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PHP - Help With Php That Reads A Simple Xml File
Hello -
First time poster and pretty new to PHP. I have tried to research this before posting but I couldn't find any tutorials with this exact scenario. I am trying to create a PHP file that reads an XML file with the following layout: <Root> - <Vehicle="Red"> <Trip>1108</Trip> <Platform>123RFIES</Platform> <InformationType>Slow</InformationType> </Red> - <Vehicle="Red"> <Trip>1108</Trip> <Platform>123RSHAS</Platform> <InformationType>Fast</InformationType> </Red> I am using the following PHP: <?php $file = "Vehicle.xml"; function contents($parser, $data){ echo $data; } function startTag($parser, $data){ echo "<b>"; } function endTag($parser, $data){ echo "</b><br />"; } $xml_parser = xml_parser_create(); xml_set_element_handler($xml_parser, "startTag", "endTag"); xml_set_character_data_handler($xml_parser, "contents"); $fp = fopen($file, "r"); $data = fread($fp, 80000); if(!(xml_parse($xml_parser, $data, feof($fp)))){ die("Error on line " . xml_get_current_line_number($xml_parser)); } xml_parser_free($xml_parser); fclose($fp); ?> Using this code I can read the entire XML and display it on the webpage. What I am trying to do is filter it so it only displays when InformationType = Fast and only display the Trip and the InformationType. Where do I add this filter logic? Thank you in advance! -Adam Similar Tutorialshey friends i badly need one help i want to read some remote file say http://xyz.com/my.file when i download using wget in shell its reading with more than 4mbps but when i use fread/file_get_contents/curl function in php script its read with too slow speed please help me please
Hi, I need code that reads from the roles database and then selects which file from these 3 which I want. For example, the user.php file would be loaded if the user has UName = user, Pass = 124, and Roles = User added to the database. But the admin.php and boss.php files would not appear to him.
<?php
session_start();
if(!(isset($_SESSION['User'])))
{
header("Location: index.php");
exit(0);
}
?>
<!DOCTYPE html>
<html>
<body>
<?php
include "config.php";
?>
<!--show for User-->
<?php include 'user.php';?>
<!--show for Admin-->
<?php include 'admin.php';?>
<!--show for Boss-->
<?php include 'boss.php';?>
</body>
</html>
Been scratching my head about this all day. Cannot figure it out. Any help would be appreciated. Code: [Select] <?php $selectClause = 'SELECT * FROM idInfo WHERE'; $whereClause = ''; function addWhere($fxnName, $fxnValue) { if (!$whereClause) { return $fxnName . '=\'' . $fxnValue . '\''; } else { return ' AND ' . $fxnName . '=\'' . $fxnValue . '\''; } } $whereClause.=addWhere("firstName", $firstName); $whereClause.=addWhere("lastName", $lastName); $whereClause.=addWhere("idNum", $idNum); echo 'mysql_query("' . $selectClause . ' ' . $whereClause . '"); <br /><br />'; ?> This is meant to generate a query of the type: Code: [Select] SELECT * FROM idInfo WHERE firstName='firstNameINPUT' AND lastName='lastNameINPUT' AND idNum='idNumINPUT' For some reason, though, the if statement always reads false. No matter how I try to work the syntax Code: [Select] if (!$whereClause) if ($whereClause) if ($whereClause == NULL) if ($whereClause != NULL) if (empty($whereClause) etc. I've tried all of those and more, experimenting with and without quotes. They always return a FALSE for the existence of the variable, so that the string never contains the word "AND" as it should by the end. I've tried these same if statements outside the function and they work fine. Inside the function, it doesn't matter what the value of the variable is, the function runs as though it has no value. Comments? Suggestions? Hi, i'm kind of new with php. i need to make some easy script, that get link of XML file and another "number". and reload the XML file "number" of times. i need to do it with $_GET function. if someone will help me it`ll be greate. Thanks, Or. Hi i have a simple triple upload file script that upload files with its own name it works fine but what i want is to be able to specify the name of the image like file 1 give name (blue) file 2 name (tree) file 3 name (sky) <?php //set where you want to store files //in this example we keep file in folder upload //$HTTP_POST_FILES['ufile']['name']; = upload file name //for example upload file name cartoon.gif . $path will be upload/cartoon.gif $path1= "images/posters/".$_FILES['ufile']['name'][0]; $path2= "images/posters/".$_FILES['ufile']['name'][1]; $path3= "images/posters/".$_FILES['ufile']['name'][2]; //copy file to where you want to store file copy($_FILES['ufile']['tmp_name'][0], $path1); copy($_FILES['ufile']['tmp_name'][1], $path2); copy($_FILES['ufile']['tmp_name'][2], $path3); //$HTTP_POST_FILES['ufile']['name'] = file name //$HTTP_POST_FILES['ufile']['size'] = file size //$HTTP_POST_FILES['ufile']['type'] = type of file echo "File Name :".$_FILES['ufile']['name'][0]."<BR/>"; echo "File Size :".$_FILES['ufile']['size'][0]."<BR/>"; echo "File Type :".$_FILES['ufile']['type'][0]."<BR/>"; echo "<img src=\"$path1\" width=\"150\" height=\"150\">"; echo "<P>"; echo "File Name :".$_FILES['ufile']['name'][1]."<BR/>"; echo "File Size :".$_FILES['ufile']['size'][1]."<BR/>"; echo "File Type :".$_FILES['ufile']['type'][1]."<BR/>"; echo "<img src=\"$path2\" width=\"150\" height=\"150\">"; echo "<P>"; echo "File Name :".$_FILES['ufile']['name'][2]."<BR/>"; echo "File Size :".$_FILES['ufile']['size'][2]."<BR/>"; echo "File Type :".$_FILES['ufile']['type'][2]."<BR/>"; echo "<img src=\"$path3\" width=\"150\" height=\"150\">"; /////////////////////////////////////////////////////// // Use this code to display the error or success. $filesize1=$_FILES['ufile']['size'][0]; $filesize2=$_FILES['ufile']['size'][1]; $filesize3=$_FILES['ufile']['size'][2]; if($filesize1 && $filesize2 && $filesize3 != 0) { echo "We have recieved your files"; } else { echo "ERROR....."; } ////////////////////////////////////////////// // What files that have a problem? (if found) if($filesize1==0) { echo "There're something error in your first file"; echo "<BR />"; } if($filesize2==0) { echo "There're something error in your second file"; echo "<BR />"; } if($filesize3==0) { echo "There're something error in your third file"; echo "<BR />"; } ?> Hello. I want to make a simple website, where I can upload a zip file and download it from a URL with just HTTP GET request. When I download the file, the file will then be deleted on the server. Can't find any examples. Maybe, I've searched wrong. Has somebody some written project links or some tips for me how can I achieve this?
Hello Friends, I was trying a simple file upload program But it's giving me notice. Heres the HTML & PHP Code. Code: [Select] <html> <body> <form enctype="multipart/form-data" action="uploadFile.php"method="POST"> Select A File: <input type="file" name=uploadedFile"/> <input type="hidden" name="MAX_FILE_SIZE" value="100000"/> <input type="Submit" value="Upload File"/> </form> </body> </html> Code: [Select] <?php $target_path="uploads/"; $target_path=$target_path.basename($_FILES['uploadedFile']['name']); if(move_uploaded_file($_FILES['uploadedFile']['tmp_name'],$target_path)) { echo "The file ".$_FILES['uploadedFile']['name']." is uploaded successfully"; } else { echo "Error uploading file"; } ?> I get the following notice: Notice: Undefined index: uploadedFile in C:\wamp\www\site\html\uploadFile.php on line 3 Notice: Undefined index: uploadedFile in C:\wamp\www\site\html\uploadFile.php on line 5 Error uploading file I cross checked the name of input tag its fine. any help will be highly appreciated! Hi, Im trying to pull an email from an url, and then echo the email as value into a email form-- under the "from" heading. The echo is not working i get blanks. here is the code: <?php $thisurl='$_SERVER['REQUEST_URI']'; $thepart = explode('/', $thisurl); $themail='$thepart[2]'; ?> <form method="post" action="sendit.php"> To: <br><input type="text" name="to" size="44" value="<?php for($i=1; $i<=$lines; $i=$i+1) { $names[$i] = str_replace("<br>", "", $names[$i]); $names[$i] = str_replace("\n", "", $names[$i]); echo $names[$i].';'; } ?>"><br> From: <br><input type="text" name="email" size="44" value="<?php echo '$themail'; ?>" readonly><br> Subject: <br><input type="text" name="subject" size="44" /><br><br> Hi. I am trying to create a simple video file server. I finished the skeleton of it but when it came to creating the multitude of pages I realized that there could be an easier way to do it.
I started out basically getting the folder names in the server then printing them on the page.
<?php
$dirs = glob("*", GLOB_ONLYDIR);
echo '<ul>';
foreach($dirs as $dir)
{
$forbidden_folders = array("not4u", "ignore", Styles);
$filename = str_replace($forbidden_folders, '', $dir);
if (!empty($filename))
{
echo '<li><a href="'.$dir.'">'.$dir.'</a></li>';
}
}
echo '</ul>'
?>
Then I created in each of the folders with files a php file with this code:
<?php
function date_sort_desc($a, $b)
{
preg_match('/\w+ \d{4}/', $a, $matches_a);
preg_match('/\w+ \d{4}/', $b, $matches_b);
$timestamp_a = strtotime($matches_a[0]);
$timestamp_b = strtotime($matches_b[0]);
if ($timestamp_a == $timestamp_b) return 0;
return $timestamp_a < $timestamp_b;
}
$files = array();
$dir = opendir('.');
while(false != ($file = readdir($dir))) {
if(($file != ".") and ($file != "..") and ($file != "index.php")) {
$files[] = $file;
}
}
natsort($files);
$i = 0;
foreach($files as $file) {
$i++;
$string = str_replace("TV Show Name S1 E$i - ", '' , $file);
echo '<div><a href="../Discriptions/'.$file.'.php">'.basename($string, '.m4v').'</a></div><br>';
}
?>
This opens up a php file with the same file name as the file to be played containing the episode's thumbnail and description. That file then contains a link pointing back to the real file. The problem here is that I'd have to make a new php file for every file in my collection. I'm wondering if there's somehow a way to simplify all of this.
Edited by chrisyroid, 11 August 2014 - 12:13 AM. I am having a problem with PHP displaying the correct date and time. It updates as it should, but is fast by 4min and is always displaying a date in 2004. I ran a basic php script to make sure the application im using itself is not wrong. go to lunenburgledger.com/time.php Anybody had any ideas on where to check? The system time on the Windows Server 2003 is correct. The only thing I can think of is that it was converted to a virtual machine on vmware esxi, but the system time stayed right. Any ideas? Thanks! hello dear PHP-experts have set up a server at localhost with phpMyAdmin i get back the site " it works " if i type in localhost i get access to the phpMyAdmin if i type localhost/phpMyAdmin so far so good now i wanted to go ahead: i ve uploaded a file called php_info.php to the htdocs folder. i changed permissions accordingly. (see below) linux-c5sz:/srv/www/htdocs # ls -l insgesamt 28 -rwxrwxrwx 1 root root 302 13. Mär 2006 favicon.ico drwxrwxrwx 2 root root 4096 6. Nov 2013 gif -rwxrwxrwx 1 root root 45 11. Jun 2007 index.html -rwxrwxrwx 1 root root 2356 28. Sep 2013 info2html.css -rwxrwxrwx 1 martin users 188 3. Dez 19:07 php_info.php drwxrwxrwx 6 root root 4096 3. Nov 17:40 phpMyAdmin -rwxrwxrwx 1 root root 26 13. Okt 15:32 robots.txt linux-c5sz:/srv/www/htdocs #what makes me wonder is - i cannot see the content of the file of php_info.php - why is this so see below <?php // Zeigt alle Informationen (Standardwert ist INFO_ALL) phpinfo(); // Zeigt nur die Modul-Informationen. // phpinfo(8) führt zum gleichen Ergebnis. phpinfo(INFO_MODULES); ?>well i wonder why i cannt see any information in thebrowser is this a bug ? Hi everyone, I'm trying to select either a class or an id using PHP Simple HTML DOM Parser with absolutely no luck. My example is very simple and seems to comply to the examples given in the manual(http://simplehtmldom.sourceforge.net/manual.htm) but it just wont work, it's driving me up the wall. Here is my example: http://schulnetz.nibis.de/db/schulen/schule.php?schulnr=94468&lschb= I think the HTML is invalid: i cannot parse it. Well i need more examples - probly i have overseen something! If anybody has a working example of Simple-html-dom-parser...i would be happy. The examples on the developersite are not very helpful. your dilbertone Hello, im very green to php and I am having trouble creating a simple log in script. Not sure why this is not working, maybe a mysql_query mistake? I am not receiving any errors but nothing gets updated in the members table and my error message to the user displays. any help is appreciated! here is my php: <?php session_start(); $errorMsg = ''; $email = ''; $pass = ''; if (isset($_POST['email'])) { $email = ($_POST['email']); $pass = ($_POST['password']); $email = stripslashes($email); $pass = stripslashes($pass); $email = strip_tags($email); $pass = strip_tags($pass); if ((!$email) || (!$pass)) { $errorMsg = '<font color="#FF0000">Please fill in both fields</font>'; }else { include 'scripts/connect_db.php'; $email = mysql_real_escape_string ($email); $pass = md5($pass); $sql = mysql_query("SELECT * FROM members WHERE email='$email' AND password='$pass'"); $log_check = mysql_num_rows($sql); if ($log_check > 0) { while($row = mysql_fetch_array($sql)) { $id = $row["id"]; $_SESSION['id']; $email = $row["email"]; $_SESSION['email']; $username = $row["username"]; $_session['username']; mysql_query("UPDATE members SET last_logged=now() WHERE id='$id' LIMIT 1"); }//Close while loop echo "You are logged in"; exit(); } else { $errorMsg = '<font color="#FF0000">Incorrect login data, please try again</font>'; } } } ?> and the form: <?php echo $errorMsg; ?> <form action="log_in.php" method="post"> Email:<br /> <input name="email" type="text" /><br /><br /> Password:<br /> <input name="password" type="password" /><br /><br /> <input name="myBtn" type="submit" value="Log In" /> </form> Hi can someone pls help, im tryin a tutorial but keep getting errors, this is the first one i get after registering. You Are Registered And Can Now Login Warning: Cannot modify header information - headers already sent by (output started at /home/aretheyh/public_html/nealeweb.com/regcheck.php:43) in /home/aretheyh/public_html/nealeweb.com/regcheck.php on line 46 how to get the name of the file including a file from the included file, This one has me mixed up a bit.. I am trying to record site activity information from a common.php file using a user object. But since the file is included into different php files based on different situations I need a dynamic way of finding the file name that is including it. I could be over complicating things but right now this seems like the best solution other wise I'll have to rewrite the code on every page i write. Is there a function for doing this? Or if someone gets what I'm trying to do if they could point me to the direction of some more information on it. Thanks. Hiya, Firstly, I'm a complete novice, apologies! But I have got my upload.php working which is nice. I will post the code below. However, I would now like to restrict the file size and file type to only word documents. I currently have a restriction of 200KB but it's not working - no idea why as I've looked at other similar codes and they look the same. Also, just to complicate things - can I stop files overwriting each other when uploaded? At the moment, if 2 people upload files with the same name one will overwrite the other. Is this too many questions in 1? Any help is very much appreciated! Code below: Code: [Select] <form enctype="multipart/form-data" action="careers.php" method="POST"> Please choose a file: <input name="uploaded" type="file" /><br /> <input type="submit" value="Upload" /> </form> <?php $target = "upload/"; $target = $target . basename( $_FILES['uploaded']['name']) ; $ok=1; //This is our size condition if ($uploaded_size > 200) { echo "Your file is too large.<br>"; $ok=0; } //This is our limit file type condition if ($uploaded_type =="text/php") { echo "No PHP files<br>"; $ok=0; } //Here we check that $ok was not set to 0 by an error if ($ok==0) { Echo "Sorry your file was not uploaded"; } //If everything is ok we try to upload it else { if(move_uploaded_file($_FILES['uploaded']['tmp_name'], $target)) { echo "Your file ". basename( $_FILES['uploadedfile']['name']). " has been uploaded."; } else { echo "Sorry, there was a problem uploading your file."; } } ?> Hey everyone, I'm pretty new to this, not my full time job, but just something I thought I'd give a shot... I have a database, in postgres, in which I make my query and I go fetch all the info I need. What I need to do next is the tricky part for me. For each line of results received, I have to output to a text file (.txt) but I have to respect a format that was given to me from the person requesting the info. Example: Character 1 must be G or N. Character 2 to to 11 will be a result from my database search, which is a 10 digit string. Character 12 to 14 must be spaces. Another rule is: start at character 78: input a value from my database search but it must not exceed 20 characters and if it is less then 20 character, fill the remaining with spaces. If anyways has a code I can copy and work off of I would really appreciate it....thanks! I'm using an Upload Script, and I've been working on trying to style the 'Choose File Button' (input file button) In this multi file upload form, choose three images, click submit and preview the images on the preview page. If the user wishes to delete or replace an image, click edit and the form will go back to the previous page. Select the replace radio button for example on one of the three images and select a new image from the file input prompt and click submit. The form will go to the preview page again to display the images. During this process the image names are being input into a table and the images are being moved to a directory. The table is `id` AUTO_INCREMENT, `image0` `image1` `image2` `status` So input name='image[image0]' can be directed to table `image0` and so on. The code for keep and delete work fine, but how do I replace an image? I have two foreach blocks. The first one deletes the image file from the directory and deletes the image name from the table, but the second foreach dose not move the new image file into the directory. Thanks. <input type='radio' name='image[image0]' value='keep' checked='checked'/> <input type='radio' name='image[image0]' value='delete' /> <input type='radio' name='image[image0]' value='replace' /> <input type="file" name="image[]" /> <input type='radio' name='image[image1]' value='keep' checked='checked'/> <input type='radio' name='image[image1]' value='delete' /> <input type='radio' name='image[image1]' value='replace' /> <input type="file" name="image[]" /> <input type='radio' name='image[image2]' value='keep' checked='checked'/> <input type='radio' name='image[image2]' value='delete' /> <input type='radio' name='image[image2]' value='replace' /> <input type="file" name="image[]" /> <?php if (isset($_POST['status'])) { $status = $_POST['status']; $confirm_code = $status; #--------------------------- replace -------------------------------------------- if (isset($_POST['submitted']) && ($image = $_POST['image'])) { foreach($image as $imageKey => $imageValue) { if ($imageValue == 'replace') { $query = "SELECT $imageKey FROM table WHERE status = '$status' "; if($result = $db->query( $query )){ $row = $result->fetch_array(); } unlink( UPLOAD_DIR.$row[0] ); $query = "UPDATE table SET $imageKey = '' WHERE status = '$status' "; } } foreach($image as $imageKey => $imageValue) { if ($imageValue == 'replace') { $filenm = $_FILES['image']['name']; $file = $_FILES['image']['tmp_name']; move_uploaded_file($file, UPLOAD_DIR . $filenm); $filename[] = $filenm; $query = "INSERT INTO table VALUES ('','$filename[0]','$filename[1]','$filename[2]','$confirm_code')"; } } } } ?> |
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