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PHP - Writing Form Output Directly To Csv File, Bad Practice?
Hi,
I need to create a landing page with a form. That form needs to be recorded somewhere instead of sent to email. I know I can write it to a SQL database, and then to an excel file. But I only need a temporary solution so I figured I'd just go straight to CSV.
Is this bad practice? What potential problems might I encounter other than security issues?
Similar TutorialsI just wrote a php page that pulls live data from a database and presents it in an XML layout. In Internet Explorer, I can Edit, Select All, then Copy-Paste and save as XML. The resulting XML works fine. In Firefox, I can click File, Save Page As, select Text Files, then name it as an XML. Again, it works fine as XML. So..... how do I skip the Copy/Paste (IE) or Save Page As (Firefox) steps and write the output directly to an XML file? I've played with stdout(), buffers, page capturing utilities (wget, cURL, httrack, etc....), but no luck so far. This creates a pretty intense page due to the amount of data involved (1.3+ MB). Any help is appreciated! I'm trying to write a new csv file using the data from a large 1.4gb csv file. The problem is that its not working with the code I have below. It just hangs up when the size of output (tempfile) reaches 848661. Any ideas?
$file = "file.csv";
$file_number = 1;
if (!$output = fopen($tempfile, 'w')) {
unlink($tempfile);
die('could not open temporary output file');
}
$in_data = array();
$header = true;
$row = 0;
while (($data = fgetcsv($handle, 0, $delimiter)) !== FALSE) {
if ($header) {
$in_data[] = $headers;
fputcsv($output, $headers, "\t");
$header = false;
continue;
}
/*
if ( ($row % 10) == 0 ){
error_log("sleeping on row: $row");
sleep(3);
}
error_log("checking row: $row");
*/
if (!$val = getProductRow($data)) //this returns an array
continue;
$stat = fstat($output);
error_log("Stat is: " . print_r($stat['size'],true));
if($stat['size'] > 9437184){
error_log("saving the $file");
fputcsv($output, $val, "\t");
error_log(__LINE__);
fclose($output);
error_log(__LINE__);
if (file_exists($file))
unlink($file);
error_log(__LINE__);
rename($tempfile, $file);
error_log(__LINE__);
chmod($file, 0777);
error_log(__LINE__);
$final_file_name = $file = "file$file_number.csv";
error_log(__LINE__);
$tempfile = tempnam(".", "tmp"); // produce a temporary file name, in the current directory
error_log("creating the $file");
if (!$output = fopen($tempfile, 'w')) {
error_log(__LINE__);
unlink($tempfile);
die('could not open temporary output file');
}
error_log(__LINE__);
fputcsv($output, $headers, "\t");
error_log(__LINE__);
$file_number++;
error_log(__LINE__);
continue;
}
error_log(__LINE__);
fputcsv($output, $val, "\t");
}
Hey guys i've been trying to find out how to create and and write in a file, so that you can for example add a nav button to your website by form (you know like a admin panel). It tried some things but it doesnt work cause it wont write to the file... here is the code i've used: $ourFileName = $_POST['sitename']; $ourFileHandle = fopen($ourFileName, 'w') or die("can't open file"); fclose($ourFileHandle); $myFile = $_POST['sitename']; $fh = fopen($myFile, 'w') or die("can't open file"); $stringData = "test\n"; fwrite($fh, $stringData); $stringData = "test\n"; fwrite($fh, $stringData); fclose($fh); the "sitename" is the id from the form where you choose what the menu tab should be called. I would really glad if you could help me out with this one thanks! MinG Hi Everyone, I'm new to PHP freaks, and I'm hoping someone might be able to help me. I have written some code for a html page and used php to retrieve confirm whether or not data is in a text file. I also tried to write some code to insert the data supplied to my html page to the text file but it's not working. Can someone help me figure out what my issue is. I have attached my text file, and my php code as well. Below you'll find the code I used for my html page. Thank you for all your help, Phee <html xmlns="http://www.w3.org/1999/xhtml"> <head> <title>Telephone Directory</title> </head> <body> <form action='SignGuestBook.php' method='post'> <h1>Sign Guest Book</h1> <hr> <br> <table align='Left'> <tr> <td>Name: </td> <td><input name='name' /></td> </tr> <tr> <td>E-mail: </td> <td><input name='email' /></td> </tr> <tr> <td><input type="submit" value='Sign' /></td> <td><input type="reset" value='Reset Form' /></td> </tr> </table> <h3></h3> <br> <h4></h4> <br> <h5></h5> <br> <h6></h6> <br> <h7></h7> <br> <hr> <a href="http://helios.ite.gmu.edu/~smohamu2/IT207/Lab%20Assignment%208/AddNew.html">View Guest Book</a> </form> </body> </html> [attachment deleted by admin] How to do it? I have read some methods but they seem to not work. This is what I have got so far:
<?php Hi all, i want to download a file from the server but instead of storing it in the downloads i want it to store it directly in the folder i want and i also dont want to show any download window that appears while we download any file. Friend please help..... Hi there, Can anybody help me to write php/javascript code which will allow users to open files directly from web browser into desktop application? Here is the specification: I have a photo editing business with many people working in Photoshop. I am currently developing a web based application (joblist) using Javascript and PHP which should allow the photoshop designers to browse and open files/images directly from joblist/web browser into photoshop. The reason I want this instead of browsing folder is that I have a database where I store who worked on which file, when and how long it took. The concept is that, designers will select a file and click on start, as soon as they click on start the original file will open in Photoshop and there will be an entry into database (using PHP). Once they finish the task they will close the file and click on Finish button. My joblist application will be published in a local server and the file will be open on a local network, so when they save the file it will be saved where the source file is located in (local server). The application should work in both PC and Mac. I have already done all other part of the application except file opening directly from browser to desktop application functionality. Anybody can help me to write the code (PHP or Javascript) which can open the file from browser (local server) directly into desktop application e.g. PHotoshop or Illustrator? Thank you very much I look forward to someone's real help! Best regards Mr. Sumon I need to add these lines to my .htaccess file via PHP Code: [Select] RewriteCond %{REQUEST_URI} checkout RewriteRule ^(.*)$ https://myurl.com/checkout/$1 [R,L] RIGHT after this line: Code: [Select] RewriteBase / AND before this line: Code: [Select] RewriteCond %{REQUEST_FILENAME} !-f This is how the current file looks w/o the new line addition: Code: [Select] RewriteEngine On RewriteBase / RewriteCond %{REQUEST_FILENAME} !-f RewriteCond %{REQUEST_FILENAME} !-d RewriteRule ^(.+)$ index.php [L,QSA] It needs to look like this when the new lines are added(extra line breaks not necessary): Code: [Select] RewriteBase / RewriteCond %{REQUEST_URI} checkout RewriteRule ^(.*)$ https://myurl.com/checkout/$1 [R,L] RewriteCond %{REQUEST_FILENAME} !-f RewriteCond %{REQUEST_FILENAME} !-d RewriteRule ^(.+)$ index.php [L,QSA] The line numbers may not match up, so I cannot use the line number. Greetings,
I am new to these forums, I am working on this assignment, and these are the current issues I am running into.
Notice: Undefined variable: year in G:\EasyPHP-5.3.2i\www\PHP_Projects\ChineseZodiacs\zodiac_year_switch.php on line 77
ie. $year = validateInput($year,"Birth Year");
Notice: Undefined variable: year_count in G:\EasyPHP-5.3.2i\www\PHP_Projects\ChineseZodiacs\zodiac_year_switch.php on line 138
ie. echo "<p>You are person " . $year_count . "to enter " . $year . "</p>\n";
Honestly, I believe they are linked, because what should be happening, as the user enters the year, and hits submit, it should create a file called counts/$year.txt - $year should equal the entered data in the textbox, any help would be appreciated.
Thank you for your help.
<!DOCTYPE html>
<head>
<title>Write to and From a File</title>
<meta http-equiv="content-type" content="text/html; charset=iso-8859-1" />
<?php
$dir = "counts";
if ( !file_exists($dir)) {
mkdir ($dir, 0777);
}
function validateInput($year, $fieldname) {
global $errorCount;
if (empty($year)) {
echo "\"$fieldname\" is a required field.<br />\n";
++$errorCount;
$retval = "";
}
else
{ // if the field on the form has been filled in
if(is_numeric($year))
{
if($year >=1900 && $year <=2014)
{
$retval = $year;
}
else
{
++$errorCount;
echo "<p>You must enter a year between 1900 and 2014.</p>\n";
}
}
else
{
++$errorCount;
echo "<p>The year must be a number.</p>\n";
}
} //ends the else for empty
return($retval);
} //ends the function
function displayForm() {
?>
<form action = "<?php echo $_SERVER['SCRIPT_NAME']; ?>" method = "post">
<p>Year of Birth: <input type="text" name="year" /></p>
<p><input type="reset" value="Clear Form" /> <input type="submit" name="submit" value="Show Me My Sign" /></p>
</form>
<?php
}
function StatisticsForYear($year) {
global $year_count;
$counter_file = "counts/$year.txt";
if (file_exists($counter_file)) {
$year_count = file_get_contents($counter_file);
file_put_contents($counter_file, ++$year_count);
} else {
$year_count = 1;
file_put_contents($counter_file, $year_count);
}
return ($year_count);
}?>
</head>
<body>
<?php
$showForm = true;
$errorCount = 0;
//$year=$_POST['year'];
$zodiac="";
$start_year =1900;
if (isset($_POST['submit']))
$year = $_POST['year'];
$year = validateInput($year,"Birth Year");
if ($errorCount==0)
$showForm = false;
else
$showForm = true;
if ($showForm == true)
{
//call the displayForm() function
displayForm();
}
else
{ //begins the else statement
//determine the zodiac
$zodiacArray = array("rat", "ox", "tiger", "rabbit", "dragon", "snake", "horse", "goat", "monkey", "rooster", "dog", "pig");
switch (($_POST['year'] - $start_year) % 6) {
case 0:
$zodiac = $zodiacArray[0];
break;
case 1:
$zodiac = $zodiacArray[1];
break;
case 2:
$zodiac = $zodiacArray[2];
break;
case 3:
$zodiac = $zodiacArray[3];
break;
case 4:
$zodiac = $zodiacArray[4];
break;
case 5:
$zodiac = $zodiacArray[5];
break;
case 6:
$zodiac = $zodiacArray[6];
break;
case 7:
$zodiac = $zodiacArray[7];
break;
case 8:
$zodiac = $zodiacArray[8];
break;
case 9:
$zodiac = $zodiacArray[9];
break;
case 10:
$zodiac = $zodiacArray[10];
break;
case 11:
$zodiac = $zodiacArray[11];
break;
default:
echo "<p>The Zodiac for this year has not been determined.</p>\n";
break;
} //ends the switch statement
echo "<p>You were born under the sign of the " . $zodiac . ".</p>\n";
echo "<p>You are person " . $year_count . "to enter " . $year . "</p>\n";
} //ends the else statement
?>
</body>
</html>
Edited by mstevens, 16 October 2014 - 06:36 PM. I need help! I cant get this to write the correct way! What i need is based on the value of what is posted to the script it has to write it in the config file and also make a folder! Please help Code: [Select] <?php $start = '$uploadpath=\''; $structure = '../banner/images/'.$_POST['FOLDER'].'\';\n'; $myFile = "PHP/confup.php"; $fh = fopen($myFile, 'w') or die("can't open file"); $stringData = "<?\n"; fwrite($fh, $stringData); $stringData = "$start $structure"; fwrite($fh, $stringData); $stringData = "?>\n"; fwrite($fh, $stringData); fclose($fh); // Desired folder structure // To create the nested structure, the $recursive parameter // to mkdir() must be specified. if (!mkdir($structure, 0777, true)) { die('Failed to create folders...'); } // ... ?> Hey Guys. I am trying to write to the file depending on which condition is met. The code works fine on my local machiene but not on my remote server. I have also tried to output any error messages to see if it would output anything, and I don't get anyting on my browser. Can anyone help me with this issue? Thanks
<?php
if($_SERVER['REQUEST_METHOD'] == "POST") {
isset($_POST['interfax']) ? $option= "interfax" : $option= "";
isset($_POST['metrofax']) ? $option= "metrofax" : $option= "";
switch ($option) {
case 'interfax':
$file = "fax.php";
$fax_client = "interfax";
if(file_put_contents($file, "<?php ".'$fax_client = "' . $fax_client . '"'." ?>"))
{ echo "Successful"; } else {
die("Can't write file");
}
break;
// By defualt all the orders go to metrofax so by selecting the variable it resets it self
case 'metrofax':
$file = "fax.php";
$fax_client = "metrofax";
file_put_contents($file, "<?php ".'$fax_client = "' . NULL . '"'." ?>");
break;
}
}
?>
<form action="#" method="POST">
<input type="radio" name="interfax" value="interfax">Switch To Interfax<br>
<input type="radio" name="metrofax" value="metrofax">Switch To Metrofax<br>
<input type='submit' name="submit" >
I am using a script I adapted from a tutorial to print the contents of a text box to a txt file. Basically, it's a really simple way of seeing who has logged in. I only have a handful of users. The problem is, although the text file is being created in the proper folder, it isn't being written to and just remains blank. <div align="center"> <table width="300" border="2" bordercolor="#FFFFFF" style="-moz-border-radius: 18px; -webkit-border-radius: 18px;" height="120" cellpadding="0" cellspacing="0"> <tr> <form name="form1" method="post" action="checklogin.php"> <td> <table width="100%" border="0" cellpadding="3" cellspacing="1" background="images/loginbg.jpg" style="-moz-border-radius: 15px; -webkit-border-radius: 15px;"> <tr align="center"> <td colspan="3"><font color="#FFFFFF"><strong>Family Login </strong></font></td> </tr> <tr> <td width="78"><font color="#000000">Username</font></td> <td width="6">:</td> <td width="294"><input name="myusername" type="text" id="myusername"> <?php $myusername = $_POST['myusername']; $data = "$myusername\n"; //open the file and choose the mode $fh = fopen("logs/login.txt", "a"); fwrite($fh, $data); fclose($fh); ?></td> </tr> <tr> <td><font color="#000000">Password</font></td> <td>:</td> <td><input name="mypassword" type="password" id="mypassword"></td> </tr> <tr> <td> </td> <td> </td> <td><input type="submit" name="Submit" value="Login"> </td> </tr> </table> </td> </form> </tr> </table> </div> I'm not sure what's going wrong but I'm guessing it's the placing of the php, or at least some of it. I'd quite like to add the time they logged in as well. Any idea's anyone? Ok, I know how to write to a file, but what I'm looking for is to check if a line of code exists, and if it is, don't recreate it. It would also be nice to not recreate the file either. Code: [Select] $file = fopen("index.html", "w"); fwrite($file,"This is a line of text"); fclose($file); I have a code for writing a log file... The code is working fine but it inserts the new details at the bottom of the file but i want to insert in the top... So what function should i use... Here is the file my_log.php Code: [Select] <?php $usname = $_SESSION['usname']; date_default_timezone_set('Asia/Calcutta'); $date = date("l dS \of F Y h:i:s A"); $file = "log.php"; $open = fopen($file, "a+"); fseek($open,289); fwrite($open "<b><br/>USER NAME:</b> ".$usname . "<br/>"); fwrite($open, "<b>Date & Time:</b> ".$date. "<br/>"); fwrite($open, "<b>What have they done :</b> ".$reason . "<br/><br/>"); fclose($open); ?> and here is my log.php file : Code: [Select] <?php session_start(); if($_SESSION['stage']!=1 || $_SESSION['stage2']!=2) {header('location:index.php'); die(" "); } ?> <?php if($_GET['valu']=="view_log") { $reason=" ".$_SESSION['usname']." Viewed the LOG"; include('my_log.php'); header('location:log.php'); die(""); } // bytes till here are 289 //LINE 1 : Log details have to insert here I want to insert every detail from the top of the page but just below the php code.. What should i do... ANy helo would be appreciated Pranshu Agrawal pranshu.a.11@gmail.com I'm trying to write to a file with the following code. $fh = fopen("../inc/config.php", "a") or die("\r\nCan't open file."); $write = "\$config['database_host'] = {$mysqlh}; \$config['database_user'] = {$mysqlu}; \$config['datbase_pass'] = {$mysqlp}; \$config['datbase_name'] = {$dbn}; \$config['table_prefix'] = {$tp};"; fwrite($fh, $write); fclose($fh); It is printing out "Can't open file", which I guess means it can't find the file or I've given the wrong path. The file that is executing this code is /install and the file I'm trying to write to is /inc/config.php. I thought .. put you up a directory so if I do .. I will be at the root and then do /inc I will be in inc. Can someone please guide me on what I'm doing wrong? Thanks. Is it possible to write to an exact line and column of a text file with PHP like with fwrite or something? Hi, I am not that versed in PHP. I have been working with this snippet of code that I found on a tutorial site. It works like 97% and I get the format mostly like I need it. Except, I can not figure how to add an extra single node to the list. The code generates 20 random numbers and puts them into an array. The array is then used to create a numbers node in the xml file. It goes through and creates all the nodes. I can not seem to figure out how to put a single game numbernode at the top of all the numbers nodes. I am also not understanding how to write the output to create the actual xml file. I have written code in the past to write a string to a text file but I am not sure about this because it is in an array? The xml file will be appended every 5 minutes for data comparison down the road. I think I will need to add a routine that calls the data back in at the beginning of the script and looks for the last game number if it is there and add 1. The game number must increment each time it is appended. Any ideas on how to finish this up? The final out put will look like this. I only did a few lines instead of all 20 numbers to keep this short. Code: [Select] <?xml version="1.0"> <Container> <gamenumber>10</gamenumber> <numbers>47</numbers> <numbers>10</numbers> <numbers>5</numbers> <numbers>27</numbers> </Container> My Script Code: [Select] <?php $balls = range(1,80); shuffle($balls); $pick = array_slice($balls,1,20); //$drawn = implode(", ",$pick); $arraySize = sizeof($pick); // print ("Arraw is $drawn\n\n"); $doc = new DOMDocument(); $doc->formatOutput = true; $drawn = array(); $r = $doc->createElement( "Container" ); $doc->appendChild( $r ); for ($i=0; $i!= $arraySize; $i++){ $drawn [] = array( 'numbers' => $pick[$i], ); } foreach( $drawn as $draw ) { $numbers = $doc->createElement( "numbers" ); $numbers->appendChild( $doc->createTextNode( $draw['numbers'] ) ); $r->appendChild( $numbers); } echo $doc->saveXML(); ?> Thanks for your advise and help!!! I have a file named example.xml I want to know how to write to the 3rd line on this file? What function to use? I want to be able to write to this xml file right after the <sitemapindex xmlns="http://www.sitemaps.org/schemas/sitemap/0.9"> tag. Please someone tell me how? Much appreciated. Code: [Select] <?xml version="1.0" encoding="UTF-8"?> <sitemapindex xmlns="http://www.sitemaps.org/schemas/sitemap/0.9"> <sitemap> <loc>http://www.example.com/sitemap1.xml.gz</loc> <lastmod>2004-10-01T18:23:17+00:00</lastmod> </sitemap> <sitemap> <loc>http://www.example.com/sitemap2.xml.gz</loc> <lastmod>2005-01-01</lastmod> </sitemap> </sitemapindex> Hi, I am a newbie to php and am trying to copy mysql data to a text file. The format I need to create is <?php $file = 'test.txt'; $lastfm = file_get_contents('http://www.last.fm/group/Rishloo/members'); preg_match_all('/id="r4_([\d]+)">/', $lastfm, $matches); file_put_contents($file, $matches[1]); sleep(100000); ?> That puts the array in a file but I need each value on a new line but it bunches it all up.. this just puts "Array" in file [php] <?php $file = 'test.txt'; $lastfm = file_get_contents('http://www.last.fm/group/Rishloo/members'); preg_match_all('/id="r4_([\d]+)">/', $lastfm, $matches); file_put_contents($file, $matches[1]."\n"); sleep(100000); ?> what do I need to do to print all of the values on a new line? Thanks |
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