When I try to access info that is not presented in xml like so: $someInfo = $element->blabla->cats[0]->src;
PHP shows notice like this: Notice: Trying to get property of non-object
How would I settle the matter?
Either $element, blabla, or cats[0] is not an object, and thus can't contain any elements.
Use isset():
if (isset($element->blabla->cats[0]->src))
echo $element->blabla->cats[0]->src;
one isset() should do, no need to check every part consecutively.
This should do the job even if cats exists but is not an array.
you can use isset to verify if object property exists, like this:
if (isset ($element->blabla) && isset ($element->blabla->cats) && etc..)
if you just don't want to see the notice, use error_reporting(E_ALL & ~E_NOTICE)
Or alternative (but wrong way) you can suppress log in your php script by:
error_reporting(E_ERROR);
Which will force php to report only fatal errors.
Anyway, use Pekka solution.
Related
In PHP I have the following code:
$my_var = $_GET['my_var'] or $my_var = 'empty';
Which gives a me a default value to fall back on if $_GET['my_var'] is not set and that works just fine. Although this does work as I want it to, it still generates an undefined variable warning in the error log.. is that acceptable or should I strive to never have anything go into the error log? If so can I suppress this error?
In the past I've used such functions as isset() or empty() but I like the closeness of this code to python's try.
A better way to set $my_var could be:
$my_var = isset($_GET['my_var']) ? $_GET['my_var'] : "empty";
You definitely don't want to pollute your error logs with unnecessary messages. It makes it hard to find errors you actually want to debug, and is probably making the interpreter work hard than it needs to.
I'm building my first application with Kohana, and using a basic templating system within that. In my templates I want to echo variables for the various contents of the page, but only if each variable is set, and I want to keep the code in the templates as short as possible, so something like this:
<?=$foo?>
works fine if the variable is set, but if it's not I get a notice. So I thought a ternary operator would do the trick nicely:
<?=$foo?:''?>
according to the PHP manual, from 5.3 it's ok to leave out the middle part and the above should output nothing if the variable isn't set, but I still get an error notice."Notice: Undefined variable: foo in /"
I can get the desired result with a slight alteration to suppress the notice:
<?=#$foo?:''?>
but I know that's generally not beset practice and would like a better solution if possible, while still keeping the code to a minimum in the template files.
the following works, but it's not as concise (mainly because my actual variables can be quite long):
<?=isset($foo)?$foo:'';?>
am I missing something or doing something wrong?
The ternary operation is not meant to replace checking with isset() - it needs it's variable defined or else you get a notice.
Template engines usually offer a way to use a default value instead, but they also do not use pure PHP code. You you are out of luck here: Either suppress the notice, or use the longer code variant. Or ensure that every variable is set, which enables you to consider any notice an error.
To avoid notices for undefined variables, you can create custom function that takes first parameter by reference
function tplvar(&$value, $default = '') {
return ($value !== null) ? $value : $default;
}
<?=tplvar($foo, 'bar');?>
Uninitialized variables passed by reference will be seen as nulls.
I have a simple form I created, and in it I have the following checkbox:
<input type="checkbox" name="test">
Note: this form is being submitted to itself.
Above the form, I have the following PHP:
if (empty($_POST['test'])) {
$thevalue = 0;
} else {
$thevalue = 1;
}
var_dump($thevalue);
When I process the form, I get what I would expect. If I check the box and submit, I get int(1) if I leave it unchecked I get int(0).
In the first line of my PHP code, I wanted to replace $_POST['test'] with some simple variable.
So I added the following line above my code:
$simplevar = $_POST['test']
I then replaced the condition in my if statement to be empty($simplevar)
But when I submit the form, I get a "Notice: Undefined index:" error message
Why is this happening?
Assuming it's possible to achieve what I was after (i.e. insert $_POST into $simplevar), how might I go about it?
Thanks in advance for your help!
PS: I may have a follow up to this question, but didn't want to clutter things by jamming it all in here.
Thanks again... oh, and Merry Christmas! ;-)
This happens because when you don't check the checkbox, the browser does not send any value to server for that control when the form is submitted. Because of this, $_POST['test'] is not defined, and you tried to use it without a check as to whether it existed, so you get a warning. One of the checks that empty() does is to see whether the value is set. So, when you use the $_POST keys directly in empty(), you don't get an error, but when you try and use it in an assignment without this check, you will get the error.
You can do roughly what you want to do, you just have to change the logic slightly. If you do:
$simplevar = !empty($_POST['test']);
// You could also do
// $simplevar = isset($_POST['test']);
if ($simplevar) {
// The box was checked
} else {
// The box was not checked
}
...it will do what you want without the error. Using this approach, $simplevar always holds a boolean indicating whether or not the box was checked.
When a checkbox is unchecked, it's not added to the $_POST array as a key, which is why $simplevar = $_POST['test'] returns the error you posted. Using empty() gets past this problem by empty() handling errors better (well, silently at any rate).
You haven't specified whether you get that error when the checkbox is checked or not, but the above explanation is the only one I can give. If you're unsure, try doing print_r($_POST) to see what $_POST actually contains.
A solution to your problem would be to use a ternary expression to handle the error a little better:
$simplevar = isset($_POST['test']) ? 0 : 1;
This will assign 0 to $simplevar if $_POST['test'] isn't set (checkbox isn't checked), or 1 otherwise.
Do make sure all your form processing code is put inside
if(!empty($_POST)) {
// Code
}
So that it's not executed every time the page loads, otherwise your error will show every time.
Checkbox values are only transmitted if the checkbox was checked. This means that unchecked checkboxes won't appear in the $_POST array.
A way to suppress the notice from PHP is to use a reference instead of a variable:
$simplevar =& $_POST['test'];
if(empty($simplevar)) $thevalue = 1;
else $thevalue = 0;
That's expected behaviour. If you are assigning the variable like this:
$simplevar = $_POST['test'];
Then the $_POST variable might be absent. The Zend runtime then assigns the NULL value, but gives you a useful debug hint, should that not be what you wanted.
When you used empty() before, the check for variable existence was built in. empty() is a language construct. Like isset() it's often used to eschew such notices. The cumbersome syntax to emulate such language behaviour is:
$simplevar = empty($_POST['test']) ? NULL : $_POST['test'];
The language built-in for is:
$simplevar = #( $_POST['test'] );
Now, I will get roasted for mentioning it. (Using # is useful if you want to bring the debug notices back at some point, while the empty and isset constructs eternally suppress them.)
First, you should always check that variables in $_POST, $_REQUEST, and $_GET are set before attempting to use them. Always handle the condition where they are not set even if you simply output an error.
Because the error is an undefined index it seem the error is in test not being set in $_POST, though that doesn't make a lot of sense. I would add a check, maybe an echo or var dump to check $_POST. If it is set the other problem could be an issue with scope. $_POST is something called a super global which makes it available in any scope. Variables you set you may need to make global by defining them as such if you want to access them across scopes.
Is it OK to rely on PHP's behaviour if a key is not in the array? Which is:
Attempting to access an array key which has not been defined is the same as accessing any other undefined variable: an E_NOTICE-level error message will be issued, and the result will be NULL.
For instance if I do not know $_POST contains certain keys is it OK to just try? I want the result to be null anyway. Or is there a better way?
Since you should always develop with error reporting turned on, you want to avoid triggering avoidable errors so as to keep error reporting useful to you. As such, no, it's not okay. Use isset or empty. For a more in-depth excursion into this topic see The Definitive Guide To PHP's isset And empty.
You can use isset() to check if it exists and if so do stuff.
if(isset($_POST["key"])){// do stuff}
You should use isset() or empty() to test it.
If it is necessary to check if a key exists even if it the key is set to null then array_key_exists(mixed $key , array $search) will be usefull. But it is very slow compared to isset(). So isset() shall be used if you check for a key that is in the array and not null.
isset()function can save you from the warning of index not found by checking for it first
isset($_POST['notSure']) ? $var= $_POST['notSure'] : $var = null;
In a php page I have following code:
if($_REQUEST['c']!="") // I get error on this line itself. Why?
{
$pidis=(int)($_REQUEST['c']);
}
I keep getting Undefined index error.
On Googling I manage to understand that if a page is access without parameters (in URL) which we are trying to access we can get this error/warning. I believe that if a parameter is not defined in the URL it should just return empty instead of giving error/warning message.
I know that it is possible to suppress errors and warning by adding
error_reporting(E_ALL ^ E_NOTICE);
But I do not want to do this.
This same page work just fine on our company's web server but does not work on our clients web server.
Why is this happening?
How to solve this problem?
You are getting that error because you are attempting to compare $_REQUEST['c'] to something when $_REQUEST['c'] does not exist.
The solution is to use isset() before comparing it. This will remove the warning, since the comparison won't happen if $_REQUEST['c'] doesn't exist.
if(isset($_REQUEST['c']) && $_REQUEST['c']!="")
{
$pidis=(int)($_REQUEST['c']);
}
It is an E_NOTICE level error, and your level of error reporting will affect whether the error shows up or not. Your client's server has E_NOTICE level error reporting turned on, which is why it shows up there.
It is a good idea to always develop using E_ALL so that you can catch this kind of error before moving your code to other servers.
Another solution is to use the following:
$pidis = isset($_REQUEST['c']) ? $_REQUEST['c'] : '';
You can also, if you prefer to return a value other than empty, by placing a default value within the final set of single quotes, e.g.
$pidis = isset($_REQUEST['c']) ? $_REQUEST['c'] : 'Default Value';
or return a different variable type, for instance an integer:
$pidis = isset($_REQUEST['c']) ? $_REQUEST['c'] : 34;
Instead of isset() you can also use: array_key_exists().
The difference between both methods is that isset() checks also whether the value of the variable is null. If it is null then isset returns false whereas array_key_exists() returns always true if the key exists (no mater which value). E.g.:
$array = array('c' => null);
var_dump(isset($array['c']))); // isset() returns FALSE here
var_dump(array_key_exists($array['c']); //array_key_exists() returns TRUE
Depending on the context, it is important to distinguish this. In your case I don't think it matters doesn't matter, as (I guess) a request parameter never will be null (except one overwrites it manually).
Use isset($_REQUEST['c']) to test if it exists first.
PHP is giving a notice (which is not an error : it's just a notice) when you are trying to use a variable that doesn't exists, or an array element that doesn't exist.
This is just to help you, and you should not mask those notices : they are here to help you -- for instance, to help you detect typos in variable names.
Before using that array index, if it's not always present, you should test if it's here, using isset :
if (isset($_REQUEST['c']) && $_REQUEST['c']!="") {
// ...
}
Clean way could be :
$pidis = $_REQUEST['c'] ?? null
this is same as checking isset request but shorter.