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Everything posted by ra_ie_darkness
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javascript function does not return true in firefox
ra_ie_darkness replied to ra_ie_darkness's topic in Javascript Help
understandable but onsubmit did not work either. I still have the same problem -
javascript function does not return true in firefox
ra_ie_darkness replied to ra_ie_darkness's topic in Javascript Help
if(checkEmpty===true && checkNum===true && checkStr===true) didn't work -
javascript function does not return true in firefox
ra_ie_darkness posted a topic in Javascript Help
I have written a fucntion in javascript for validation. The submit button is inside a form <input type="submit" name="submit_request" value="Submit" id="submit_req" onclick="return checkQ(event);" style="backgroundColor:Transparent;border:0;color:blue;width:100;"/> This is the function function checkQ() { //Validation var checkEmpty; //check empty text field var checkNum; //check whether authorized value is greater than the requested value var checkStr //check whether string is entered var qLength = document.getElementsByName("pQuantity[]").length; for(i=0;i<qLength;i++) { var pValue = document.getElementsByName("pQuantity[]")[i].value; //authorized quantity var reQnty = document.getElementsByName("quantity[]")[i].value; //requested quantity if(pValue != "") { checkEmpty = true; } else { alert("Quantity missing"); checkEmpty = false; return false; } if(Number(pValue)>Number(reQnty)) { alert("greater value"); checkNum = false; return false; } else { checkNum = true; } if(!Number(pValue)) { alert("You are only allowed to enter a number"); checkStr = false; return false; } else { checkStr = true; } } if(checkEmpty==true && checkNum==true && checkStr==true) { alert("working"); return true; } } It works fine in chrome and IE but for some reason even when every condition is satifisfied the form is not submitted in firefox. I get the alert dialogue box saying "working" but nothing happens after that. how to fix it -
what I am trying to do is something similar to a shopping cart. I am using ajax to display the search results. when i click on one item from the results it gets added to another paragraph. The problem is that when i select the second item it is appended as string to the last item. For example if I select books and pens. It shows bookspens I want each item to be shown separately so that i can send the seperate values to a php script for inserting it as requested items. how can i achieve this without jquery This is the code search.php $keyword = mysql_real_escape_string($_POST['search_res']); $search_q = mysql_query("Select * from products where pname like '%$keyword%'"); if(mysql_num_rows($search_q)!=0) { while($result = mysql_fetch_array($search_q)) { $productid = $result['id']; $name = $result['pname']; echo "<input type='button' name='resultname' id='$productid' value='$name' onclick='throwval(this)'><br/>"; } } index.php <html> <head> <script language="javascript"> function showresult() { var product_name = document.getElementById("searchval").value; if(window.XMLHttpRequest) { XMLHttpRequestObject = new XMLHttpRequest(); } else if(window.ActiveXObject) { XMLHttpRequestObject = new ActiveXObject("Microsoft.XMLHTTP"); } XMLHttpRequestObject.open("POST", "search.php", true); XMLHttpRequestObject.setRequestHeader("Content-Type", "application/x-www-form-urlencoded"); XMLHttpRequestObject.send("search_res=" + product_name); XMLHttpRequestObject.onreadystatechange = function() { if(XMLHttpRequestObject.readyState == 4) { if(XMLHttpRequestObject.status == 200) { document.getElementById("displayresult").innerHTML = XMLHttpRequestObject.responseText; } } } } function throwval(obj) { var sent_id = obj.id; var v = document.getElementById(sent_id).value; var content = document.createTextNode(v); document.getElementById("newdiv").appendChild(content); } function sendvalues() { var data = document.getElementById("newdiv").textContent; alert(data); } </script> </head> <body> <!--Search Form--> <form method="post" name ="searchform" id="idsearchform" > <input type="text" name="search" id="searchval" /> <input type="button" name="starts" id="btn" value="startsearch" onclick="showresult()"/> </form> <div id="displayresult"> <!--Search results will be displayed here--> </div> Selected : <p id='newdiv'> <!--Selected values will be displayed here--> </p> <form name="form1" method="post" action="send_data.php"> <input type="button" id="sendtophp" name="sendingval" value="next step" onclick="sendvalues()"> </form> <p id='fetched'> </p> </body> </html>
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To validate email you can use filter_input() $email = filter_input(INPUT_GET,'email',FILTER_VALIDATE_EMAIL); You can check the username by sending a query to the database mysql_num_rows will return the number of rows containing the username. $q = mysql_query("Select * from table where username='$username'"); if(mysql_num_rows($q)!=0) { echo "Username exists"; }
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correction <form method="post" name ="searchform" id="idsearchform" > <input type="text" name="search"/> <input type="button" name="starts" value="search" onclick="showresult(search)"/> </form>
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Hello, I am new to ajax and javascript and am trying to create a search engine This is the javascript code var XMLHttpRequestObject = false; if(window.XMLHttpRequest) { XMLHttpRequestObject = new XMLHttpRequest(); } else if(window.ActiveXObject) { XMLHttpRequestObject = new ActiveXObject("Microsoft.XMLHTTP"); } function showresult(search) { if(XMLHttpRequestObject) { var obj = document.getElementById("showresult"); XMLHttpRequestObject.open("POST","index.php",true); XMLHttpRequestObject.setRequestHeader("Content-type","application/x-www-form-urlencoded"); XMLHttpRequestObject.onreadystatechange = function() { if(XMLHttpRequestObject.readystate == 4 $$ XMLHttpRequestObject.status == 200) { obj.innerHTML = XMLHttpRequestObject.responseText; } } XMLHttpRequestObject.send("search=" + search); } //alert("You clicked me"); } PHP + HTML: <form method="post" name ="searchform" id="idsearchform" > <input type="text" name="search"/> <input type="button" name="starts" value="search" onclick="fun(search)"/> </form> <?php if(isset($_POST['starts'])) { $keyword = $_POST['search']; $search_q = mysql_query("Select * from products where pname like '%$keyword%'"); if(mysql_num_rows($search_q)!=0) { while($result = mysql_fetch_array($search_q)) { ?> <div id="showresult"> <?php $name = $result['pname']; echo "$name<br/>"; } } }?> </div> I don't get the search results when i click the search button. What is the issue here
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Hello, I am trying to retrieve some data through my php file from an ms access data base $sql = "Select AM_CheckInOut.[CheckInOut_Joint] as nr,AM_CheckInOut.[CheckTime] from AM_CheckInOut where AM_CheckInOut.[Person_Joint]=100170 and AM_CheckInOut.[DeviceID]=2 and mid(AM_CheckInOut.[CheckTime],1,10)='$fixdate';"; echo $sql.'<br/>'; $rs = odbc_exec($odbc_con,$sql) or die("query is not working"); if(!$rs) { echo "unable to execute query"; } while ($rs1= odbc_fetch_array($rs)) { $maxid = $rs1['nr']; echo 'Maximum ID'.$maxid.'<br/>'; } The column AM_CheckInOut.[CheckTime] contains timestamps. That looks something like 2010-06-22 12:42:50. The problem is that when i run the code it does not give me any result however if i paste the same query in msaccess it does return some values. The query that is shown on the browser is Select AM_CheckInOut.[CheckInOut_Joint] as nr,AM_CheckInOut.[CheckTime] from AM_CheckInOut where AM_CheckInOut.[Person_Joint]=100170 and AM_CheckInOut.[DeviceID]=2 and mid(AM_CheckInOut.[CheckTime],1,9)='2011-01-11'; What could be the issue
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Problem with uploading huge files
ra_ie_darkness replied to ra_ie_darkness's topic in PHP Coding Help
I changed the settings in my php.ini file and it is working now. Thank you -
I'm trying to create a function to upload files on my server It is working fine but the problem is that it doesn't work when I try to upload files larger than 15/20 mb function uploadfile() { global $id; $getid = mysql_query("Select * from user where email='$id'"); while($result = mysql_fetch_array($getid)) { $userid = $result['id']; } $dir = "C:\\xampp\\htdocs\\filehost\\"; $dir.=$userid."\\"; if($_FILES['usr']['size']>2097152) //2megabytes { echo "You cannot upload files greater than 2mb"; } else { if(move_uploaded_file($_FILES['usr']['tmp_name'],$dir.$_FILES['usr']['name'])) { $fname = mysql_real_escape_string($_FILES['usr']['name']); $fsize = $_FILES['usr']['size']; $date = date("Y-m-d"); $dlurl = substr(md5(uniqid()), 0,10); $newdir = mysql_real_escape_string($dir); $fquery = mysql_query("SELECT * from folder where name ='$userid'"); while($row = mysql_fetch_array($fquery)) { $folderid = $row['name']; } echo "File uploaded"; if(!mysql_query("INSERT INTO myfile (fileid,userid,fid,name,size,date,location,dlurl) VALUES ('','$userid','$folderid','$fname','$fsize','$date','$newdir','$dlurl')")) { echo "Identify"; } else { echo "File uploaded"; } } } //echo "File moved"; } I have given a condition which does not allow files larger than 2 mb but instead of getting the message "You cannot upload files greater than 2mb" I get php error messages How can I fix this
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that worked, thank you
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okay i am now using move_uploaded_file here is how it looks like <?php class upload { function fupload() { echo "My name is: ".$_FILES['usr']['name']."<br/>"; echo "Location ".$_FILES['usr']['tmp_name']."<br/>"; echo "Size:".($_FILES['usr']['size'] / 1024)." kb <br/>"; echo "File type".$_FILES['usr']['type']."<br/>"; if(file_exists("htdocs/".$_FILES['usr']['name'])) { echo $_FILES['usr']['name']."already exists"; } else { move_uploaded_file($_FILES['usr']['tmp_name'],"/learning/".$_FILES['usr']['name']); echo "File moved"; } } } $up = new upload(); $up->fupload(); ?> is it possible that i am not using the function properly? learning is a folder inside htdocs(xampp) now I am getting this message Warning: move_uploaded_file(/learning/) [function.move-uploaded-file]: failed to open stream: Invalid argument in C:\xampp\htdocs\learning\uploadfile.php on line 19 Warning: move_uploaded_file() [function.move-uploaded-file]: Unable to move 'C:\xampp\tmp\phpA56E.tmp' to '/learning/' in C:\xampp\htdocs\learning\uploadfile.php on line 19
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changing the file name did not work still getting the same warning message
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Thanks, i will rename the file but does that mean there is no problem in my code other than the file name is there a problem if there is a space in the file name if so then how will i solve this problem considering a user wants to upload a file that has spaces in its name
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I'm using xampp on windows vista I am a beginner in php and am working on creating a file hosting website got a form to upload the file from the client in uploadform.php <html> <body> <form method="post" action="uploadfile.php" enctype="multipart/form-data"> <input type="file" name="usr" id="usr"/><br /> <input type="submit" value="Upload"/> </form> </body> </html> the file handling code is in uploadfile.php <?php class upload { function fupload() { $dir = "C:\\xampp\htdocs\learning\fileshare database.txt"; echo "My name is: ".$_FILES['usr']['name']."<br/>"; echo "Location ".$_FILES['usr']['tmp_name']."<br/>"; echo "Size:".($_FILES['usr']['size'] / 1024)." kb <br/>"; echo "File type".$_FILES['usr']['type']."<br/>"; if(file_exists("htdocs/".$_FILES['usr']['name'])) { echo $_FILES['usr']['name']."already exists"; } else { copy($_FILES['usr']['name'],$dir); echo "File moved"; } } } $up = new upload(); $up->fupload(); ?> The problem is that it does not copy the file from the tmp directory to the specified one the error i get is: Warning: copy(fileshare databse.txt) [function.copy]: failed to open stream: No such file or directory in C:\xampp\htdocs\learning\uploadfile.php on line 20 I am facing the same problem if i try move_uploaded_file () function Could any one help with this issue, thanks
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I am trying to create an index page which contains registration and login field the problem that i get is on successful login a warning is displayed session_start() [function.session-start]: Cannot send session cookie - headers already sent by (output started at C:\xampp\htdocs\Eventz.com\index.php:116) in C:\xampp\htdocs\Eventz.com\index.php on line 235 This is the login part of my index.php this tag is inside an html table below the login form I also have a registration form and its php code above the login form <?php if (isset($_REQUEST['pass'])) { $id=$_POST['id']; $pass=$_POST['pass']; $conn =mysql_connect("localhost","root",""); if (!$conn) { die('Could not connect: ' . mysql_error()); } /* checking connection....success! */ $e=mysql_select_db('test', $conn); if(!$e) { die(''.mysql_error()); } else { echo 'database selected successfully'; } if (isset($_REQUEST['id']) || (isset($_REQUEST['pass']))) { if($_REQUEST['id'] == "" || $_REQUEST['pass']=="") { echo "login fields cannot be empty"; } else { $sql=mysql_query("Select email,password from login where email='$id' AND password='$pass'"); $count=mysql_num_rows($sql); if($count==1) /* $count checks if username and password are in same row */ { session_start(); $_SESSION['id']=$id; echo "</br>Login Successful</br>"; } else { echo "</br>invalid</br>"; echo "please try to login again</br>"; } } } } ?> Any help or suggestion would be appreciated
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that worked thank you
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I'm using xampp and am trying to create a login session...these are the my php files login.php <form action='login1.php' method='post'> Email: <input type='text' name='id'></br> Pass : <input type='password' name='pass'></br> <input type='submit' name='login' value='login'> login1.php <?php $id=$_POST['id']; $pass=$_POST['pass']; $conn=mysql_connect("localhost","root",""); if (!$conn) { die('Could not connect: ' . mysql_error()); } $e=mysql_select_db('test', $conn); if(!$e) { die(''.mysql_error()); } else { echo 'database selected successfully'; } $sql=mysql_query("Select email,password from login where email='$id' AND password='$pass'"); $count=mysql_num_rows($sql); if($count==1) { session_start(); echo "</br>Login Successful</br>"; echo "Please wait 5 seconds "; /* redirct to the specified page */ header("refresh:5;url=empty.php"); } else { echo "please try to login again</br>"; echo "you will be redirected to the login page in 5 seconds"; /* redirct to the specified page */ header("refresh:5;url=login.php"); } mysql_close(); ?> empty.php <?php session_start(); $logi=$_SESSION['id']; echo 'welcome '.$logi; ?> empty.php is supposed to display the email id from login.php I checked it about an hour ago and it was working fine but now it gives me an error Undefined index: id in C:\xampp\htdocs\littleprogress\empty.php on line 4 I didn't change anything after checking it. only cleared the history and cookies of my firefox browser what could be the problem
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I need to send mail through localhost
ra_ie_darkness replied to ra_ie_darkness's topic in PHP Coding Help
yes I have posted those in sendmail.ini and have displayed that in this post (not the real password) -
I need to send mail through localhost
ra_ie_darkness replied to ra_ie_darkness's topic in PHP Coding Help
Thank you for the reply. Here is the detail of what i've tried so far I am trying to use google's smtp server these are the changes that i have made in php.ini [mail function] ; For Win32 only. ; http://php.net/smtp SMTP = smtp.gmail.com ; http://php.net/smtp-port smtp_port = 465 ; For Win32 only. ; http://php.net/sendmail-from ;sendmail_from = postmaster@localhost ; For Unix only. You may supply arguments as well (default: "sendmail -t -i"). ; http://php.net/sendmail-path sendmail_path = "\"C:\xampp\sendmail\sendmail.exe\" -t" Below is a part of send mail.ini smtp_server=smtp.gmail.com ; smtp port (normally 25) smtp_port=465 ; the default domain for this server will be read from the registry ; this will be appended to email addresses when one isn't provided ; if you want to override the value in the registry, uncomment and modify smtp_ssl=auto default_domain=google.com ; log smtp errors to error.log (defaults to same directory as sendmail.exe) ; uncomment to enable logging ;error_logfile=error.log ; create debug log as debug.log (defaults to same directory as sendmail.exe) ; uncomment to enable debugging ;debug_logfile=debug.log ; if your smtp server requires authentication, modify the following two lines [email protected] auth_password=mygmailpassword Finally this is the php file that i am trying to run <?php $to='[email protected]'; $subject='mysub'; $message='messagebody.'; $headers = 'From: [email protected]' . "\r\n" . 'Reply-To: [email protected]' . "\r\n" . 'X-Mailer: PHP/' . phpversion(); mail($to,$subject,$message,$headers); echo 'see'; ?> when i use the port 586 I only get "see" (from the last line of my php file) but the recipients have not received any mail but when i use port 465 the page keeps on loading forever and nothing happens. I need to know what i'm doing wrong and how to fix it -
Hello, I am trying to create a registration module where an email would be sent to the user after registration. I am using xampp and have tried php mailer http://sourceforge.net/projects/phpmailer/files/phpmailer%20for%20php4/0.90/ but couldn't get it to work... I've been told that it is possible to send an email from localhost through smtp... I need to know how to accomplish that. Am a beginner. suggestions and help would be appreciated