I need help return everything after "I can read." I understand that this will search the string and find whats in the array $check but how do I make it check and then return all whats found after "I can read"?
$string = "I can read. I can count. I can spell.";
$check = array("I can count.", "I can't count.");
$find = stristr($string, $check, true);
echo $find;
$find = 'I can read.';
$string = '[...]';
echo substr($string,stripos($string,$find)+strlen($find));
Here is what the docs say the second parameter (needle) of stistr:
If needle is not a string, it is converted to an integer and applied as
the ordinal value of a character.
Then you have to define $check as follow:
$check = "I can read";
you do not need the third parameter if you want to get the substring after $check, documentation states:
If TRUE, stristr() returns the part of the haystack before the
first occurrence of the needle.
That's why you have to call stristr with only two arguments:
$find = stristr($string, $check);
Related
I want check if a string contains a variable amount of generated words. This is the code I came up with, it stores the pattern in a variable which I put in the preg_match.
Result of this code: it returns true;
$string = "What a beatiful world";
$pieces = "#\b(" . implode("|",$pieces_arr). ")\b#";
$pieces outputs for example:
#\b(hello|you)\b#
into:
if(preg_match($pieces, $string)) {
echo "piece found in string!!";
}
Sorry, a bit new to php. Have tried all I know, but failed and seeking expert help.
I have a string similar to this..
/%mydocs%/%myfolder%/%date%/%filename%.html
Basically the %vars% represents different strings and seperated by /, like a URL.
I need to find %date% (or any given variable) if it's present and it's position in the string, like it's at position 3 in the example above.
I tried this,
$date = preg_match( '%date%', '/%mydocs%/%myfolder%/%date%/%filename%.html');
But it's neither returning anything nor the position.
Any help, even some heads up to play with, will be appriciated.
strpos
<?php
$mystring = '/%mydocs%/%myfolder%/%date%/%filename%.html';
$findme = '%date%';
$pos = strpos($mystring, $findme);
if ($pos === false) {
echo "The string '$findme' was not found in the string '$mystring'";
} else {
echo "The string '$findme' was found in the string '$mystring'";
echo " and exists at position $pos";
}
?>
Result:
The string '%date%' was found in the string '/%mydocs%/%myfolder%/%date%/%filename%.html' and exists at position 21
How about using preg_split:
$var = '%date%';
$str = '/%mydocs%/%myfolder%/%date%/%filename%.html';
$list = preg_split('#/#', $str, -1, PREG_SPLIT_NO_EMPTY);
for($i=0; $i<count($list); $i++) {
if ($list[$i] == $var) {
echo "Found $var at position ",$i+1,"\n";
}
}
output:
Found %date% at position 3
preg_match returns a boolean for whether there was a match or not, but you can pass a third parameter - $matches, which will contain the matches according to the regular expression provided.
You don't look to be using a regular expressions - for what you are doing you probably want to look at strpos instead. http://uk3.php.net/strpos
I want to check the first word of some sentences. If the first word are For, And, Nor, But, Or, etc, I want to skip the sentence.
Here's the code :
<?php
$sentence = 'For me more';
$arr = explode(' ',trim($sentence));
if(stripos($arr[0],'for') or stripos($arr[0],'but') or stripos($arr[0],'it')){
//doing something
}
?>
Blank result, Whats wrong ? thank you :)
Here, stripos will return 0 if the word is found (found at position 0).
It returns false if the word is not found.
You should write :
if(stripos($arr[0],'for') !== false or stripos($arr[0],'but') !== false or stripos($arr[0],'it') !== false){
//skip
}
Stripos returns the position on the first occurrence of the needle in the haystack
The first occurrence is at position 0, which evaluates to false.
Try this as an alternative
$sentence = 'For me more';
// make all words lowercase
$arr = explode(' ', strtolower(trim($sentence)));
if(in_array($arr[0], array('for', 'but', 'it'))) {
//doing something
echo "found: $sentence";
} else {
echo 'failed';
}
Perhaps use preg_filter if you are going to know what the string to be evaluated is (i.e. you don't need to parse out sentences).
$filter_array = array(
'/^for\s/i',
'/^and\s/i',
'/^nor\s/i',
// etc.
}
$sentence = 'For me more';
$result = preg_filter(trim($sentence), '', $filter_array);
if ($result === null) {
// this sentence did not match the filters
}
This allows you to determine a set of filter regex patterns to see if you have a match. Note that in this case I just used '' as "replacement" value, as you don't really care about actually making a replacement, this function just gives you a nice way to pas in an array of regular expressions.
How do I find exact 2, in a string using strpos? Is it possible using strpos? The example below returns "Found" even though the match is NOT exact to what I need. I understand 2, is matching with 22,. It should return "Not Found". I am matching ID's in this example.
$string = "21,22,23,26,";
$find = "2,";
$pos = strpos($string, $find);
if ($pos !== false) {
echo "Found";
} else {
echo "Not Found";
}
Unless the string is enormous, make an array and search it:
$string = "21,22,23,26,";
$arr = explode(",", $string);
// array_search() returns its position in the array
echo array_search("2", $arr);
// null output, 2 wasn't found
Actually, in_array() is probably faster:
// in_array() returns a boolean indicating whether it is found or not
var_dump(in_array("2", $arr));
// bool(false), 2 wasn't found
var_dump(in_array("22", $arr));
// bool(true), 22 was found
This will work as long as your string is a comma-delimited list of values. If the string is really long, making an array may be wasteful of memory. Use a string manipulation solution instead.
Addendum
You didn't specify, but if by some chance these strings came from a database table, I would just add that the appropriate course of action would be to properly normalize it into another table with one row per id rather than store them as a delimited string.
Try with explode and in_array
Example:
$string = "21,22,23,26,";
$string_numbers = explode(",", $string);
$find = 2;
if (in_array($find, $string_numbers)) {
echo "Found";
} else {
echo "Not Found";
}
You can use preg_match if you want to avoid arrays.
$string = "21,22,23,26,";
$find = '2';
$pattern = "/(^$find,|,$find,|,$find$)/";
if (0 === preg_match($pattern, $string)) {
echo "Not Found";
} else {
echo "Found";
}
This will find your id at beginning, middle or at the end of the string. Of course, I am assuming $string does not contain characters other than numbers and commas (like spaces).
I have a string
8,7,13,14,16
Whats the easiest way to determine if a given number is present in that string?
$numberA = "13";
$string = "8,7,13,14,16";
if($string magic $numberA){
$result = "Yeah, that number is in there";
} else {
$result = "Sorry.";
}
Looking for magic.
<?php
in_array('13', explode(',', '8,7,13,14,16'));
?>
…will return whether '13' is in the string.
Just to elaborate: explode turns the string into an array, splitting it at each ',' in this case. Then, in_array checks if the string '13' is in the resulting array somewhere.
Another way, that might be more efficient for laaaaaaaarge strings, is using a regexp:
$numberA = "13";
$string = "8,7,13,14,16";
if(preg_match('/(^|,)'.$numberA.'($|,)/', $string)){
$result = "Yeah, that number is in there";
} else {
$result = "Sorry.";
}
if (strpos(','.$string.',' , ','.$numberA.',') !== FALSE) {
//found
}
Notice guard ',' chars, they will help to deal with '13' magic '1, 2, 133' case.
Make sure you match the full number in the string, not just part of it.
function numberInList($num, $list) {
return preg_match("/\b$num\b/", $list);
}
$string = "8,7,13,14,16";
numberInList(13, $string); # returns 1
numberInList(8, $string); # returns 1
numberInList(1, $string); # returns 0
numberInList(3, $string); # returns 0
A simple string search should do if you are just checking to find existence of the string. I dont speak php but i think this is how it could be done.
$mystring = '8,7,13,14,16';
$findme = '13';
if (preg_match('/(?>(^|[^0-9])'.$findme.'([^0-9]|$))/', $mystring)) {
$result = "Yeah, that number is in there";
} else {
$result = "Sorry.";
}