|
PHP - How To Echo Target Path Into File Box?
I have an edit page but whenever I click edit the pathname of the original file is not in the file box so everytime I hit the save button , it will be considered as no file so my previous image or file will be overwritten by "nothing".
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. Hi! So I'm working for someone, and they want me to fix this error in a PHP file.. Here is the code: <?php
include_once('config.php');
$online = mysql_query("SELECT * FROM bots WHERE status LIKE 'Online'");
$offline = mysql_query("SELECT * FROM bots WHERE status LIKE 'Offline'");
$dead = mysql_query("SELECT * FROM bots WHERE status LIKE 'Dead'");
$admintrue = mysql_query("SELECT * FROM bots WHERE admin LIKE 'True'");
$adminfalse = mysql_query("SELECT * FROM bots WHERE admin LIKE 'False'");
$windows8 = mysql_query("SELECT * FROM bots WHERE so LIKE '%8%'");
$windows7 = mysql_query("SELECT * FROM bots WHERE so LIKE '%7%'");
$windowsvista = mysql_query("SELECT * FROM bots WHERE so LIKE '%vista%'");
$windowsxp = mysql_query("SELECT * FROM bots WHERE so LIKE '%xp%'");
$unknown = mysql_query("SELECT * FROM bots WHERE so LIKE 'Unknown'");
$totalbots = mysql_num_rows(mysql_query("SELECT * FROM bots"));
$onlinecount = 0;
$offlinecount = 0;
$deadcount = 0;
$admintruecount = 0;
$adminfalsecount = 0;
$windows8count = 0;
$windows7count = 0;
$windowsvistacount = 0;
$windowsxpcount = 0;
$unknowncount = 0;
while($row = mysql_fetch_array($online)){
$onlinecount++;
}
while($row = mysql_fetch_array($offline)){
$offlinecount++;
}
while($row = mysql_fetch_array($dead)){
$deadcount++;
}
while($row = mysql_fetch_array($admintrue)){
$admintruecount++;
}
while($row = mysql_fetch_array($adminfalse)){
$adminfalsecount++;
}
while($row = mysql_fetch_array($windows8)){
$windows8count++;
}
while($row = mysql_fetch_array($windows7)){
$windows7count++;
}
while($row = mysql_fetch_array($windowsvista)){
$windowsvistacount++;
}
while($row = mysql_fetch_array($windowsxp)){
$windowsxpcount++;
}
while($row = mysql_fetch_array($unknown)){
$unknowncount++;
}
$statustotal = $onlinecount + $offlinecount + $deadcount;
$admintotal = $admintruecount + $adminfalsecount;
$sototal = $windows7count + $windowsvistacount + $windowsxpcount + $unknowncount;
?>
Can anyone tell me the error here, can how to fix it? Hi
I try to echo out random lines of a html file and want after submit password to whole content of the same html file. I have two Problems.
1st Problem When I echo out the random lines of the html file I don't get just the text but the code of the html file as well. I don't want that. I just want the text. How to do that?
for($x = 1;$x<=40;$x++) {
$lines = file("$filename.html");
echo $lines[rand(0, count($lines)-1)]."<br>";
}
I tried instead of "file("$filename.html");" "readfile("$filename.html");" But then I get the random lines plus the whole content. Is there anything else I can use instead of file so that I get the random lines of text without the html code?P.S file_get_contents doesn't work either have tried that one.
2nd Problem:
As you could see in my first problem I have a file called $filename.html. After I submit the value of a password I want the whole content. But it is like the program did forget what $filename.html is. How can I make the program remember what $filename.html is? Or with other words how to get the whole content of the html file?
My code:
if($_POST['submitPasswordIT']){
if ($_POST['passIT']== $password ){
$my_file = file_get_contents("$filename.html");
echo $my_file;
}
else{
echo "You entered wrong password";
}
}
If the password isn't correct I get: You entered wrong password. If the password is correct I get nothing. I probably need to create a path to the file "$filename.html", but I don't know exactly how to do that.
I have a made a section on my site where I can upload files to a specific directory, rename them and delete them. So I don't have to go through a FTP program. When renaming and deleting a file I connect using FTP commands with PHP, except when uploading. When uploading it gets the file that has been uploaded from the form on the page before which has its own code using <input type="file"> to search for a local file, then when submit is clicked it uploads it. Then on my PHP script it moves that file from the temp upload folder to the actual site. Bottom line is, is there any way I can make a button that opens a file browser on the users PC and returns the filepath chosen? So I can use FTP to upload the file rather than the form? What does this code mean with respect to the file path?? include ('./includes/header.html'); TomTees Hello am new to php and am working with a config file. But it always give me a wrong path to file /homedomains/xxxxx/public_html/ no / between home an domains.. am frustrated on how to solve this... please help me in this an attaching the script file.. config file too Hi Everyone i am looking for a way to automatically upload a file but skip the browse and choose file process. So i set the location of the file somehow via the form and the client just has to click submit and it will automatcially upload the file from the path i have set. skipping the browse for file process. I would like to intergrate it into the following undesigns S3 Class http://undesigned.org.za/2007/10/22/amazon-s3-php-class/documentation <?php S3::setAuth(awsAccessKey, awsSecretKey); $bucket = "upload-bucket"; $path = "myfiles/"; // Can be empty "" $lifetime = 3600; // Period for which the parameters are valid $maxFileSize = (1024 * 1024 * 50); // 50 MB $metaHeaders = array("uid" => 123); $requestHeaders = array( "Content-Type" => "application/octet-stream", "Content-Disposition" => 'attachment; filename=${filename}' ); $params = S3::getHttpUploadPostParams( $bucket, $path, S3::ACL_PUBLIC_READ, $lifetime, $maxFileSize, 201, // Or a URL to redirect to on success $metaHeaders, $requestHeaders, false // False since we're not using flash ); $uploadURL = "https://{$bucket}.s3.amazonaws.com/"; ?><!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> <html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en"> <head> <title>S3 Form Upload</title> </head> <body> <form method="post" action="<?php echo $uploadURL; ?>" enctype="multipart/form-data"> <?php foreach ($params as $p => $v) echo " <input type=\"hidden\" name=\"{$p}\" value=\"{$v}\" />\n"; ?> <input type="file" name="file" />&#160;<input type="submit" value="Upload" /> </form> </body> </html> at what point does it grab the file path ? is it grabbing it out of the name here <input type="file" name="file" /> can i do something with $_FILES ??? Can someone please give me a point on where to start i would really appriciate this. thanks hello. im having trouble getting my images to show my navigation (which i downloaded from http://www.dynamicdrive.com/dynamicindex17/ddaccordionmenu.htm) im pulling in the navigation dynamically. i can get it to show the navigation and css but not the images its quite a long destination url... if i put this code on the ddAccordian.php page it works Code: [Select] <link href="http://localhost:8888/djsonrotation/admin/pages/contElements/Navigation/Vertical/ddAccordion/css/styles.css" rel="stylesheet" type="text/css" /> if i try and shorten the url like below the css works but i get no images Code: [Select] <?PHP require_once("../includes/initialize.php"); ?> <style type="text/css" rel="stylesheet"> <?php include(NAVIGATION.DS."styles.css"); ?> </style> the Quote include(NAVIGATION.... is being created in my initialize file Code: [Select] defined('NAVIGATION') ? null : define('NAVIGATION', $_SERVER['DOCUMENT_ROOT'].DS.'djsonrotation'.DS.'admin'.DS.'pages'.DS.'contElements'.DS.'Navigation'.DS.'Vertical'.DS.'ddAccordion'.DS.'css'); in my css file i have tried all of these (1 at a time) but none work: Code: [Select] background: black url(titlebar.png) repeat-x center left; background: black url(/titlebar.png) repeat-x center left; background: black url(./titlebar.png) repeat-x center left; background: black url(../titlebar.png) repeat-x center left; background: black url(images/titlebar.png) repeat-x center left; background: black url(/images/titlebar.png) repeat-x center left; background: black url(./images/titlebar.png) repeat-x center left; background: black url(../images/titlebar.png) repeat-x center left; my folders are like this /ddAccordion/css/styles.css /ddAccordion/images/ /ddAccordion/js/ /ddAccordion/ddAccordion.php any thoughts. thanks ricky Hi, I need to echo a file name excluding the full path. i.e.: if the path stored in my DB is '../uploads/none.jpg', I need to echo 'none.jpg' only. How can I do that?. Thanks Need Suggestion for my codes, I want msg1.php to be open in the frame bottomframe. if($username=="" || $password=="") { echo "<form method=\"post\" action=\"msg1.php\" target=\"bottomframe\" >"; } Is there a way to get the source path for an uploaded file? It seems that's a function of the browser and the two I've tried FF and Safari don't return it in $_FILES['file']['name']. I am using glob() to see a list of .zip files on my website. I wanted to display them on a drop down list. I got that working fine, but the file path shows up with it. Any idea how to remove it? Here is the code I am using. Code: [Select] <?php $dir = "/downloads/"; $files = glob($dir."*.zip", GLOB_NOSORT); echo'<select name="Files">'; foreach($files as $file){ echo'<option value="'.$file.'">'.$file.'</option>'; } echo'</select>'; ?> What is an equivalent super global to $_SERVER['PHP_SELF'] that gives me JUST the currently opened document name without the path? e.g. profile.php but NOT: directory1/directory2/profile.php What I mean by that is, if I echo out $_SERVER['PHP_SELF'], it will give me the directories as well, but I'm just interested in the document name. I've looked for the $_SERVER super global in the PHP manual, but no results come up. Hi, i have problem with libraries, with fpdf and PHPEcel, problem is same... Can't open file C:\xampp\htdocs\skripta\.... - this is path to php script on my computer everything work ok on localhost when i test, but when i upload to web server, pdf script try to create file with old windows path: Can't open file C:\xampp\htdocs\skripta\, and not with linux path... does anyone have solution, and why is that? I want to save an image with a unique name that is created from a hash function. How can I cretae the path with the variable... Code: [Select] $hash=$_GET['hash']; $path ="/Applications/MAMP/htdocs/test_upload/images/imagename2.png"; //becomes $path ="/Applications/MAMP/htdocs/test_upload/images/'.$hash.'.png";? Hello Pros... Hi Guys, I need help for in storing data from PHP from array in mysql. I'm very new to PHP/Mysql and have started learing it just few weeks back. I'm tryting to build a website for myself. I'm having tough time trying to insert the data from Form in mysql. I have a form where user update the company name and insert there product name with there image. I want to rename the image with the corresponding text field value and insert the upload path in the table. my FORM Code: [Select] <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Untitled Document</title> <style type="text/css"> body,td,th { color: #000; font-family: Tahoma, Geneva, sans-serif; font-size: 80%; } body { background-color: #9CF; } table { background: #CCF; } th { font: bold normal 16px/normal "Times New Roman", Times, serif; text-transform: capitalize; color: #00F; background: #FFC; } td { font: bold 14px Georgia, "Times New Roman", Times, serif; text-transform: capitalize; color: #3A00FF; background: #FF9; } </style> </head> <body> <form action="upload-file1.php" method="post" enctype="multipart/form-data" name="form"> <strong>Company Name:</strong> <input name="product" type="text" /> <br /> <table> <tr bgcolor="#FF9900"> <th colspan="3" bgcolor="#CCFFFF">filter </th> <th colspan="5" bgcolor="#CCFFFF">heater</th> </tr> <tr><td>filter1</td><td><input name="filter[]" type="text" id="filter"/> <td><input name="img_filter[]" type="file" /></td><td>heater1: </td> <td><input name="heater[]" type="text" id="heater"/></td> <td><input name="img_heater[]" type="file" /></td> </tr> <tr><td>filter2</td><td><input name="filter[]" type="text" id="filter"/> <td><input name="img_filter[]" type="file" /></td><td>heater2: </td> <td><input name="heater[]" type="text" id="heater"/></td> <td><input name="img_heater[]" type="file" /></td> </tr> <tr><td>filter3</td><td><input name="filter[]" type="text" id="filter"/> <td><input name="img_filter[]" type="file" /></td><td>heater3: </td> <td><input name="heater[]" type="text" id="heater"/></td> <td><input name="img_heater[]" type="file" /></td> </tr> <tr><td>filter4</td><td><input name="filter[]" type="text" id="filter"/> <td><input name="img_filter[]" type="file" /></td><td>heater4: </td> <td><input name="heater[]" type="text" id="heater"/></td> <td><input name="img_heater[]" type="file" /></td> </tr> <tr><td>filter5</td><td><input name="filter[]" type="text" id="filter"/> <td><input name="img_filter[]" type="file" /></td><td>heater5: </td> <td><input name="heater[]" type="text" id="heater"/></td> <td><input name="img_heater[]" type="file" /></td> </tr> </table> <br /> <br /> <input name="submit" type="submit" value="submit" /> </form> </body> </html> my PHP Code Code: [Select] <?php require("connect.php"); ?> <?php $product = $_POST['product']; echo $product; if(isset($_POST["submit"])){ $sql = "INSERT INTO company (product) VALUES ('$product')"; $query = mysql_query($sql) OR DIE(mysql_error()); $comp_id = mysql_insert_id(); echo $sql; } ?> <?php function filter() { if(isset($_POST['submit'])) { foreach($_POST['filter'] as $key => $val) { if(trim($val) != '') $filter[] = $val; } $total_records_filter = count($filter); for($i = 0; $i < $total_records_filter; $i++) { $prod_val = $filter[$i]; $sql_prod = "INSERT INTO filter(filter_id, filter) VALUES('', '$prod_val')"; echo $sql_prod.'<br>'; mysql_query($sql_prod) OR die(mysql_error()); } } } filter(); function heater() { if(isset($_POST['submit'])) { foreach($_POST['heater'] as $key => $val) { if(trim($val) != '') $heater[] = $val; } $total_records_heater = count($heater); for($i = 0; $i < $total_records_heater; $i++) { $dir_val = $heater[$i]; $sql_dir = "INSERT INTO heater(heater_id, heater) VALUES('', '$dir_val')"; echo $sql_dir.'<br>'; mysql_query($sql_dir) OR die(mysql_error()); } } } heater(); ?> <?php mysql_close($link)?> My Table structure company comp_id int(5) PK product varchar(50) No filter filter_id int(5) PK filter varchar(25) No filter_path varchar(100) No heater heater_id int(5) PK heater varchar(25) No heater_path varchar(100) No company_filter id int(5) PK comp_id int(5) PK FK (From company table) filter_id int(5) PK FK (From filter table) company_heater id int(5) PK comp_id int(5) PK FK (From company table) heater_id int(5) PK FK (From heater table) Regards BW Given a directory - folder1/subfolder1, I would like to loop through all subdirectories under the given directory and return the file path relative to the given directory. Example: Code: [Select] Given Directory FolderA FolderB SubFolderA SubFolderB SubSubFolderA File1 FolderC the file path for File1 should be FolderB/SubFolderB/SubSubFolderA/ Any ideas? Hi: I'm a newby regarding uploading files to MySQL, turning report output into an HREF, and getting MySQL data via a hyperlink. I have successfully uploaded, files to MySQL, I have also been able to display filename information as a hyperlink in report output, but when I click on the hyperlink, I get the following message format on a 404 page: The requested URL /current_dir_of_requesting_page/filename.filetype was not found on this server. My reporting page has teh following line of code in a report table to create the hyperlink: <td style="width:225px"><? echo "<a href=".$row['name'].">".$row['name']."</a>";?></td> Can anyone assist me with this? Hi, I have a form that allows a user to enter details and also to upload a image. The form works great and inserts into the database if i dont include the file input. What i would like happen is: The user enters details and selects a file The file then gets uploaded to ../images and the path is then written to bedroom1 variable The addresses and the bedroom1 variable that now holds the uploaded file path e.g. images/test.jpg is written to the database Form Code: [Select] <form action="admin.php" method="POST"> <table> <tr><td> Address 1: </td><td> <input type="text" name="address1" id="address1" onfocus="selected(this)" onblur="notselected(this)"> </td><td> <div id="dynamicText1"> </div> </td></tr> <tr><td> Address 2: </td><td> <input type="text" name="address2" id="address2" onfocus="selected(this)" onblur="notselected(this)"> </td><td> <div id="dynamicText1"> </div> </td></tr> <tr><td> County: </td><td> <input type="text" name="county" id="county" onfocus="selected(this)" onblur="notselected(this)"> </td><td> <div id="dynamicText1"> </div> </td></tr> <tr><td> Bedroom 1: </td><td> <input type="file" name="bedroom1" id="bedroom1" input name="uploadedfile"> </td><td> <div id="dynamicText1"> </div> </td></tr> <tr><td> <input type="submit" name="submitaddapartment" value="Add" id="registerbutton"> </td></tr> </table </form> And the php code to add it to the database Code: [Select] <?php //get variables $submitaddapartment = $_POST['submitaddapartment']; $address1 = strip_tags($_POST['address1']); $address2 = strip_tags($_POST['address2']); $county = strip_tags($_POST['county']); $bedroom1 = strip_tags($_POST['bedroom1']); if ($submitaddapartment) //if submit button was pressed { if ($address1&&$address2&&$county&&$bedroom1&&) //if fields arn't blank { include 'dbase.php'; //connect to database if ((($_FILES["bedroom1"]["type"] == "image/gif") || ($_FILES["bedroom1"]["type"] == "image/jpg") || ($_FILES["bedroom1"]["type"] == "image/pjpeg")) && ($_FILES["bedroom1"]["size"] < 20000000)) { if ($_FILES["bedroom1"]["error"] > 0) { echo "Return Code: " . $_FILES["bedroom1"]["error"] . "<br />"; } else { echo "Upload: " . $_FILES["bedroom1"]["name"] . "<br />"; echo "Type: " . $_FILES["bedroom1"]["type"] . "<br />"; echo "Size: " . ($_FILES["bedroom1"]["size"] / 1024) . " Kb<br />"; echo "Temp file: " . $_FILES["bedroom1"]["tmp_name"] . "<br />"; if (file_exists("../images/" . $_FILES["bedroom1"]["name"])) { echo $_FILES["bedroom1"]["name"] . " already exists. "; } else { move_uploaded_file($_FILES["bedroom1"]["tmp_name"], "/" . $_FILES["bedroom1"]["name"]); echo "Stored in: " . "../images/" . $_FILES["bedroom1"]["name"]; $bedroom1=".'images/.' $_FILES["bedroom1"]["name"]"; } } } else { echo "Invalid file"; } mysql_query ("INSERT INTO user VALUES(NULL,'$address1','$address2','$county','$bedroom1')"); } else { } } ?> Any help would be greatly appreciated. Thanks Fred |
|||||||