PHP - Php Var Inside File.sql
im trying to create a table for a site admin. bsicaly all he has to do is fill out a form and the values of the form are turned into the table name.
im using a .sql file as a template but to create the table i need the php genarated table name to be inserted into the .sql file. <?php include("config.php"); $con details $tbl_name = "static_name_for_now"; if (mysql_query("seat_plan.sql")) { echo "success"; } else{ echo "fail"; } ?> sql file snippet INSERT INTO `<?php $tbl_name ?>` VALUES('A', 1, 0, 'Y'); also tried INSERT INTO `<?php echo $tbl_name ?>` VALUES('A', 1, 0, 'Y'); thanks Similar Tutorials<td><label for='images'> <b>File to upload:</b> </label></td> <td><input type='file' name = 'drama_image' '<?php echo $row['drama_image']; ?>'/></ </tr> <?php $target_path = "images/"; $target_path = $target_path . basename( $_FILES['images']['name']); if(move_uploaded_file($_FILES['images']['tmp_name'], $target_path)) { echo "The file ". basename( $_FILES['images']['name']). " has been uploaded"; } else{ echo $row['drama_image']; } ?> ['drama_image'] is the name of the file I wanna echo it out in the box of file upload so when I save , the default picture will still be there instead of being overwritten as the box does not have any value in it. I have a php file that generates a string that I need to use in a .js (javascript)file. Being that php developers sometimes using javascript with php, Im hoping someone can help me with this, cause i dont know any javascript. Code: [Select] //This is the varible inside the .js file var suggestionText = "I need to be able to include my string generated by the php file here..."; So say I have a file with the contents.. Code: [Select] ; this is a comment var2 : variable ; another comment var3 : file.txt how would I be able to do <?php echo $var2; ?> That would echo "variable" and <?php $file = file($var3); ?> So its reading the file to get the variable from the text file and if the line starts with ";" its disregarded.. Thanks hi i have this form that runs with ASP. [this is a must since i didn't create that form] and must run in the wordpress header.php... is this even possible? Hello. Trying to learn MVC better by creating my own little framework to understand how it works. Things were going OK til now. I have a base class: <?php /** * Base class most classes will extend from. * Simply put, this class just has methods that * most, if not all, classes will need. */ class Application { public function includer($path) { if (is_readable($path) == true) { include_once($path); } else { die("404 not found =["); } } } ?> So the above method 'includer' just sees if a file exists/is readable and if so, include it. Here's where I am using said method. class Index extends Application { function __construct($method = 'view') { // load the index model include(ROOT . '/app/models/model.index.php'); // Invoke requested method. $this->$method(); } public function view() { $name = 'Smith'; $this->includer(ROOT . '/app/views/view.index.php'); # Problem here, i think } } view.index.php just contains <?php echo $name; ?> All that is called from the index page, with this line of code Application::includer($controller_path); Now, in line $this->includer(ROOT . '/app/views/view.index.php'); # Problem here, i think If i get rid of $this->includer, the script will work and say 'Smith'. If i have $this->includer(...) or parent::includer(...), it doesn't work. why? I am working PHP project on localhost using wamp, for debug process we using error_log() function, every time error_log() stored details of the error in wamp/logs, how can I view my error in the same project folder like wamp/www/goodgoal/ eg: wamp/www/goodgoal/logs/error_log.log Note : Pls recommend PHP Quick Debug Tricks Edited July 5, 2019 by aveevaI am trying to pass a variable from inside of a function in a file that is included eg include.php Code: [Select] <? function noms(){ $foobar = "apples"; } ?> main.php Code: [Select] <? include("include.php"); noms(); echo "these ". $foobar. " are most delicious... OM NOM NOM"; ?> I am essentially using a include file for a mysql connection and based on the connection outcome i am either setting a value to true or false. It prints "Connected succesfully" and what not just fine, but, after that it wont pass on the variables data. I know its the function because, it passes the data if i put the variable before the function... i just dont get it.. any help would be great. Thanks. Ansel The Script:
$desired_width = 110; if (isset($_POST['submit'])) { $j = 0; //Variable for indexing uploaded image for ($i = 0; $i < count($_FILES['file']['name']); $i++) {//loop to get individual element from the array $target_path = $_SERVER['DOCUMENT_ROOT'] . "/gallerysite/multiple_image_upload/uploads/"; //Declaring Path for uploaded images $validextensions = array("jpeg", "jpg", "png"); //Extensions which are allowed $ext = explode('.', basename($_FILES['file']['name'][$i]));//explode file name from dot(.) $file_extension = end($ext); //store extensions in the variable $new_image_name = md5(uniqid()) . "." . $ext[count($ext) - 1]; $target_path = $target_path . $new_image_name;//set the target path with a new name of image $j = $j + 1;//increment the number of uploaded images according to the files in array if (($_FILES["file"]["size"][$i] < 100000) //Approx. 100kb files can be uploaded. && in_array($file_extension, $validextensions)) { if (move_uploaded_file($_FILES['file']['tmp_name'][$i], $target_path)) {//if file moved to uploads folder echo $j. ').<span id="noerror">Image uploaded successfully!.</span><br/><br/>'; $tqs = "INSERT INTO images (`original_image_name`, `image_file`, `date_created`) VALUES ('" . $_FILES['file']['name'][$i] . "', '" . $new_image_name . "', now())"; $tqr = mysqli_query($dbc, $tqs); // Select the ID numbers of the last inserted images and store them inside an array. // Use the implode() function on the array to have a string of the ID numbers separated by commas. // Store the ID numbers in the "image_file_id" column of the "thread" table. $tqs = "SELECT `id` FROM `images` WHERE `image_file` IN ('$new_image_name')"; $tqr = mysqli_query($dbc, $tqs) or die(mysqli_error($dbc)); $fetch_array = array(); $row = mysqli_fetch_array($tqr); $fetch_array[] = $row['id']; /* * This prints e.g.: Array ( [0] => 542 ) Array ( [0] => 543 ) Array ( [0] => 544 ) */ print_r($fetch_array); // Goes over to create the thumbnail images. $src = $target_path; $dest = $_SERVER['DOCUMENT_ROOT'] . "/gallerysite/multiple_image_upload/thumbs/" . $new_image_name; make_thumb($src, $dest, $desired_width); } else {//if file was not moved. echo $j. ').<span id="error">please try again!.</span><br/><br/>'; } } else {//if file size and file type was incorrect. echo $j. ').<span id="error">***Invalid file Size or Type***</span><br/><br/>'; } } }Hey, sorry that I am posting this darn image upload script again, I have this almost finished and I am not looking to ask more questions when it comes to this script specifically. With the script above I have that part where the script should store the ID numbers (the auto_increment column of the table) of the image files inside of one array and then the "implode()" function would get used on the array and then the ID numbers would get inserted into the "image_file_id" column of the "thread" table. As you can see at the above part the script prints the following: Array ( [0] => 542 ) Array ( [0] => 543 ) Array ( [0] => 544 )And I am looking to insert into the column of the table the following: 542, 543, 544I thought of re-writing the whole image upload script since this happens inside the for loop, though I thought maybe I could be having this done with the script as it is right now. Any suggestions on how to do this? This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=316454.0 I'm trying to pull results from a database (using php) to populate a javascript "top news rotation" script I found. The problem is that I'm getting the "java is disabled" message that is in the code...instead of the results I'm expecting. Here's the code: Code: [Select] <!-- create a element in your HTML like the following --> <div id="quotetext" > Text will go here. Be sure to add initial text here for users with JavaScript disabled. </div> <!-- The easiest way is to place the below JavaScript code after the above HTML. The better way would be to add in the <head> section of the document and call the rotatequote() function through the window.onload event. However this can cause problems if you have other scripts that use the window onLoad settings --> <script type="text/javascript" > var myquotes = new Array( <?php $link = mysql_pconnect($host, $username, $password); mysql_select_db('briansch_brn',$link); $sql = "SELECT * FROM story WHERE ORDER BY date DESC LIMIT 4"; $rs = mysql_query($sql,$link); $matches = 0; while ($row = mysql_fetch_assoc($rs)) { $matches++; echo "'<strong>$row[headline]</strong><br />(posted $row[date]) - $row[short_story]<br /><a href='/pages/$row[keyword]'>READ MORE</a>'"; if($matches < 4) { echo ','; } if($matches == 4) { echo ''; } } if (! $matches) { echo (""); } echo ""; ?> ); function rotatequote() { thequote = myquotes.shift(); //Pull the top one myquotes.push(thequote); //And add it back to the end document.getElementById('quotetext').innerHTML = thequote; // This rotates the quote every 10 seconds. // Replace 10000 with (the number of seconds you want) * 1000 t=setTimeout("rotatequote()",10000); } // Start the first rotation. rotatequote(); </script> I know that javascript inside php is hard to make work...and I know that php inside javascript is hard to make work. Any ideas? Thanks! 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. 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! 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( $_FILES['uploadedfile']['name']). " has been uploaded."; } else { echo "Sorry, there was a problem uploading your file."; } } ?> I'm using an Upload Script, and I've been working on trying to style the 'Choose File Button' (input file button) In this multi file upload form, choose three images, click submit and preview the images on the preview page. If the user wishes to delete or replace an image, click edit and the form will go back to the previous page. Select the replace radio button for example on one of the three images and select a new image from the file input prompt and click submit. The form will go to the preview page again to display the images. During this process the image names are being input into a table and the images are being moved to a directory. The table is `id` AUTO_INCREMENT, `image0` `image1` `image2` `status` So input name='image[image0]' can be directed to table `image0` and so on. The code for keep and delete work fine, but how do I replace an image? I have two foreach blocks. The first one deletes the image file from the directory and deletes the image name from the table, but the second foreach dose not move the new image file into the directory. Thanks. <input type='radio' name='image[image0]' value='keep' checked='checked'/> <input type='radio' name='image[image0]' value='delete' /> <input type='radio' name='image[image0]' value='replace' /> <input type="file" name="image[]" /> <input type='radio' name='image[image1]' value='keep' checked='checked'/> <input type='radio' name='image[image1]' value='delete' /> <input type='radio' name='image[image1]' value='replace' /> <input type="file" name="image[]" /> <input type='radio' name='image[image2]' value='keep' checked='checked'/> <input type='radio' name='image[image2]' value='delete' /> <input type='radio' name='image[image2]' value='replace' /> <input type="file" name="image[]" /> <?php if (isset($_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]','$confirm_code')"; } } } } ?> Hello, all: been trying to convert this little single-file upload to multiple by naming each file form-field as "userfile[]" as it's supposed to automatically treat them as an array.. but no luck! Can you guide me as to what am I doing wrong?? appreciate the help! Code: [Select] <?php if (!isset($_REQUEST["seenform"])) { ?> <form enctype="multipart/form-data" action="#" method="post"> Upload file: <input name="userfile[]" type="file" id="userfile[]"> Upload file: <input name="userfile[]" type="file" id="userfile[]"> <input type="submit" value="Upload"> <input type="hidden" name="seenform"> </form> <?php } else { // upload begins $userfiles = array($_FILES['userfile']); foreach ($userfiles as $userfile) { // foreach begins $uploaded_dir = "uploads/"; $userfile = $_FILES['userfile']["name"]; $path = $uploaded_dir . $userfile; if (move_uploaded_file($_FILES['userfile']["tmp_name"], $path)) { print "$userfile file moved"; // do something with the file here } else { print "Move failed"; } } // foreach ends } // upload ends ?> I am using apache web server on linux. I am using PHP for web designing. On web server, i want to show the configuration data by reading the ini file. I am creating this ini file from one php code itself. If this php code i run through linux terminal, the file is created with file and group owner as root.(i am having sudo rights on machine) Then if i try to read the ini file from my apache web server, it gives warning as failed to open stream: permission denied. I have tried changing the owner, and permissions to 777 of the file. Still it is not readable.
On the other hand, if i run the php code of ini file creation through web server, ini file is created with file and group owner as apche. and web server is able to read/ write the file.
But i want to create that file from root or some other user and later read/written by apache.
How to give this access permission?
I havent included the whole title as it wouldnt let me but I was wondering if someone could help me on this? I know this is possible as torrentflux caters for this but unsure of where to start. I dont want to allow file or directory uploads or creation in my /etc/php.ini file (this is turned off). Yet then torrentflux allows me to link a torrent from an external source (using legal downloads of course ) but then it uploads it on my server and creates folders on a per user basis. How is this possible can someone give me some pointers please? I look forward to any replies, Jeremy. I have a page using forms to help build listing templates for eBay. I have a folder where I have hundreds of logos stored. I know the logo names but not their extensions. . . . I have to test each potential (jpg, jpeg, gif, png, etc.) until I guess right. Here is an example code for the web form:
<form action="extension_test2.php" method="post"> <p>Logo: <input name="e" value="" type="text" size="15" maxlength="30" /><p> <input name="Submit" type="Submit"/> </form> and the form's result: <? $e =$_POST['e']; if(!empty($e)) { echo '<img src="http://www.gbamedica...ebayimg/logos/'.$e.'">'; }; ?> Here is a link to the example: http://www.gbamedica...ension_test.php Use "olympus.jpg" for test. I am looking for code that can determine the file type and dynamically add the extension. Can it be done? Hi everybody. I use flash 8 and actionscript 2.0. I am trying to create a small program that needs communication between php / HTML and flash. I have found ( after much frustration and having wasted days on this ) that no matter what - if i make a change to my program and re publish the HTMl AND SWF files after I have deleted the old HTML and swf files , even then when i run the NEW PUBLISHED html / php file, the movie that runs is the old one. I have tried all that I know, like checking paths and stuff to ensure that everything is ok. I am unable to shed the chached old swf file and so the old movie continues to run. Is there any one who has encountered anything like this? Please help me. This is driving me nuts. Thanks loads in anticipation of a reply from the nerds on this ! |