the code

    maybe conditions has problem .. this is another one...

    original script is working..but when I want to seperate..its not working :s

    who can help me for the coding part...

    <?php
$name = $_POST[name];
$surname = $_POST[surname];
$email = $_POST[email];
$msg_title = "alert";
$to="alert@xxx.com";
$file = $_POST[fileatt];
$check = trim($file);

if (empty($check))
{
just send email........ mail();
}

else {

$message= '
<table width="522" height="235" border="1">
  <tr>
    <td>Name Surname </td>
    <td>'.$name.' '.$surname.'</td>
  </tr>
</table>';

  // Obtain file upload variables
            $fileatt      = $_FILES['fileatt']['tmp_name'];
            $fileatt_type = $_FILES['fileatt']['type'];
            $fileatt_name = $_FILES['fileatt']['name'];

            $headers = "From: $email  \n";
//upload file type
$FILE_EXTS = array('.txt'); 
$file_name = $_FILES['fileatt']['name'];
$file_ext = strtolower(substr($file_name,strrpos($file_name,".")));
if (!in_array($file_ext, $FILE_EXTS)){
include 'problem.html';
exit();
}

           // if($_FILES['fileatt']['size'] > 0)
           if (is_uploaded_file($fileatt)) {
           // Read the file to be attached ('rb' = read binary)
            $file = fopen($fileatt,'rb');
            $data = fread($file,filesize($fileatt));
            fclose($file);
            // Generate a boundary string
            $semi_rand = md5(time());
            $mime_boundary = "==Multipart_Boundary_x{$semi_rand}x";
            // Add the headers for a file attachment
            $headers .= "MIME-Version: 1.0\n" .
                        "Content-Type: multipart/mixed;\n" .
                        " boundary=\"{$mime_boundary}\"";
            // Add a multipart boundary above the  message
            $message = "This is a multi-part message in MIME format.\n\n" .
            "--{$mime_boundary}\n" .
            "Content-Type: text/html; charset=\"iso-8859-1\"\n" .
            "Content-Transfer-Encoding: 7bit\n\n" .
            $message . "\n\n";

            // Base64 encode the file data
            $data = chunk_split(base64_encode($data));
             // Add file attachment to the message
            $message .= "--{$mime_boundary}\n" .
            "Content-Type: {$fileatt_type};\n" .
            " name=\"{$fileatt_name}\"\n" .
            //"Content-Disposition: attachment;\n" .
            //" filename=\"{$fileatt_name}\"\n" .
            "Content-Transfer-Encoding: base64\n\n" .
            $data . "\n\n" .
            "--{$mime_boundary}--\n";
             }else echo "File error!  ";

            if(mail($to,$msg_title,$message,$headers)) 
            {
            	include'ok.html';
            }
            else
            { 
            	include 'problem.html';
            }
}
?>

    php check empty string ??

    in PHP Coding Help

    Posted

    I have a form like that ( email form with attachment)

    <?php
$name = $_POST[name];
$surname = $_POST[surname];
$email = $_POST[email];
$msg_title = "alert";
$to="alert@xxx.com";

$message= '
<table width="522" height="235" border="1">
  <tr>
    <td>Name Surname </td>
    <td>'.$name.' '.$surname.'</td>
  </tr>
</table>';

  // Obtain file upload variables
            $fileatt      = $_FILES['fileatt']['tmp_name'];
            $fileatt_type = $_FILES['fileatt']['type'];
            $fileatt_name = $_FILES['fileatt']['name'];

            $headers = "From: $email  \n";
//upload file type
$FILE_EXTS = array('.txt'); 
$file_name = $_FILES['fileatt']['name'];
$file_ext = strtolower(substr($file_name,strrpos($file_name,".")));
if (!in_array($file_ext, $FILE_EXTS)){
include 'problem.html';
exit();
}

           // if($_FILES['fileatt']['size'] > 0)
           if (is_uploaded_file($fileatt)) {
           // Read the file to be attached ('rb' = read binary)
            $file = fopen($fileatt,'rb');
            $data = fread($file,filesize($fileatt));
            fclose($file);
            // Generate a boundary string
            $semi_rand = md5(time());
            $mime_boundary = "==Multipart_Boundary_x{$semi_rand}x";
            // Add the headers for a file attachment
            $headers .= "MIME-Version: 1.0\n" .
                        "Content-Type: multipart/mixed;\n" .
                        " boundary=\"{$mime_boundary}\"";
            // Add a multipart boundary above the  message
            $message = "This is a multi-part message in MIME format.\n\n" .
            "--{$mime_boundary}\n" .
            "Content-Type: text/html; charset=\"iso-8859-1\"\n" .
            "Content-Transfer-Encoding: 7bit\n\n" .
            $message . "\n\n";

            // Base64 encode the file data
            $data = chunk_split(base64_encode($data));
             // Add file attachment to the message
            $message .= "--{$mime_boundary}\n" .
            "Content-Type: {$fileatt_type};\n" .
            " name=\"{$fileatt_name}\"\n" .
            //"Content-Disposition: attachment;\n" .
            //" filename=\"{$fileatt_name}\"\n" .
            "Content-Transfer-Encoding: base64\n\n" .
            $data . "\n\n" .
            "--{$mime_boundary}--\n";
             }else echo "File error!  ";

            if(mail($to,$msg_title,$message,$headers)) 
            {
            	include'ok.html';
            }
            else
            { 
            	include 'problem.html';
            }
//}
?>

     

    I need to control if 

    <input name="fileatt" type="file" id="fileatt" size="60">

    "fileatt" is posted empty or not...

     

    if "fileatt" is empty or null send email without upload...

     

    but if fileatt has some value  send it with attachment...

     

    I dont know how many if else combination I tested....like

     

    $a = $_POST[fileatt];

     

    if($a=="" | $a==NULL)

    send email without attachment

    else

    send with attachment

     

    I want to do this...Pls help :s

    php mail and headers problem???

    in PHP Coding Help

    Posted 

    $to = $_REQUEST["email"];
$title = $_REQUEST["title"];
$from= "info@xxx.com";
$content = "Advertisement";

$headers  = "MIME-Version: 1.0\r\n";
$headers .= "Content-type: text/html; charset=iso-8859-1\r\n";
$headers .= "To: $to\r\n";
$headers .= "From: $from\r\n";

mail($to,$subject,$content,$headers);

    script is getting the receiver email ($to) and

    inside of the headers $to(josh@asd.com) and $from(nfo@xxx.com)

     

    Problem is that......when I send the email josh@asd.com also script is automaticaly sending the email to me info@xxx.com..!!!!

     

    I dont want to take a copy of this email in this email box info@xxx.com..but sender must see the From: part....what must I do....???

    php get page error....

    in PHP Coding Help

    Posted

    Error is

    Not Found

    The requested URL /add.php was not found on this server.

    Additionally, a 404 Not Found error was encountered while trying to use an ErrorDocument to handle the request.

     

    İm trying to learnd $_GET method....Whats the error, I didnt see...Pls help

    a.php

    if($add_person)
{	header('Location: http://www.xxx.com/add.php?cpage='.$count_commit.'     ');
}

    b.php

    <?php
include("dbconfig.php");

switch($_GET['cpage']) {
    case "1":
        check(1);
        break;
    case "2":
      check(2);
        break;
    default:
    	echo "error";
}

function check($number){
$search = $number;

$data = mysql_query("select p.*
                     from join_person_committee j, person p
                     where j.person_id = p.person_id AND j.person_id='$search'
                     ORDER BY p.person_id ASC LIMIT $baslangic,$limit"); 

while($info=mysql_fetch_array($data))
{
	echo"			{$info['university']}<br>";
}

}}
?>

    how to make php link

    in PHP Coding Help

    Posted

    How can I do that link...

    paged=2 & s=xxxx

    I need to make this, than I will control with switch but what is the name of this ......

    how will I search on the google for learn...or if you have sample code, could you share it...

    like : www.2checkout.com/community/?paged=2&s=xxxxxx

    simple SESSION problem

    in PHP Coding Help

    Posted

    sorry but

    i didnt understand the solution way?

     

    $dd_name=$_SESSION['PassWord'];
$sql = "select d_name from doctor where password='$dd_name'  ";
$id = mysql_query($sql);
echo $id;

    $dd_name=$_SESSION['PassWord'];
$sql = "select d_id from doctor where password='$dd_name'  ";
$id = mysql_query($sql);
echo $id;

    if I write like that....Output is only Resource id #4

     

    What must I write for see the true result...??

     

    simple SESSION problem

    in PHP Coding Help

    Posted 

    $sql="SELECT d_name,password FROM doctor WHERE d_name='$myusername' and password='$mypassword'";
    $result=mysql_query($sql);
    // Mysql_num_row is counting table row
    $count=mysql_num_rows($result);
    // If result matched table row must be 1 row
    if($count==1){
	$_SESSION['loggedInD'] = true;
	$_SESSION['UserName'] = $myusername;
	$_SESSION['PassWord'] = $mypassword;
	header('location:doctor.php?'.SID.'');
}

    There is a problem

    <?
session_start();
include("dbconfig.php");
if (!$_SESSION['loggedInD']) 
{
header("location:login.php"); 
die ();
}
else
{ 
$dd_name=$_SESSION['PassWord'];
$sql = "select d_id from doctor where password like  '%$dd_name%'  ";
echo mysql_query($sql);
echo "\n";
echo $sql;
}

    output is like that

    Resource id #5 select d_id from doctor where password like '%lukeW%'

    but LUKE ID is 1...

     

    $dd_name=$_SESSION['PassWord']; //this is working..

    but there is a problem here, $sql = "select d_id from doctor where password like  '%$dd_name%'  ";

    im my opinion, we can use the session variables in the SQL querys..

     

    What can be the error?

    MySQL UPDATE problem ???

    in PHP Coding Help

    Posted

    .sql

    CREATE TABLE department (
   dept_id INT NOT NULL,
   dept_name TEXT,
   PRIMARY KEY (dept_id)
);

DROP TABLE IF EXISTS proficiency;

CREATE TABLE proficiency (
   p_id INT NOT NULL, 
   p_name TEXT NOT NULL,
   PRIMARY KEY (p_id)
);

DROP TABLE IF EXISTS doctor;

CREATE TABLE doctor (
   d_id INT NOT NULL PRIMARY KEY,
   d_name varchar(50) NOT NULL,
   d_surname VARCHAR(50) NOT NULL,
   d_add TEXT NOT NULL,
   SSN INT(4) NOT NULL, 
   d_tel INT(4) NOT NULL,
   d_bdate DATETIME NOT NULL,  # 2003-03-31 11:22:12
   p_id INT NOT NULL,          
   dept_id INT NOT NULL,
   password VARCHAR(50) NOT NULL,
   FOREIGN KEY (dept_id) REFERENCES department (dept_id),
   FOREIGN KEY (p_id) REFERENCES proficiency (p_id)    
);

     

    I want to update the doctor information, like proficiency or password

    trick point is multiple querys :)

    this is my query...whats wrong...

    $_u ="UPDATE doctor dr, proficiency p, department d 
					SET  dr.d_id = '$id',
					     dr.d_name='$_d_name', 
					     dr.d_surname='$_d_surname', 
					     dr.d_add='$_d_add',
					     dr.SSN='$_SSN', 
					     dr.d_tel='$_d_tel', 
                                                     dr.d_bdate='$_d_bdate',
                                                     p.p_id='$prof_id',
                                                    dr.dept_id='$dept_id',
                                                    dr.password = '$password'
                            
					WHERE dr.d_id ='$id'    AND 
					d.dept_id = '$dept_id'  AND
					p.p_id = '$prof_id'     AND dr.dept_id = d.dept_id AND dr.p_id = p.p_id ;";

$query=mysql_query($_u);

    I test the $query....there is no error.....but it is not recording the data

    mysql_fetch_array ERROR ?

    in PHP Coding Help

    Posted 

    $query = "select * from patient where pt_name like '%".$name."%' AND pt_surname like '%".$surname."%' ";

     

    I delete this line 

    AND pt_surname like '%".$surname."%'

     

    now, the script is working correctly but only searching the name

     

    but  

    $query = "select * from patient where pt_name like '%".$name."%' AND pt_surname like '%".$surname."%' ";  
echo $query; // new line

    output like that select * from patient where pt_name=laura AND pt_surname=

    surname is blank...

     

    php combobox selected ID problem ?

    in PHP Coding Help

    Posted

    hımm..I need :)

     

    everything is possible, if you are good programmer:) but I didnt find the solution to GET ID..

     

    Maybe there is a another way to compare the strings in the _update.php page...

    If the strings are equal, try to get the ID :) try:)

     

    But also I want to learn this way, by the help of simple form, send the SELECTED ID....???? 

    mysql_fetch_array ERROR ?

    in PHP Coding Help

    Posted 

    $name      = $HTTP_POST_VARS['name'];
$surname   = $HTTP_SERVER_VARS['surname'];
  
$query = "select * from patient where pt_name like '%".$name."%' AND pt_surname like '%".$surname."%' ";  


Print "<table border cellpadding=1>";

$data = mysql_query($query); 
while($info=mysql_fetch_array($data))
{
Print "<tr>";
Print "<th>Patient ID:</th> <td>{$info['pt_id']}</td> ";
Print "<th>Name:</th> <td>{$info['pt_name']}</td> ";
Print "<th>Surname:</th> <td>{$info['pt_surname']}</td> ";
Print "<th>SSN:</th> <td>{$info['pt_SSN']}</td> ";
Print "<th>Add:</th> <td>{$info['pt_add']} </td>";
Print "<th>Tel:</th> <td>{$info['pt_tel']} </td>";
Print "<th>Blood:</th> <td>{$info['pt_blood']} </td>";
Print "<th>Visit Date:</th> <td>{$info['pt_visit']} </td>";
echo "<td><a href=\"_edit.php?id=".$info['pt_id']."\"> EDIT </a>";
Print"</tr>";
}

Print "</table>";

    it gives an error like mysql_fetch_array() supplied ..... why ?

    mysql_fetch_array ERROR ?

    in PHP Coding Help

    Posted

    if I write this sentence

    select * from patient  where pt_name like '%laura%' AND pt_surname like '%sams%'

    php script is working...

     

    but if I write

    $name      = $HTTP_POST_VARS['name'];

    $surname  = $HTTP_POST_VARS['surname'];

     

    $name = addslashes($name);

    $surname = addslashes($surname);

     

    $query = "select * from patient where pt_name like '%".$name."%' AND pt_surname like '%".$surname."%' ";

    page shows nothing...why ?

    mysql_fetch_array ERROR ?

    in PHP Coding Help

    Posted 

    $name      = $HTTP_POST_VARS['name'];
$surname   = $HTTP_SERVER_VARS['surname'];
  
$query = "select * from patient where pt_name like '%".$name."%' AND pt_surname like '%".$surname."%' ";  


Print "<table border cellpadding=1>";

$data = mysql_query($query); 
while($info=mysql_fetch_array($data))
{
Print "<tr>";
Print "<th>Patient ID:</th> <td>{$info['pt_id']}</td> ";
Print "<th>Name:</th> <td>{$info['pt_name']}</td> ";
Print "<th>Surname:</th> <td>{$info['pt_surname']}</td> ";
Print "<th>SSN:</th> <td>{$info['pt_SSN']}</td> ";
Print "<th>Add:</th> <td>{$info['pt_add']} </td>";
Print "<th>Tel:</th> <td>{$info['pt_tel']} </td>";
Print "<th>Blood:</th> <td>{$info['pt_blood']} </td>";
Print "<th>Visit Date:</th> <td>{$info['pt_visit']} </td>";
echo "<td><a href=\"_edit.php?id=".$info['pt_id']."\"> EDIT </a>";
Print"</tr>";
}

Print "</table>";

    it gives an error like mysql_fetch_array() supplied ..... why ?

    php combobox selected ID problem ?

    in PHP Coding Help

    Posted 

    <?php 
<form action="_update.php" method="post" id="_editform" >
   echo'<select name="proficiency">';
   $res=mysql_query("select * from proficiency");
   if(mysql_num_rows($res)==0)
   {
   echo "there is no data in table..";
   }
   else
   {
   while ($row = mysql_fetch_assoc($res) )
   {
   $id = $row['p_id'];
$name = $row['p_name'];
print "<option id='$id'>$name</option>\n";
    }
    }
echo'</select>';

<input type="submit" name="Submit" value="SAVE ">
</form>
?>

    _update.php

    $_POST['proficiency']; it only shows the selected name...not for me ! ....

     

    How can I take the SELECTED INDEX NUMBER..I used form..This form will send only the selected ID...

    this is my proficiency table.....

    1 PROF. DR

    2 DOC. DR

    3 ASS. PROF. DR

    4 DR

     

     

    [SOLVED] PHP COMBOBOX problem SOLVED THNX!!!

    in PHP Coding Help

    Posted 

    <?php 
<form action="_update.php" method="post" id="_editform" >
  	echo'<select name="proficiency">';
  	$res=mysql_query("select * from proficiency");
  	if(mysql_num_rows($res)==0)
  	{
  		echo "there is no data in table..";
  	}
  	else
  	{
  		while ($row = mysql_fetch_assoc($res) )
  		{
  		$id = $row['p_id'];
	$name = $row['p_name'];
	print "<option id='$id'>$name</option>\n";
    }
    }
echo'</select>';

<input type="submit" name="Submit" value="SAVE ">
</form>
?>

     

    _update.php

    $_p_id    = $_POST['proficiency'];    // it show the selected name           

    How can I take the OPTON ID  

    $_p_id    = $_POST['id']; //it is not working

       

    [SOLVED] PHP COMBOBOX problem SOLVED THNX!!!

    in PHP Coding Help

    Posted

    Dbase department

    CREATE TABLE department (
   dept_id INT NOT NULL,
   dept_name TEXT,
   PRIMARY KEY (dept_id)
);

    combobox code

    <?php 
  echo'<select name="department">';
  $res=mysql_query("select * from department");
  if(mysql_num_rows($res)==0) 
  echo "there is no data in table..";
  else
  for($i=1;$i<=mysql_num_rows($res);$i++) {
  $row=mysql_fetch_assoc($res);
  echo"<option>$row[dept_name]</option>";
  }
echo'</select>';
?>

     

    it is listing the department names....but I need to take IDs...

     

    1 PEDIATRICS

    2 DIABETES

    3 CARDIOLOGY

     

    How to use selectedIndex or how can I get the number of selected ındex

    [SOLVED] PHP COMBOBOX problem SOLVED THNX!!!

    in PHP Coding Help

    Posted

    Dbase department

    CREATE TABLE department (
   dept_id INT NOT NULL,
   dept_name TEXT,
   PRIMARY KEY (dept_id)
);

    combobox code

    <?php 
  echo'<select name="department">';
  $res=mysql_query("select * from department");
  if(mysql_num_rows($res)==0) 
  echo "there is no data in table..";
  else
  for($i=1;$i<=mysql_num_rows($res);$i++) {
  $row=mysql_fetch_assoc($res);
  echo"<option>$row[dept_name]</option>";
  }
echo'</select>';
?>

     

    it is listing the department names....but I need to take IDs...

     

    1 PEDIATRICS

    2 DIABETES

    3 CARDIOLOGY

     

    How to use selectedIndex or how can I get the number of selected ındex

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