I've some strange issues with some php code.
if ($user->userType=='admin'){
If I use the above command, the php engine just stop interpreting and display the code in plain text on my browser. On the other hand if I use the below method it works:
if ($user['userType']=='admin'){
Again here also:
$_SESSION['currentUser']->id
If I use the above code it just displays the rest of code as plain text:
id); // fail user }else{ $authentication="failed"; $noAuthPresentation="loginForm"; }
Why this is happening? It's a big project and I don't want to change every line where there is an occurrence of ->.
Do I need to change some setting somewhere? I'm using WAMP server with php 5.5.12.
Any help ? Thanks!
You're mixing up types, user is an array, and not an object. Something in your php config is doing something strange to your error display it seems. Right click on the page that has the errors, and view source if possible.
Does login.php contain html and php code by chance?
Related
I have a slightly frustrating problem...
I'm sending a form value to PHP via AJAX and that seem to work fine.
When I do var_dump in PHP I see my values I can also set a variable and echo it correctly.
However, the line
$prod_id=$_POST['product'];
causes an uncaught type error in the browser.
If I just set the variable with text in PHP everything works fine.
To conclude, this piece of code works fine:
$prod_id=("Slab Skate");
$selected_customers = mysqli_query($link,"SELECT * FROM customers
INNER JOIN cust_products on cust_products.cust_id=id
INNER JOIN products on products.prod_id=cust_products.product_id
WHERE products.prod_name='$prod_id'");
This code causes uncaught typerror:
$prod_id=$_POST['product']
Same SQL statement as above.
If I do
var_dump ($prod_id);
after setting it with $_POST I get:
string(10) "Slab Skate"
My form data in network headers tab of Chrome developer tools say:
product:Slab Skate
I don't get it...
Thanks in advance for any tips.
Update and some clarifications.
The error I get is this: "Uncaught TypeError: Cannot read property 'documentElement' of null" which is a Javascript error coming from a function later in the code. However, since the whole thing works if I "hardcode" the variable instead of setting it from $_POST my assumption was that the error must reside in PHP.
But maybe that's not the case...
What I'm doing is the following: posting a form value to PHP -> use value to select from my_sql and prepare an XML output. So far so good, (I can see the xml output in Chrome dev tools) but then I go back to a javascript to fetch the xml output from my PHP file and then it fails.
When thinking about it, it's rather obvious why it works with a "hardoded" variable and not with the $_POST set one.
So, I see two solutions either set the PHP variable in my_sql or using javascript more intelligently.
Do anyone have a smart solution? I could post all the code, but it's quite long.
Second update:
I solved the issue by writing an xml file to the server instead of trying to download it from the php file. Then my java function can process the xml correctly.
It does work, but I'm not sure how well it scales? It must be better to process the xml output from PHP directly rather then saving it to file first and then process. But, I have no insight on how big the difference is...
/Tim
I wouldn't use SELECT * FROM when using Inner Join if i were you. that would cause all sort of problems , simply give the names of columns.
Hey everybody, this issue has had me stumped for the last week or so, here's the situation:
I've got a site hosted using GoDaddy hosting. The three files used in this issue are index.html , milktruck.js , and xml_http_request.php all hosted in the same directory.
The index.html file makes reference to the milktruck.js file with the following code:
<script type="text/javascript" src="milktruck.js"></script>
The milktruck.js file automatically fires when the site is opened. The xml_http_request.php has not fired at this point.
On line 79 out of 2000 I'm passing the variable "simple" to a function within the milktruck.js file with:
placem('p2','pp2', simple, window['lla0_2'],window['lla1_2'],window['lla2_2']);
"simple" was never initialized within the milktruck.js file. Instead I've included the following line of code in the xml_http_request.php file:
echo "<script> var simple = 'string o text'; </script>";
At this point I have not made any reference whatsoever to the xml_http_request.php file within the milktruck.js file. I don't reference that file until line 661 of the milktruck.js file with the following line of code:
xmlhttp.open('GET',"xml_http_request.php?pid="+pid+"&unLoader=true", false);
Everything compiles (I'm assuming because my game runs) , however the placem function doesn't run properly because the string 'string o text' never shows up.
If I was to comment out the line of code within the php file initializing "simple" and include the following line of code just before I call the function placem, everything works fine and the text shows up:
var simple = 'string o text';
Where do you think the problem is here? Do I need to call the php file before I try using the "simple" variable in the javascript file? How would I do that? Or is there something wrong with my code?
So, we meet again!
Buried in the question comments is the link to the actual Javascript file. It's 2,200 lines, 73kb, and poorly formatted. It's also derived from a demo for the Google Earth API.
As noted in both the comments here and in previous questions, you may be suffering from a fundamental misunderstanding about how PHP works, and how PHP interacts with Javascript.
Let's take a look at lines 62-67 of milktruck.js:
//experiment with php and javascript interaction
//'<?php $simpleString = "i hope this works"; ?>'
//var simple = "<?php echo $simpleString; ?>";
The reason this never worked is because files with the .js extension are not processed by PHP without doing some bizarre configuration changes on your server. Being on shared hosting, you won't be able to do that. Instead, you can rename the file with the .php extension. This will allow PHP to process the file, and allow the commands you entered to actually work.
You will need to make one more change to the file. At the very top, the very very top, before anything else, you will need the following line:
<?php header('Content-Type: text/javascript'); ?>
This command will tell the browser that the file being returned is Javascript. This is needed because PHP normally outputs HTML, not Javascript. Some browsers will not recognize the script if it isn't identified as Javascript.
Now that we've got that out of the way...
Instead I've included the following line of code in the xml_http_request.php file: <a script tag>
This is very unlikely to work. If it does work, it's probably by accident. We're not dealing with a normal ajax library here. We're dealing with some wacky thing created by the Google Earth folks a very, very long time ago.
Except for one or two in that entire monolithic chunk of code, there are no ajax requests that actually process the result. This means that it's unlikely that the script tag could be processed. Further, the one or two that do process the result actually treat it as XML and return a document. It's very unlikely that the script tag is processed there either.
This is going to explain why the variable never shows up reliably in Javascript.
If you need to return executable code from your ajax calls, and do so reliably, you'll want to adopt a mature, well-tested Javascript library like jQuery. Don't worry, you can mix and match the existing code and jQuery if you really wanted to. There's an API call just to load additional scripts. If you just wanted to return data, that's what JSON is for. You can have PHP code emit JSON and have jQuery fetch it. That's a heck of a lot faster, easier, and more convenient than your current unfortunate mess.
Oh, and get Firebug or use Chrome / Safari's dev tools, they will save you a great deal of Javascript pain.
However...
I'm going to be very frank here. This is bad code. This is horrible code. It's poorly formatted, the commenting is a joke, and there are roughly one point seven billion global variables. The code scares me. It scares me deeply. I would be hesitant to touch it with a ten foot pole.
I would not wish maintenance of this code on my worst enemy, and here you are, trying to do something odd with it.
I heartily encourage you to hone your skills on a codebase that is less archaic and obtuse than this one before returning to this project. Save your sanity, get out while you still can!
perhaps init your values like this:
window.simple = 'blah blah blah'
then pass window.simple
You could try the debugger to see what is going on, eg. FireBug
I'am building simple Ajax application (via jquery). I have strange issue. I found where the problem is, but I don't know how to solve it.
This is simple server-side php code:
<?php
require('some.php');
$return['pageContent'] = 'test';
echo(json_encode($return));
?>
On the client side, the error "Invalid JSON" is thrown.
I have discovered that if I delete require function, everything work fine.
Just for information, the "some.php" is an empty php file. There is no error when I open direct php files.
So, conclusion: I cant use require or include function if I want to use ajax?
Use Firebug to see what you're actually getting back during the AJAX call. My guess is that there's a PHP error somewhere, so you're getting more than just JSON back from the call (Firebug will show you that). As for your conclusion: using include/require by itself has absolutely no effect on the AJAX call (assuming there are no errors).
Try changing:
<?php
require('some.php');
$return['pageContent'] = 'test';
echo(json_encode($return));
?>
To:
<?php
$return = array(
'pageContent' => 'test'
);
echo json_encode($return);
?>
The problem might have to do with $return not being declared as an array prior to use.
Edit: Alright, so that might not be the problem at all. But! What might be happening is you might be echoing out something in the response. For example, if you have an error echoing out prior to the JSON, you'd be unable to parse it on the client.
if the "some.php" is an empty php file, why do you require it at all?!
require function throws a fatal error if it could't require the file. try using include function instead and see what happens, if it works then you probably have a problem with require 'some.php';
A require call won't have any effect. You need to check your returned output in Firebug. If using Chrome there is a plugin called Simple REST Client. https://chrome.google.com/extensions/detail/fhjcajmcbmldlhcimfajhfbgofnpcjmb through which you can quickly query for stuff.
Also, it's always good to send back proper HTTP headers with your response showing the response type.
It's most likely the BOM as has been discussed above. I had the same problem multiple times and used Fiddler to check the file in hex and noticed an extra 3 bytes that didn't exist in a prior backup and in new files I created. Somehow my original files were getting modified by Textpad (both in Windows). Although when I created them in Notepad++ I was fine.
So make sure that you have your encoding and codepages set up properly when you create, edit, and save your files in addition to keeping that consistent across OSes if you're developing on windows let's say and publishing to a linux environment at your hosting provider.
After making a javascript variable accesable into a php file the $_GET['something'] keeps saying undefined index.
Although the proper result gets displayed and being written into the xml.
If i try to initialise it my score no longer shows up for some reason.
How can i initalise this without it showing anything at al on my screen?
Regards.
If you do this with javascript:
window.location.href="index.php?scoreresult="+score
you have to access the variable in PHP using
$_GET['scoreresult']
instead of
$_GET['score']
If you make use of XSL to transform the XML, XSLT::setParameter may be interesting to you. It allows you to register (PHP)-variables for use inside a XSL-stylesheet.
I have a strange behavior with PHP system function. In this php script there are only 2 instructions (the rest being pure html):
<?php echo system('cgi-bin/gallery2/galleryheaderview.cgi'); ?>
<?php echo system('cgi-bin/gallery2/galleryview.cgi'); ?>
The first cgi just returns a single line as you can check here
http://reboltutorial.com/cgi-bin/gallery2/galleryheaderview.cgi
It returns
My Gallery
But the whole php script returns My Gallery Twice:
My Gallery
My Gallery
http://reboltutorial.com/gallery2.php
Is there a reason (I don't use My Gallery in second cgi script of course see http://reboltutorial.com/cgi-bin/gallery2/galleryview.cgi) and how to prevent this ?
Thanks.
Update: The system function will do two things. The first is, it will run a command and pass its output through to the browser and/or output buffer. The second is, it will return the last line of output. So when you're saying
echo system('/...');
You're saying "Hey system, output the results of this command" and then "Hey ehco, output whatever system returns". Removing the echo
system('/...');
will fix your problem.
A few other things to check
Are you sure its galleryheaderview.cgi that's returning things twice? Comment out the include to make sure its actually the script that's echoing My Gallery twice
Is your PHP page/program included/constructed in such a way that galleryheaderview.cgi is being called twice?
Are you sure that calling the URL http://reboltutorial.com/cgi-bin/gallery2/galleryheaderview.cgi is calling the same command line as cgi-bin/gallery2/galleryheaderview.cgi?
If you've checked out the three items above, you'll need to drop into the source of galleryheaderview.cgi and see why its outputting the header twice.
Are you absolutely sure that nothing else is outputting the My Gallery before this line? You should try removing it, and see if it goes completely away or if there is still is one "My Gallery"
<?php echo system('cgi-bin/gallery2/galleryheaderview.cgi'); ?>
If this doesn't bring you any further, maybe you have included some php file twice?