How do I make the die() message to echo in a certain place in the HTML section of the same page?
$files = array();
$upload = $_FILES['upload']['tmp_name'];
foreach($upload as $uploaded){
if(!empty($uploaded)) {
if(isset($uploaded)){
$finfo = finfo_open(FILEINFO_MIME_TYPE);
$mime= finfo_file($finfo, $uploaded);
switch($mime) {
case 'application/pdf':
break;
default:
die('pdf file only.');
break;
}
}
}
}
die will immediately stop execution and send anything in buffers to the browser.
Personally, I like to do something like this:
function halt($str="") {
if( $str) echo "<div class=\"server_notice\">".$str."</div>";
require("template/foot.php");
exit;
}
How about don't just use die, do something to insert html there and then die.
Alternatively, you have register_shutdown_function (Documentation: http://php.net/manual/es/function.register-shutdown-function.php) which will allow you to do things right after the script is ended with die;
There is no way to do that. die() or exit() will stop executing of your script.
Thouh, you can make some error reporting system.
$lastError = null
Then do what you want and set this error.
Then you can check it in some place:
if ($lastError == 2){
echo "The file is no in PDF format" ;
} //and so on.
Also you could create some constants like:
define("ERROR_WRONG_FORMAT", 2); //Make the error clear.
You'll need to echo out div tags and then position them using CSS.
die('<div id="error">pdf file only.</div>');
Then add the following text to your CSS:
#error{position:absolute;top:10;left:10;}
You'll need to change the top and left values depending on where you wnat them to be.
If you don't know what CSS is, I suggest you watch TheNewBoston's tutorials on YouTube!
Related
I am developing a PHP script that allows me to modify tags in an XML file and move them once done.
My script works correctly but I would like to add error handling: So that if the result of my SQL query does not return anything display an error message or better, send a mail, and not move the file with the error and move to the next.
I did some tests but the code never displays the error and it moves the file anyway.
Can someone help me to understand why? Thanks
<?php
}
}
$xml->formatOutput = true;
$xml->save($source_file);
rename($source_file,$destination_file);
}
}
closedir($dir);
?>
Give this one a try
$result = odbc_fetch_array($exec);
if ($result === false || $result['GEAN'] === null) {
echo "GEAN not found for $SKU_CODE";
// continue;
}
$barcode = (string) $result['GEAN'];
echo $barcode; echo "<br>"; //9353970875729
$node->getElementsByTagName("SKU")->item(0)->nodeValue = "";
$node->getElementsByTagName("SKU")->item(0)->appendChild($xml->createTextNode($result[GEAN]));
I'm very new to PHP, and I can't figure out why this is happening.
For some reason, when exit fires the entire page stops loading, not just the PHP script. Like, it'll load the top half of the page, but nothing below where the script is included.
Here's my code:
$page = $_GET["p"] . ".htm";
if (!$_GET["p"]) {
echo("<h1>Please click on a page on the left to begin</h1>\n");
// problem here
exit;
}
if ($_POST["page"]) {
$handle = fopen("../includes/$page", "w");
fwrite($handle, $_POST["page"]);
fclose($handle);
echo("<p>Page successfully saved.</p>\n");
// problem here
exit;
}
if (file_exists("../includes/$page")) {
$FILE = fopen("../includes/$page", "rt");
while (!feof($FILE)) {
$text .= fgets($FILE);
}
fclose($FILE);
} else {
echo("<h1>Page "$page" does not exist.</h1>\n");
// echo("<h1>New Page: $page</h1>\n");
// $text = "<p></p>";
// problem here
exit;
}
Even if you have HTML code following your PHP code, from the web server's perspective it is strictly a PHP script. When exit() is called, that is the end of it. PHP will output process and output no more HTML, and the web server will not output anymore html. In other words, it is working exactly as it is supposed to work.
If you need to terminate the flow of PHP code execution without preventing any further HTML from being output, you will need to reorganize your code accordingly.
Here is one suggestion. If there is a problem, set a variable indicating so. In subsequent if() blocks, check to see if previous problems were encountered.
$problem_encountered = FALSE;
if (!$_GET["p"]) {
echo("<h1>Please click on a page on the left to begin</h1>\n");
// problem here
// Set a boolean variable indicating something went wrong
$problem_encountered = TRUE;
}
// In subsequent blocks, check that you haven't had problems so far
// Adding preg_match() here to validate that the input is only letters & numbers
// to protect against directory traversal.
// Never pass user input into file operations, even checking file_exists()
// without also whitelisting the input.
if (!$problem_encountered && $_GET["page"] && preg_match('/^[a-z0-9]+$/', $_GET["page"])) {
$page = $_GET["p"] . ".htm";
$handle = fopen("../includes/$page", "w");
fwrite($handle, $_GET["page"]);
fclose($handle);
echo("<p>Page successfully saved.</p>\n");
// problem here
$problem_encountered = TRUE;
}
if (!$problem_encountered && file_exists("../includes/$page")) {
$FILE = fopen("../includes/$page", "rt");
while (!feof($FILE)) {
$text .= fgets($FILE);
}
fclose($FILE);
} else {
echo("<h1>Page "$page" does not exist.</h1>\n");
// echo("<h1>New Page: $page</h1>\n");
// $text = "<p></p>";
// problem here
$problem_encountered = TRUE;
}
There are lots of ways to handle this, many of which are better than the example I provided. But this is a very easy way for you to adapt your existing code without needing to do too much reorganization or risk breaking much.
In PHP 5.3+ you can use the goto statement to jump to a label just before the ?> instead of using exit in the example given in the question.
It would'n work well with more structured code (jumping out of functions), tough.
Maybe this should be a comment, who knows.
I was following this tutorial.
I need to use a php file's ouput in my HTML file to dynamically load images into a gallery. I call
function setOutput()
{
if (httpObject.readyState == 4)
document.getElementById('main').src = httpObject.responseText;
alert("set output: " + httpObject.responseText);
}
from
function doWork()
{
httpObject = getHTTPObject();
if (httpObject != null) {
httpObject.open("GET", "gallery.php?no=0", true);
httpObject.send(null);
httpObject.onreadystatechange = setOutput;
}
}
However, the alert returns the php file, word for word. It's probably a really stupid error, but I can't seem to find it.
The php file:
<?php
if (isset($_GET['no'])) {
$no = $_GET['no'];
if ($no <= 10 && $no >1) {
$xml = simplexml_load_file('gallery.xml');
echo "images/" . $xml->image[$no]->src;
}
else die("Number isn't between 1 and 10");
}
else die("No number set.");
?>
If the alert is returning the contents of the PHP file instead of the results of executing it, then the server is not executing it.
Test by accessing the URI directly (instead of going via JavaScript).
You probably need to configure PHP support on the server.
Your Server doesn't serve/parse PHP files! You could test your JavaScript code by setting the content of gallery.php to the HTML code you want to receive.
I have got this JavaScript code for uploading files to my server (named it "upload.js"):
function startUpload(){
document.getElementById('upload_form').style.visibility = 'hidden';
return true;
}
function stopUpload(success){
var result = '';
if (success == 1){
result = '<div class="correct_sms">The file name is [HERE I NEED THE VARIABLE FROM THE EXTERNAL PHP FILE]!</div>';
}
else {
result = '<div class="wrong_sms">There was an error during upload!</div>';
}
document.getElementById('upload_form').innerHTML = result;
document.getElementById('upload_form').style.visibility = 'visible';
return true;
}
And I've got a simple .php file that process uploads with renaming the uploaded files (I named it "process_file.php"), and connects again with upload.js to fetch the result:
<?php
$file_name = $HTTP_POST_FILES['myfile']['name'];
$random_digit = rand(0000,9999);
$new_file_name = $random_digit.$file_name;
$path= "../../../images/home/smsbanner/pixels/".$new_file_name;
if($myfile !=none)
{
if(copy($HTTP_POST_FILES['myfile']['tmp_name'], $path))
{
$result = 1;
}
else
{
$result = 0;
}
}
sleep(1);
?>
<script language="javascript" type="text/javascript">window.top.window.stopUpload(<?php echo $result; ?>);</script>
What I need is inside upload.js to visualize the new name of the uploaded file as an answer if the upload process has been correct? I wrote inside JavaScript code above where exactly I need to put the new name answer.
You have to change your code to the following.
<?php
$file_name = $HTTP_POST_FILES['myfile']['name'];
$random_digit=rand(0000,9999);
$new_file_name=$random_digit.$file_name;
$path= "../../../images/home/smsbanner/pixels/".$new_file_name;
if($myfile !=none)
{
if(copy($HTTP_POST_FILES['myfile']['tmp_name'], $path))
{
$result = 1;
}
else
{
$result = 0;
}
}
sleep(1);
?>
<script language="javascript" type="text/javascript">window.top.window.stopUpload(<?php echo $result; ?>, '<?php echo "message" ?>');</script>
And your JavaScript code,
function stopUpload(success, message){
var result = '';
if (success == 1){
result = '<div class="correct_sms">The file name is '+message+'!</div>';
}
else {
result = '<div class="wrong_sms">There was an error during upload!</div>';
}
document.getElementById('upload_form').innerHTML = result;
document.getElementById('upload_form').style.visibility = 'visible';
return true;
}
RageZ's answer was just about what I was going to post, but to be a little more specific, the last line of your php file should look like this:
<script language="javascript" type="text/javascript">window.top.window.stopUpload(<?php echo $result; ?>, '<?php echo $new_file_name ?>');</script>
The javascript will error without quotes around that second argument and I'm assuming $new_file_name is what you want to pass in. To be safe, you probably even want to escape the file name (I think in this case addslashes will work).
A dumb man once said; "There are no stupid questions, only stupid answers". Though he was wrong; there are in fact loads of stupid questions, but this is not one of them.
Besides that, you are stating that the .js is uploading the file. This isn't really true.
I bet you didn't post all your code.
You can make the PHP and JavaScript work together on this problem by using Ajax, I recommend using the jQuery framework to accomplish this, mostly because it has easy to use functions for Ajax, but also because it has excellent documentation.
How about extending the callback script with:
window.top.window.stopUpload(
<?php echo $result; ?>,
'<?php echo(addslashes($new_file_name)); ?>'
);
(The addslashes and quotes are necessary to make the PHP string come out encoded into a JavaScript string literal.)
Then add a 'filename' parameter to the stopUpload() function and spit it out in the HTML.
$new_file_name=$random_digit.$file_name;
Sorry, that is not sufficient to make a filename safe. $file_name might contain segments like ‘x/../../y’, or various other illegal or inconsistently-supported characters. Filename sanitisation is much harder than it looks; you are better off making up a completely new (random) file name and not relying on user input for it at all.
when I'm trying to getimagesize($img) and the image doesn't exist, I get an error. I don't want to first check whether the file exists, just handle the error.
I'm not sure how try catch works, but I want to do something like:
try: getimagesize($img) $works = true
catch: $works = flase
Like you said, if used on a non-existing file, getimagesize generates a warning :
This code :
if ($data = getimagesize('not-existing.png')) {
echo "OK";
} else {
echo "NOT OK";
}
will get you a
Warning: getimagesize(not-existing.png) [function.getimagesize]:
failed to open stream: No such file or directory
A solution would be to use the # operator, to mask that error :
if ($data = #getimagesize('not-existing.png')) {
echo "OK";
} else {
echo "NOT OK";
}
As the file doesn't exist, $data will still be false ; but no warning will be displayed.
Another solution would be to check if the file exists, before using getimagesize ; something like this would do :
if (file_exists('not-existing.png') &&
($data = getimagesize('not-existing.png'))
) {
echo "OK";
} else {
echo "NOT OK";
}
If the file doesn't exist, getimagesize is not called -- which means no warning
Still, this solution is not the one you should use for images that are on another server, and accessed via HTTP (if you are in this case), as it'll mean two requests to the remote server.
For local images, that would be quite OK, I suppose ; only problem I see is the notice generated when there is a read error not being masked.
Finally :
I would allow errors to be displayed on your developpement server,
And would not display those on your production server -- see display_errors, about that ;-)
Call me a dirty hacker zombie who will be going to hell, but I usually get around this problem by catching the warning output into an output buffer, and then checking the buffer. Try this:
ob_start();
$data = getimagesize('not-existing.png');
$resize_warning = ob_get_clean();
if(!empty($resize_warning)) {
print "NOT OK";
# We could even print out the warning here, just as PHP would do
print "$resize_warning";
} else {
print "OK"
}
Like I said, not the way to get a cozy place in programmer's heaven, but when it comes to dysfunctional error handling, a man has to do what a man has to do.
I'm sorry that raise such old topic. Recently encountered a similar problem and found this topic instead a solution. For religious reasons I think that '#' is bad decision. And then I found another solution, it looks something like this:
function exception_error_handler( $errno, $errstr, $errfile, $errline ) {
throw new Exception($errstr);
}
set_error_handler("exception_error_handler");
try {
$imageinfo = getimagesize($image_url);
} catch (Exception $e) {
$imageinfo = false;
}
This solution has worked for me.
try {
if (url_exists ($photoUrl) && is_array (getimagesize ($photoUrl)))
{
return $photoUrl;
}
} catch (\Exception $e) { return ''; }
Simple and working solution based on other answers:
$img_url = "not-existing.jpg";
if ( is_file($img_url) && is_array($img_size = getimagesize($img_url)) ) {
print_r($img_size);
echo "OK";
} else {
echo "NOT OK";
}