PHP super weird warning : illigal string offset while creating a key - php

I have a very strange problem.
I am running through a foreach loop to compile an array but I receive an error.
I reveive the following warning :
Warning: Illegal string offset 'clientaccount_id' in
For this line of code:
$this->PreparedData[$table][$field] = 0;
I would say this to be logic if I would be doing something like:
$testVariable = $this->PreparedData[$table][$field];
Then the variable $field filled with 'clientaccount_id' would not exist.
But I am CREATING the field 'clientaccount_id' so to ME this is almost impossible to give an error.
The code
private function AssignData(){
foreach($this->FieldKeys as $table => $value){
///######## IF THE PREPARED DATA ARRAY DOES NOT EXIST
if(isset($this->PreparedData[$table]) === false){
///######## SET THE ARRAY KEY
$this->PreparedData[$table] = array();
}
///######## RUN THROUGH ALL SET SUB DATA
foreach($value as $field){
///######## IF THE FIELD EXISTS
if(isset($this->AccountData[$field]) === true){
///######## ASSIGN THE DATA
///$this->PreparedData[$table][$field] = $this->AccountData[$field];
///$this->PreparedData[$field] = $this->AccountData[$field];
$this->PreparedData[$table][$field] = 0;
}
}
}
exit('GOT THROUGH!!');
}
Could anyone see the error I am overlooking?
Solved!!
Thanks to VMcreator
Changed :
isset($this->PreparedData[$table]) === false
to this:
is_array($this->PreparedData[$table]) === false
Please read the explanation below WHY

Try to change this line:
isset($this->PreparedData[$table]) === false
to this:
!is_array($this->PreparedData[$table])
I saw this explanation here:
It just boils down to PHP's crazy type system.
$fruits['response']['errormessage'] is the string 'banana', so you're
attempting to access a character in that string by the ['orange']
index.
The string 'orange' is converted to an integer for the purposes of
indexing, so it becomes 0, as in
$fruits['response']['errormessage'][0]. The 0th index of a string is
the first character of the string, so for non-empty strings it's
essentially set. Thus isset() returns true.
You might be curious why your situation is comparable to that quoted statement even if $this->PreparedData[$table] seems a single dimensional array only, well its not a single dimensional array only, because you are accessing a class object, its just like doing this $this["PreparedData"][$table].

Related

While loop: only Variables can be passed by reference error

I am running into a "Only variables should be passed by reference" error, because on the code I am using there is a line that does not put the explode() result into a variable. As required when using strict PHP standards.
However because the explode() function is used in a While loop I can't think of a appropriate solution.
My code looks like
function user_exists($username) {
rewind($this->fp);
while(!feof($this->fp) && trim($lusername = array_shift(explode(":",$line = rtrim(fgets($this->fp)))))) {
if($lusername == $username)
return 1;
}
return 0;
}
Any suggestions on how to solve this?
I think maybe you need to sit back and break your code apart a bit and take a look at what is happening.
First, condition is while !feof($this->fp)
From the manual:
feof — Tests for end-of-file on a file pointer
One thing you will notice here is that feof() is only a test which returns true or false. It does not advance the pointer position while looping over, so while using this function, somewhere else in your while loop there needs to be something that advances the pointer or else you will have an infinite loop.
Second condition is:
trim($lusername = array_shift(explode(":",$line = rtrim(fgets($this->fp)))))
First function from left to right is trim(), which returns a string. From our handy dandy comparison table we see that when doing if ((String) $var) it evaluates to false if and only if the string is empty ("") or the number zero as a string ("0"), otherwise it returns true. Personally I tend to really hate using if ((String) $var) (first because it's slightly unclear to newbies unless you know your comparison table well and second because 99% of the time people are doing that they are actually checking for string length, in which case I would want it to return true for the string "0"). So assuming that you don't need it to return false for "0" we could change this to strlen($var) > 0 and then manipulate the variable within the loop. That should greatly simplify things here.
So now we have:
while (!feof($this->fp) && strlen($var) > 0) { /*...*/ }
This will loop over until either we are at the end of the file or $var is an empty line. Everything else can be offloaded into the body of the while loop, so it is much easier to break apart.
So this is what we have now:
$line = rtrim(fgets($this->fp));
$lusername = array_shift(explode(":",$line)));
Uh-oh! There's that "nasty" error:
Strict Standards: Only variables should be passed by reference in /path/to/file.php on line x.
So we can see from here, the part producing the error is not explode(), but array_shift(). See also: Strict Standards: Only variables should be passed by reference
What this means is that since array_shift() modifies the array, it requires it to be by reference. Since you are not passing an actual variable but instead the result of a function, PHP is unable to modify it. It's similar to doing something like function($var) = 3;. Of course you can't do that. Instead you need to save the value to a temporary variable. So now we have:
$line = rtrim(fgets($this->fp));
$split = explode(":",$line);
$lusername = array_shift($split);
Woo hoo! No more warning message.
So putting this together, we now have:
while (!feof($this->fp) && strlen($lusername) > 0) {
$line = rtrim(fgets($this->fp));
$split = explode(":",$line);
$lusername = array_shift($split);
if($lusername == $username) {
return 1;
}
}
Also, as mentioned earlier, the fgets() will advance the pointer, which allows the !feof($this->fp) part in the while statement to vary.

isset() returns true from a string variable accessed as an array with any key

I face a problem like this:
$area="Dhaka";
isset($area); //returns true which is OK
isset($area['division']); //returns true why?
// actually, any array key of area returns true
isset($area['ANY_KEY']);//this is my question 1
isset($area['division']['zilla');//now it returns false.
//as I know it should returns false but why previous one was true.
Now if I do this:
$area['division'] = "Dhaka";
isset($area); // returns true which is OK
isset($area['division']); // returns true it's also OK
isset($area['ANY_KEY']); // returns false. I also expect this
isset($area['division']['ANY_KEY']); // returns true why? question #2
Basically both of my questions are the same.
Can anyone explain this?
As with every programming language in existence, a string is stored as an array of characters.
If I did:
$area = "Dhaka";
echo $area[0];
It would return D.
I could also echo the whole string by doing:
echo $area[0].$area[1].$area[2].$area[3].$area[4];
PHP will also type juggle a string into 0 when passed in a manner that accepts only integers.
So by doing:
echo $area['division'];
You would essentially be doing:
echo $area[0];
and again, getting D.
That's why isset($area['division']) returns a true value.
Why doesn't $area['foo']['bar'] (aka $area[0][0]) work? Because $area is only a single-dimension array.
The best approach to handle this problem when you're working with a variable that could either be a string or an array is to test with is_array() before trying to treat your variable as an array:
is_array($area) && isset($area['division'])
PHP lets you treat a string as an array:
$foo = 'bar';
echo $foo[1]; // outputs 'a'
So
$area['division']
will be parsed/executed as
$area[0];
(the keys cannot be strings, since it's not REALLY an array, so PHP type-converts your division string by its convert-to-int rules, and gives 0), and evaluate to the letter D in Dhaka, which is obviously set.
Okay, here's a solution rather than explaining why isset isn't going to work properly.
You want to check if an array element is set based on it's index string. Here's how I might do it:
function isset_by_strkey($KeyStr,$Ar)
{
if(array_key_exists($KeyStr,$Ar))
{
if(strlen($Ar[$KeyStr]) > 0 || is_numeric($Ar[$KeyStr] !== FALSE)
{
return TRUE;
}
return FALSE;
}
}
isset_by_strkey('ANY_KEY',$area); // will return false if ANY_KEY is not set in $area array and true if it is.
The best way to access a linear array in php is
// string treated as an linear array
$string= "roni" ;
echo $string{0} . $string{1} . $string{2} . $string{3};
// output = roni
It is expected behaviour.
PHP Documentation covers this
You can try empty() instead.
If it is returning true for keys that do not exist there's nothing you can do; however, you can make sure that it doesn't have a negative effect on your code. Just use array_key_exists() and then perform isset() on the array element.
Edit: In fact, using array_key_exists() you shouldn't even need isset if it is misbehaving just use something like strlen() or check the value type if array_key_exists returns true.
The point is, rather than just saying isset($Ar['something']) do:
if(array_key_exists('something',$Ar) )
and if necessary check the value length or type. If you need to check the array exists before that of course use isset() or is_array() on just the array itself.

Illegal string offset warning when getting value out of array

There must be an obvious bug in this code but I'm not seeing it. Mind taking a look?
The below code returns
string
fleet
Warning: Illegal offset type (line 6)
The taskforces subroutine just pulls an .ini file, reads it into an array, and returns the array, which the foreach then iterates through. In relevant part, the array looks like this.
; this is an INI file
[scout]
type = "fleet"
Here is the code:
foreach($_SESSION['ini']->taskforces() as $key => $val)
{
echo gettype($val["type"]);
echo $val["type"];
if($val["type"] == "fleet") {
$commanderData[$val] = "BLOB";
$commanderData["sc$val"] = "INT NOT NULL";
}
}
I'd like to not have the illegal offset type, because I want the code to go through to the if condition. What obvious thing am I missing?
Thanks.
Instead of this:
echo $val["type"];
you should have simply:
echo $val;
Just because $val is not an array, it's a string. You've made a foreach through an array, so on each iteration you get an array key and an array value (which is, obviously, the string "fleet").
I'm not sure why this caused the problem, but I realized that the result of the if statement was not correct. The code
$commanderData[$val] = "BLOB";
attempts to use the matrix $val as the key for the $commanderData array. It should use the string $key from the iteration through the ini file. Once fixed, I stopped getting the warning, but it's not clear why this would have thrown the error on the proceeding line.

PHP - Array fatal error

I've got an odd error in my PHP code regarding dynamic arrays.
The error outputted is:
Fatal error: Cannot use string offset as an array ... on line 89
This is a portion of my code, it is within a foreach loop, which is looping through settings in a database:
foreach($query->fetchAll() as $row)
{
if($site!=CURRENT_SITE_TEMPLATE)
{
$property = 'foreignSettings';
$propertyType = 'foreignSettingsTypes';
} else {
$property = 'settings';
$propertyType = 'settingTypes';
}
$this->$property[$row['variable_section']][$row['variable_name']] = $row['variable_value'];
settype($this->$property[$row['variable_section']][$row['variable_name']],$row['variable_type']);
$this->$propertyType[$row['variable_section']][$row['variable_name']] = $row['variable_type'];
}
For the sake of the example code, $site is 'admin' and CURRENT_SITE_TEMPLATE is 'admin'.
In addition, $foreignSettings, $foreignSettingsTypes, $settings, and $settingTypes are all defined as arrays in the class scope
The error is on line 89, which is:
$this->$property[$row['variable_section']][$row['variable_name']] = $row['variable_value'];
I originally thought it was because of the $property variable accesing the array, however, this looks like valid legal code in the PHP documentation ( http://php.net/manual/en/language.variables.variable.php in example #1)
Any help on this error would be appreciated.
Thanks
In your given example $property is a string. You are then trying to use that as an array. Strings only has numeric indexes (if you need to use as an array).
The problem is as follows: $this->$property[0] means you access the 0th place of $property which in your case would be the first letter of the string $property. Thus you end up with $this->f or $this->s.
with $this->$property[0][0] you would be trying to access the 0th place of the 0th place of the $property string what results in an error because you try to access the 0th place of the char s what is not possible since the char s can not be referenced as an array.
what you want is $this->{$propperty}[0][0] what means that you try to access the 0th place of the 0th place of the variable that has the name $propperty.

How to check if a PHP array has any value set?

I am working on a signup form, I am using PHP and on my processing part I run some code, if a submitted item fails I then add it to an errors array.
Below is a snip of the code, I am at the point where I need to find the best method to determine if I should trigger an error.
So if there is a value set in the error array then I need to redirect and do some other stuff.
I was thinking of using isset or else is_array but I don't think that is the answer since I set the array using **$signup_errors = array()** wouldn't this make the is_array be true?
Can anyone suggest a good way to do this?
//at the beginning I set the error array
$signup_errors = array();
// I then add items to the error array as needed like this...
$signup_errors['captcha'] = 'Please Enter the Correct Security Code';
if ($signup_errors) {
// there was an error
} else {
// there wasn't
}
How does it work? When converting to boolean, an empty array converts to false. Every other array converts to true. From the PHP manual:
Converting to boolean
To explicitly convert a value to
boolean, use the (bool) or (boolean)
casts. However, in most cases the cast
is unncecessary, since a value will be
automatically converted if an
operator, function or control
structure requires a boolean argument.
See also Type Juggling.
When converting to boolean, the
following values are considered FALSE:
the boolean FALSE itself
the integer 0 (zero)
the float 0.0 (zero)
the empty string, and the string "0"
an array with zero elements
an object with zero member variables (PHP 4 only)
the special type NULL (including unset variables)
SimpleXML objects created from empty tags
Every other value is considered TRUE (including any resource).
You could also use empty() as it has similar semantics.
Perhaps empty()?
From Docs:
Return Values
Returns FALSE if var has a non-empty
and non-zero value.
The following things are considered to
be empty:
"" (an empty string)
0 (0 as an integer)
"0" (0 as a string)
NULL
FALSE
array() (an empty array)
var $var; (a variable declared, but without a value in a class)
Check if...
if(count($array) > 0) { ... }
...if it is, then at least one key-value pair is set.
Alternatively, check if the array is not empty():
if(!empty($array)) { ... }
Use array_filter if you already have keys, but want to check for non-boolean evaluated values.
<?php
$errors = ['foo' => '', 'bar' => null];
var_dump(array_filter($errors));
$errors = ['foo' => 'Oops', 'bar' => null];
var_dump(array_filter($errors));
Output:
array(0) {
}
array(1) {
["foo"]=>
string(4) "Oops"
}
Use:
<?php
if(array_filter($errors)) {
// Has errors
}
You could check on both the minimum and maximum values of the array, in this case you can have a large array filled with keys and empty values and you don't have to iterate through every key-value pair
if(!min($array) && !max($array)) { ... }
The language construct isset(), is for testing to see if variables and array elements are set and not NULL. Using is_array() would tell you if the argument you supply to it is an array. Thus, I do not think using isset() or is_array() would give you the correct and desired result that you are seeking.
The code:
$signup_errors = array();
means that ...
is_array($signup_errors);
would return true. However, this does not mean that the Boolean language rules of PHP would evaluate....
if($signup_errors)
{
//*Do something if $signup_errors evaluates to true*;
}
as true, unless some elements are added to it. When you did this,
$signup_errors['captcha'] = 'Please Enter the Correct Security Code';
you fulfilled the PHP language requirement for the array above to evaluate to true.
Now, if for some reason you wanted, or needed, to use isset() on the array elements in the future, you could. But, the conditional statement above is enough for you this case.
I should add an obvious answer here. If you initialise your error array as an empty array. And later want to check if it is no longer an empty array:
<?php
$errors = [];
if($errors !== [])
{
// We have errors.
}

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