plodos
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Posts posted by plodos
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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?
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thnx for reply...but I want that
by the help of basic anchor tag that would move the reader to another page on the same website or to another website...
i want to control the HREF with anchor tag
but the coding is proglemm..
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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...
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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...
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<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
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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'; } } ?>
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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=="" | $a==NULL)
send email without attachment
else
send with attachment
I want to do this...Pls help :s
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$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....???
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yes you are right...problem was the file name sorry
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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>"; } }} ?>
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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
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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 $id;
if I write like that....Output is only Resource id #4
What must I write for see the true result...??
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$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?
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.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 $query....there is no error.....but it is not recording the data
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$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 $query; // new line
output like that select * from patient where pt_name=laura AND pt_surname=
surname is blank...
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hımm..I need
everything is possible, if you are good programmer:) but I didnt find the solution to GET ID..
Maybe there is a another way to compare the strings in the _update.php page...
If the strings are equal, try to get the ID try:)
But also I want to learn this way, by the help of simple form, send the SELECTED ID....????
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Who knows ?
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$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 ?
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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 '%".$surname."%' ";
page shows nothing...why ?
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$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 ?
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<?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
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<?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
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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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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
combobox selection problem
in PHP Coding Help
Posted
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