I am tring this method to find the common characters in two strings namely, $a and $r, but the first character isn't getting printed . Moreover the $already collects the common characters and prevents them from being printed for multiple times( I need each character to be printed once only) but it isn't doing so. Please tell me what errors I am making.
<?php
$a="BNJUBCI CBDIDIBO";
$r="SBKJOJLBOU";
$already="";
for($i=0;$i<strlen($r);$i++)
{
if (stripos($a,$r[$i])!=FALSE)
{
if (stripos($already,$r[$i])==FALSE)
{
$already=$already.$r[$i];
echo "already=".$already."<br>";
echo $r[$i]."<br>";
}
}
}
?>
Use !==FALSE instead of !=FALSE. The problem is that stripos returns 0 if the needle is at the start of the haystack, and 0 is falsy. By using !== you are forcing it to ensure the result is actually false, and not just 0.
This is actually listed in the docs. An "RTM" might be appropriate here.
Warning
This function may return Boolean FALSE, but may also return a non-Boolean value which evaluates to FALSE. Please read the section on Booleans for more information. Use the === operator for testing the return value of this function.
The simplest way to find the intersection of the two strings in PHP is to turn them into arrays and use the built-in functions for that purpose.
The following will show all the unique and common characters between the two strings.
<?php
$a="BNJUBCI CBDIDIBO";
$r="SBKJOJLBOU";
$a_arr = str_split($a);
$r_arr = str_split($r);
$common = implode(array_unique(array_intersect($a_arr, $r_arr)));
echo "'$common'";
?>
I would think a much simpler solution to this would be to make the strings into arrays and compare those no?
Something like:
<?php
$a="BNJUBCI CBDIDIBO";
$r="SBKJOJLBOU";
$shared = implode( '' , array_intersect( str_split($a) , str_split($r) ) );
?>
That should return you a string of all the characters in $a that are present in $r
Related
<?php
$a = 'abc';
if($a among array('are','abc','xyz','lmn'))
echo 'true';
?>
Suppose I have the code above, how to write the statement "if($a among...)"?
Use the in_array() function.
Manual says:
Searches haystack for needle using loose comparison unless strict is set.
Example:
<?php
$a = 'abc';
if (in_array($a, array('are','abc','xyz','lmn'))) {
echo "Got abc";
}
?>
Like this:
if (in_array($a, array('are','abc','xyz','lmn')))
{
echo 'True';
}
Also, although it's technically allowed to not use curly brackets in the example you gave, I'd highly recommend that you use them. If you were to come back later and add some more logic for when the condition is true, you might forget to add the curly brackets and thus ruin your code.
There is in_array function.
if(in_array($a, array('are','abc','xyz','lmn'), true)){
echo 'true';
}
NOTE:
You should set the 3rd parameter to true to use the strict compare.
in_array(0, array('are','abc','xyz','lmn')) will return true, this may not what you expected.
Try this:
if (in_array($a, array('are','abc','xyz','lmn')))
{
// Code
}
http://php.net/manual/en/function.in-array.php
in_array — Checks if a value exists in an array
bool in_array ( mixed $needle , array $haystack [, bool $strict =
FALSE ] ) Searches haystack for needle using loose comparison unless
strict is set.
I understand that, with a sting assigned to a variable, individual characters can be expressed by using the variable as an indexed array, but why does the code below, using an associative array, not just die with missing required? Why does 'isset' not throw FALSE on an array key that definitely doesn't exist?
unset($a);
$a = 'TESTSTRING';
if(!isset($a['anystring'])){
die('MISSING REQUIRED');
}else{
var_dump($a['anystring']);
}
The above example will output:
string(1) "T"
EDIT:
As indicated by Jelle Keiser, this is probably the safer thing to do:
if(!array_key_exists('required',$_POST)){
die('MISSING REQUIRED');
}else{
echo $_POST['required'];
}
What PHP is doing is using your string as a numeric index. In this case, 'anystring' is the equivalent of 0. You can test this by doing
<?php
echo (int)'anystring';
// 0
var_dump('anystring' == 0);
// bool(true)
PHP does a lot of type juggling, which can lead to "interesting" results.
$a is a string not an associative array.
If you want to access it that way you have to do something like this.
$a['anystring'] = 'TESTSTRING';
You need to use array_key_exists() to test if a key exists
The working of isset is correct in your case.
Because $a is a string, the index-operator will give you the specified char in the string at the declared position. (like a "Char-Array")
A small example:
$a = 'TESTSTRING';
echo $a[0]; // Output: T
echo $a[1]; // Output: E
// ...
This will output the first and the second character at index 0 and 1 of the string.
And because the index-operator always expects an integer value on strings. The given value will be automatically casted to an integer. You can see this, when you cast the string to an integer, like this:
echo (int) 'TESTSTRING'; // Output: 0
For char-access on strings, also see the PHP-Manual.
Try enabling PHP to show all errors by using error_reporting(E_ALL);
This should give you a warning saying you are using an illegal offset. PHP therefore automatically assumes you are looking for the first element in the array or letter in this case.
it works as expected for... it returned false... but when I force it to return true ... itz throws an error saying illegal offset somekind.... but still output the first string.... as anystring casted as int equals to 0.. check the version of php you are using bro... I used notepad++ to create the php file... no special ide...
I've been working with PHP for quite a while now, but this was always a mystery to me, the correct use of the exclamation mark (negative sign) in front of variables.
What does !$var indicate? Is var false, empty, not set etc.?
Here are some examples that I need to learn...
Example 1:
$string = 'hello';
$hello = (!empty($string)) ? $string : '';
if (!$hello)
{
die('Variable hello is empty');
}
Is this example valid? Would the if statement really work if $string was empty?
Example 2:
$int = 5;
$count = (!empty($int)) ? $int : 0;
// Note the positive check here
if ($count)
{
die('Variable count was not empty');
}
Would this example be valid?
I never use any of the above examples, I limit these if ($var) to variables that have boolean values only. I just need to know if these examples are valid so I can broaden the use of the if ($var) statements. They look really clean.
Thanks.
if(! $a) is the same as if($a == false). Also, one should take into account that type conversion takes place when using == operator.
For more details, have a look into "Loose comparisons with ==" section here. From there it follows, that for strings "0" and "" are equal to FALSE ( "0"==false is TRUE and ""==false is TRUE, too).
Regarding posted examples:
Example 1
It will work, but you should note, that both "0" and "" are 'empty' strings.
Example 2
It will work
It's a boolean tester. Empty or false.
It's the not boolean operator, see the PHP manual for further detail.
<?php
$a = 'abc';
if($a among array('are','abc','xyz','lmn'))
echo 'true';
?>
Suppose I have the code above, how to write the statement "if($a among...)"?
Use the in_array() function.
Manual says:
Searches haystack for needle using loose comparison unless strict is set.
Example:
<?php
$a = 'abc';
if (in_array($a, array('are','abc','xyz','lmn'))) {
echo "Got abc";
}
?>
Like this:
if (in_array($a, array('are','abc','xyz','lmn')))
{
echo 'True';
}
Also, although it's technically allowed to not use curly brackets in the example you gave, I'd highly recommend that you use them. If you were to come back later and add some more logic for when the condition is true, you might forget to add the curly brackets and thus ruin your code.
There is in_array function.
if(in_array($a, array('are','abc','xyz','lmn'), true)){
echo 'true';
}
NOTE:
You should set the 3rd parameter to true to use the strict compare.
in_array(0, array('are','abc','xyz','lmn')) will return true, this may not what you expected.
Try this:
if (in_array($a, array('are','abc','xyz','lmn')))
{
// Code
}
http://php.net/manual/en/function.in-array.php
in_array — Checks if a value exists in an array
bool in_array ( mixed $needle , array $haystack [, bool $strict =
FALSE ] ) Searches haystack for needle using loose comparison unless
strict is set.
I've seen a lot of php code that does the following to check whether a string is valid by doing:
$str is a string variable.
if (!isset($str) || $str !== '') {
// do something
}
I prefer to just do
if (strlen($str) > 0) {
// something
}
Is there any thing that can go wrong with the second method? Are there any casting issues I should be aware of?
Since PHP will treat a string containing a zero ('0') as empty, it makes the empty() function an unsuitable solution.
Instead, test that the variable is explicitly not equal to an empty string:
$stringvar !== ''
As the OP and Gras Double and others have shown, the variable should also be checked for initialization to avoid a warning or error (depending on settings):
isset($stringvar)
This results in the more acceptable:
if (isset($stringvar) && $stringvar !== '') {
}
PHP has a lot of bad conventions. I originally answered this (over 9 years ago) using the empty() function, as seen below. I've long since abandoned PHP, but since this answer attracts downvotes and comments every few years, I've updated it. Should the OP wish to change the accepted answer, please do so.
Original Answer:
if(empty($stringvar))
{
// do something
}
You could also add trim() to eliminate whitespace if that is to be considered.
Edit:
Note that for a string like '0', this will return true, while strlen() will not.
You need isset() in case $str is possibly undefined:
if (isset($str) && $str !== '') {
// variable set, not empty string
}
Using !empty() would have an important caveat: the string '0' evaluates to false.
Also, sometimes one wants to check, in addition, that $str is not something falsy, like false or null[1]. The previous code doesn't handle this. It's one of the rare situations where loose comparison may be useful:
if (isset($str) && $str != '') {
// variable set, not empty string, not falsy
}
The above method is interesting as it remains concise and doesn't filter out '0'. But make sure to document your code if you use it.
Otherwise you can use this equivalent but more verbose version:
if (isset($str) && (string) $str !== '') {
// variable set, not empty string, not falsy
}
Of course, if you are sure $str is defined, you can omit the isset($str) from the above codes.
Finally, considering that '' == false, '0' == false, but '' != '0', you may have guessed it: PHP comparisons aren't transitive (fun graphs included).
[1] Note that isset() already filters out null.
This will safely check for a string containing only whitespace:
// Determines if the supplied string is an empty string.
// Empty is defined as null or containing only whitespace.
// '0' is NOT an empty string!
function isEmptyString($str) {
return !(isset($str) && (strlen(trim($str)) > 0));
}
What about this:
if( !isset($str[0]) )
echo "str is NULL or an empty string";
I found it on PHP manual in a comment by Antone Roundy
I posted it here, because I did some tests and it seems to work well, but I'm wondering if there is some side effect I'm not considering. Any suggestions in comments here would be appreciated.
According to PHP empty() doc (http://ca1.php.net/empty):
Prior to PHP 5.5, empty() only supports variables; anything else will result in a parse error. In other words, the following will not work: empty(trim($name)). Instead, use trim($name) == false.
This simple old question is still tricky.
strlen($var) works perfectly ONLY if you're absolutely sure the $var is a string.
isset($var) and empty($var) result are based on type of the variable, and could be tricky at some cases (like empty string ""). View the table in this page for more details.
UPDATE
There are actually 2 cases for this question:
Case 1: You're sure that your variable is always going to be a "string":
In this case, just test the length:
if(strlen($str) > 0) {
// do something..
}
Case 2: Your variable may and may not be a "string":
In this case, it depends on what you want to do. For me (most of the time), if it's not a string then I validate it as "false". You can do it this way:
if(is_string($var) && $var !== '') {// true only if it's a string AND is not empty
// do something ...
}
And to make it shorter and in 1 condition instead of 2 (specially useful if you're testing more than 1 string in same if condition), I made it into function:
function isNonEmptyString($var) {
return is_string($var) && $var !== '';
}
// Somewhere else..
// Reducing conditions to half
if(isNonEmptyString($var1) && isNonEmptyString($var2) && isNonEmptyString($var3)) {
// do something
}
If your variable $str is not defined then your strlen() method will throw an exception. That is the whole purpose of using isset() first.
trimming the string will also help if there are string with white spaces.
if (isset($str) && trim($str) !== '') {
// code
}
I think not, because strlen (string lenght) returns the lenght (integer) of your $str variable.
So if the variable is empty i would return 0. Is 0 greater then 0. Don't think so.
But i think the first method might be a little more safer. Because it checks if the variable is init, and if its not empty.