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PHP - Need Help With Writing To Files..
im trying to make a program that changes some files from 0 to 1,but im having some trouble...
heres my code, Code: [Select] $id = $_GET["id"]; $file1 = "http://mysite.co.cc/users/".$id."/file1.txt"; $fh = fopen($file1, 'w'); fwrite($fh, "1"); fclose($fh); I followed this tut: http://www.tizag.com/phpT/filewrite.php I cant find anything wrong with the code,but it will not make the file have 1 in it. Similar TutorialsI'm trying this simple script on my windows server 2008 standard edition but keep getting errors. Code: [Select] <?php error_reporting(-1); ini_set( 'display_errors' , 1 ); $file=fopen("2.txt","a") ; fwrite($file,"Hai"); ?> Code: [Select] Warning: fopen(): It is not safe to rely on the system's timezone settings. You are *required* to use the date.timezone setting or the date_default_timezone_set() function. In case you used any of those methods and you are still getting this warning, you most likely misspelled the timezone identifier. We selected 'Europe/Paris' for '1.0/no DST' instead in C:\inetpub\wwwroot\10.php on line 3 Warning: fopen(2.txt): failed to open stream: Permission denied in C:\inetpub\wwwroot\10.php on line 3 Warning: fwrite(): It is not safe to rely on the system's timezone settings. You are *required* to use the date.timezone setting or the date_default_timezone_set() function. In case you used any of those methods and you are still getting this warning, you most likely misspelled the timezone identifier. We selected 'Europe/Paris' for '1.0/no DST' instead in C:\inetpub\wwwroot\10.php on line 4 Warning: fwrite() expects parameter 1 to be resource, boolean given in C:\inetpub\wwwroot\10.php on line 4 Any help is appreciated. Hi All, I need some help blocking files that don't pass authentication from being uploaded to the server. Here is my script (below). It correctly throws errors, but regardless of error / no-error, the file is uploaded. For example, the only files allowed to be uploaded should be .png, .jpeg, .gif -- but I see .flv .docx etc in the destination folder. You'll see below a variable for each of the error messages. They basically say that the file name is too long, too big, not an allowed file type, or English characters only allowed in file title name. Again the errors work properly, but regardless files are uploaded to the "quiz_images" folder. What do I need to do to stop the upload if an error is triggered? Thank you for your help... here is the code snip: Code: [Select] $directory_self = str_replace(basename($_SERVER['PHP_SELF']), '', $_SERVER['PHP_SELF']); $uploadsDirectory = $_SERVER['DOCUMENT_ROOT'] . $directory_self . 'quiz_images/'; $fieldname = 'file'; if ($_POST[submit]) { if ($_POST[title] <> "") { $a = TRUE; } else { $a = FALSE; $content .= "<p>$enter_title</p>\n"; } if ($_POST[video] <> "") { $b = TRUE; } else { $b = FALSE; $content .= "<p>$enter_embed</p>\n"; } if ($_POST[description_text] <> "") { $c = TRUE; } else { $c = FALSE; $content .= "<p>$enter_description</p>\n"; } if ($_POST[submit] <> "") { $f = TRUE; } else { $f = FALSE; $content .= "<p>$sorry_error</p>\n"; } if (is_uploaded_file($_FILES[$fieldname]['tmp_name'])) { $g = TRUE; } else { $g = FALSE; $content .= "<p>$sorry_error</p>\n"; } if (getimagesize($_FILES[$fieldname]['tmp_name'])){ $h = TRUE; } else { $h = FALSE; $content .= "<p>$error2</p>\n"; } if ((($_FILES["file"]["type"] == "image/gif") || ($_FILES["file"]["type"] == "image/jpeg") || ($_FILES["file"]["type"] == "image/png") || ($_FILES["file"]["type"] == "image/pjpeg")) && ($_FILES["file"]["size"] < 50000)) { $i = TRUE; } else { $i = FALSE; $content .= "<p>$error2</p>\n"; } if(preg_match('#^[a-z0-9_\s-\.]*$#i', $pic) ) { $k = TRUE; } else { $k = FALSE; $content .= "<p>$error3</p>\n"; } $desc_length = strlen($pic); $limit = 40; if ($desc_length <= $limit) { $l = TRUE; } else { $l = FALSE; $content .= "<p>$error4</p>\n"; } $now = time(); while(file_exists($uploadFilename = $uploadsDirectory.$now.'-'.$_FILES[$fieldname]['name'])) { $now++; } if (move_uploaded_file($_FILES[$fieldname]['tmp_name'], $uploadFilename)){ $j = TRUE; } else { $j = FALSE; $content .= "<p>$enter_thumb</p>\n"; } $pic=$now++.'-'.$_FILES[$fieldname]['name']; if ($a AND $b AND $c AND $f AND $g AND $h AND $i AND $j AND $k AND $l) {............ So far I have managed to create an upload process which uploads a picture, updates the database on file location and then tries to upload the db a 2nd time to update the Thumbnails file location (i tried updating the thumbnails location in one go and for some reason this causes failure) But the main problem is that it doesn't upload some files Here is my upload.php
<?php
include 'dbconnect.php';
$statusMsg = '';
$Title = $conn -> real_escape_string($_POST['Title']) ;
$BodyText = $conn -> real_escape_string($_POST['ThreadBody']) ;
// File upload path
$targetDir = "upload/";
$fileName = basename($_FILES["file"]["name"]);
$targetFilePath = $targetDir . $fileName;
$fileType = pathinfo($targetFilePath,PATHINFO_EXTENSION);
$Thumbnail = "upload/Thumbnails/'$fileName'";
if(isset($_POST["submit"]) && !empty($_FILES["file"]["name"])){
// Allow certain file formats
$allowTypes = array('jpg','png','jpeg','gif','pdf', "webm", "mp4");
if(in_array($fileType, $allowTypes)){
// Upload file to server
if(move_uploaded_file($_FILES["file"]["tmp_name"], $targetFilePath)){
// Insert image file name into database
$insert = $conn->query("INSERT into Threads (Title, ThreadBody, filename) VALUES ('$Title', '$BodyText', '$fileName')");
if($insert){
$statusMsg = "The file ".$fileName. " has been uploaded successfully.";
$targetFilePathArg = escapeshellarg($targetFilePath);
$output=null;
$retval=null;
//exec("convert $targetFilePathArg -resize 300x200 ./upload/Thumbnails/'$fileName'", $output, $retval);
exec("convert $targetFilePathArg -resize 200x200 $Thumbnail", $output, $retval);
echo "REturned with status $retval and output:\n" ;
if ($retval == null) {
echo "Retval is null\n" ;
echo "Thumbnail equals $Thumbnail\n" ;
}
}else{
$statusMsg = "File upload failed, please try again.";
}
}else{
$statusMsg = "Sorry, there was an error uploading your file.";
}
}else{
$statusMsg = 'Sorry, only JPG, JPEG, PNG, GIF, mp4, webm & PDF files are allowed to upload.';
}
}else{
$statusMsg = 'Please select a file to upload.';
}
//Update SQL db by setting the thumbnail column to equal $Thumbnail
$update = $conn->query("update Threads set thumbnail = '$Thumbnail' where filename = '$fileName'");
if($update){
$statusMsg = "Updated the thumbnail to sql correctly.";
echo $statusMsg ; }
else
{ echo "\n Failed to update Thumbnail. Thumbnail equals $Thumbnail" ; }
// Display status message
echo $statusMsg;
?>
And this does work on most files however it is not working on a 9.9mb png file which is named "test.png" I tested on another 3.3 mb gif file and that failed too? For some reason it returns the following Updated the thumbnail to sql correctly.Updated the thumbnail to sql correctly. Whereas on the files it works on it returns REturned with status 0 and output: Retval is null Thumbnail equals upload/Thumbnails/'rainbow-trh-stache.gif' Failed to update Thumbnail. Thumbnail equals upload/Thumbnails/'rainbow-trh-stache.gif'The file rainbow-trh-stache.gif has been uploaded successfully. Any idea on why this is? Hello I have a simple question about file handling... Is it possible to list all files in directories / subdirectories, and then read ALL files in those dirs, and put the content of their file into an array? Like this: array: [SomePath/test.php] = "All In this php file is being read by a new smart function!"; [SomePath/Weird/hello.txt = "Hello world. This is me and im just trying to get some help!";and so on, until no further files exists in that rootdir. All my attempts went totally crazy and none of them works... therefore i need to ask you for help. Do you have any ideas how to do this? If so, how can I be able to do it? Thanks in Advance, pros I am using WPSQT plugin in my blog site .I code some files in PHP also.how to add that files in plugin files.
I'm writing a script that takes user input from a html form and updates the database with the new data. In this example the user is updating the data about the university they attend. The problem with this is, the script to update the database with the new information appears to be not working - the new data is not being added to the database. The SQL queries used in the script work in phpMyadmin, and the variables - $uni and $username - contain the data the correct data. This has me stumped, could someone look over this for me please and tell me what is going wrong here. Code: [Select] <?php include 'connect.php'; session_start(); $_SESSION['username']; if(!(isset($_SESSION['login']) && $_SESSION['login']!= " ")){ header("Location: login.php"); } $username = $_SESSION['username']; $tablename = 'usr_test'; $uni = $_POST['uni']; $uni = stripslashes($uni); $uni = mysql_real_escape_string($uni); $uni = trim($uni); if(isset($uni)) { $username = $_SESSION['username']; $loc = mysql_query("SELECT * FROM usr_test WHERE usr = '$username'"); if (mysql_num_rows($loc) == 0) header("Location:notlogged.php"); else { extract(mysql_fetch_array($loc)); mysql_query("UPDATE usr_test SET uni = ('$uni') WHERE usr = '$username'") or die (mysql_error()); header("Location:profile.php"); } } ?> 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" >
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($file); I am using a cache script, well it writes a array to a file like this: Code: [Select] fwrite($fh, '<?php'."\n\n".'define(\'PUN_LOTTERY_LOADED\', 1);'."\n\n".'$lottery = '.var_export($output2, true).';'."\n\n".'?>'); output2 is Code: [Select] $result2 = $db->query('MY QUERY '); $output2 = array(); while ($cur_donors = $db->fetch_assoc($result2)) $output2[] = $cur_donors; Now, I want to ditch the mysql and I want to use this script with 7 variables that I already have loaded, so I dont need to use the mysql, how do I add my 7 variables to my var_export function instead of using mysql to loop them? I have a web app hosted on Just Host that I have nearly finished writing but still needs to have coded the ability to write a text file to user's computer. the user should be able to specify where on their computer they would like the file to be stored. Is this possible ? And can you tell me how it's done, or point me to a souce that can tell me I rarely ever ask for help regarding programming, but this has flumoxed me. If it is not possible to do this, then would I have to generate this file on the Host's server (Just Host in my case), then download it ? If this can be done then can you please tell me how, as any info I have found related to file downloads seems a bit obscure thanks in advance I don't think the start of my code is right?! Code: [Select] echo '<button onclick=\"gohere(viewpub.php?PubID='.$row['PubID'].')" id="button" type="button" class="ui-button ui-widget ui-state-default ui-corner-all ui-button-text-only" role="button" aria-disabled="false"><span class="ui-button-text">View Pub</span></button>'; Please help?! 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...'); } // ... ?> Hi everyone. Still new to PHP and to Object-Oriented Programming. My goal for today is to write a few dummy Classes in PHP that do enough so that I can see something on my webpage. I am wondering if there is someone here who would be willing to help in either the forums or one-on-one. It may sound like a silly request, but this is all new to me as a Procedural Programmer who has actively programmed in about 10 years!! Thanks, TomTees 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 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 have written some code to past information from one page to a diffrent page however all the variables are sent over but they are not writting to the database could anyone help me out please .... $ud_P_Id = $_POST['ud_P_Id']; $ud_LastName = $_POST['ud_LastName']; $ud_FirstName = $_POST['ud_FirstName']; $ud_GroupCode=$_POST['ud_GroupCode']; $ud_P_Unit1=$_POST['ud_P_Unit1']; $ud_P_Unit2=$_POST['ud_P_Unit2']; $ud_P_Unit3=$_POST['ud_P_Unit3']; $ud_P_Unit6=$_POST['ud_P_Unit6']; $ud_P_Unit14=$_POST['ud_P_Unit14']; $ud_P_Unit20=$_POST['ud_P_Unit20']; $ud_P_Unit27=$_POST['ud_P_Unit27']; $ud_P_Unit28=$_POST['ud_P_Unit28']; $ud_P_Unit42=$_POST['ud_P_Unit42']; $ud_P_Unit10=$_POST['ud_P_Unit10']; $ud_P_Unit11=$_POST['ud_P_Unit11']; $ud_P_Unit12=$_POST['ud_P_Unit12']; $ud_P_Unit13=$_POST['ud_P_Unit13']; $ud_P_Unit1454=$_POST['ud_P_Unit1454']; $ud_P_Unit15=$_POST['ud_P_Unit15']; $ud_P_Unit16=$_POST['ud_P_Unit16']; $ud_P_Unit17=$_POST['ud_P_Unit17']; $ud_P_Unit18=$_POST['ud_P_Unit18']; $con = mysql_connect('localhost', 'lccstude_progre', '********'); $db= "lccstude_pro"; if (! $con) die("Couldn't connect to MySQL"); mysql_select_db($db , $con) or die("Couldn't open $db: ".mysql_error()); mysql_query("UPDATE BTECL31113 SET FirstName='$ud_FirstName' , LastName='$ud_LastName' , GroupCode='$ud_GroupCode' , Unit1='$ud_P_Unit1' , Unit2='$ud_P_Unit2' , Unit3='$ud_P_Unit3' , Unit6='$ud_P_Unit6' , Unit14='$ud_P_Unit14' , Unit20='$ud_P_Unit20' , Unit27='$ud_P_Unit27' , Unit28='$ud_P_Unit28' , Unit42='$ud_P_Unit42' , Unit10='$ud_P_Unit10' , Unit11='$ud_P_Unit11' , Unit12='$ud_P_Unit12' , Unit13='$ud_P_Unit13' , Unit1454='$ud_P_Unit1454' , Unit15='$ud_P_Unit15' , Unit16= $ud_P_Unit16' , Unit17='$ud_P_Unit17' , Unit18='$ud_P_Unit18' WHERE P_Id='$ud_P_Id'"); echo "Record Updated"; mysql_close($con); Please any help would be great 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. 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? Greetings, I am trying to write XML using SimpleXML for a web service call. I was having success until the XML got a little more complicated. Here is the XML format where I am running into problems: Code: [Select] <Agent> <Person Last='Smith' First='John'> <Addresses /> <PhoneNumbers> <Phone Type='1' Number='888-555-1212'> </PhoneNumbers> </Person> </Agent> Here is my php: Code: [Select] $xmlOutput = new SimpleXMLElement('<?xml version="1.0"?><ReportRequest> </ReportRequest>'); $xmlOutput->addAttribute('CID','9200'); $xmlOutput->addAttribute('Diagram','1'); $xmlOutput->addAttribute('DueDate','2011-11-15'); $xmlOutput->addAttribute('NumPhotos','6'); $xmlOutput->addAttribute('InspectAfter',''); $xmlOutput->addAttribute('PolicyNumber','JTC0004425'); $reportType = $xmlOutput->addChild('ReportType'); $reportType->addAttribute('CPType','Commercial'); $reportType->addAttribute('SectionIDs',''); $reportType->addAttribute('Description',''); $reportType->addAttribute('ReportTypeID','123'); $locations = $xmlOutput->addChild('Locations')->addChild('Addresses')->addChild('Address'); $locations->addAttribute('Zip','91216'); $locations->addAttribute('City','Chatsworth'); $locations->addAttribute('Line1','123 Main St'); $locations->addAttribute('Line2','Suite 201'); $locations->addAttribute('State','CA'); $locations->addAttribute('Latitude',''); $locations->addAttribute('Longitude',''); $agent = $xmlOutput->addChild('Agent')->addChild('Person')->addChild('Addresses')->addChild('PhoneNumbers')->addChild('Phone'); $agent->addAttribute('Last','Smith'); $agent->addAttribute('Email',''); $agent->addAttribute('First','John'); $agent->addAttribute('Title',''); $agent->addAttribute('Type','1'); $agent->addAttribute('Number','888-555-1212'); $agent->addAttribute('TypeName','Office'); $agent->addAttribute('Extension',''); Header('Content-type: text/xml'); echo $xmlOutput->asXML(); Everything outputs as it should until the line that starts with $agent. I want the output to match the section above. I have tried several variations but I cannot seem to figure it out. I know the current PHP sample doesn't work. Help?? Thanks in advance, John Ok I need to write an array to a file here is the code if($place[2] < 10){ echo "Account Invalid"; $account = var_dump($account_data[$i][0]); echo $account; $fp3=fopen('invalid.txt', 'a+'); fwrite($fp3, $account); fclose($fp3);} when I try if($place[2] < 10){ echo "Account Invalid"; $account = var_dump($account_data[$i][0]); echo $account; $fp3=fopen('invalid.txt', 'a+'); fwrite($fp3, $account_data[$i][0]); fclose($fp3);} I just get "Array[1]" written to the file.. What can I do to write the array to the file? Thanks |
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