I am trying to show temperature on my website. I have a forecast in json that I get from api and there is this field that I need to get the temperature from:
"fcttext_metric":"A mainly sunny sky. High near 30 ° C. Winds SE at 10 to 15 km/h."
I tried to parse only float numbers which works well, but it fails when there is more text behind it like in the case above. I tried to show only first 2 chars but when the temperature is negative it fails again. What can I do about this?
$str = 'A mainly sunny sky. High near 30 ° C. Winds SE at 10 to 15 km/h.';
$deg = filter_var($str, FILTER_SANITIZE_NUMBER_FLOAT);
$res = substr($deg, 0, 2);
print_r($res);
I would use this regex. It should match integer values before ° (°) sign. Like 30 or -30
preg_match('/(\-?\d+)\s*°/', $str, $matches);
echo $matches[1];
Related
I'm struggling with some regular expressions. What I want to do is find money amounts in a string, remove the €,$, or £ but keep the number, and then look to see if there is a 'b' or an 'm' - in which case write 'million platinum coins' or 'million gold coin' respectively but otherwise just put 'gold coins'.
I have most of that as a hack (see below) with the small problem that my regex does not seem to work. The money amount comes out unchanged.
Desired behaviour examples
I intend to leave the decimal places and thousands separators as is
$12.6m ==> 12.6 million gold coins
£2b ==> 2 million platinum coins
€99 ==> 99 gold coins
My code
Here is my non-working code (I suspect my regex might be wrong).
protected function funnymoney($text){
$text = preg_replace('/[€$£]*([0-9\.,]+)([mb])/i','\0 %\1%',$text);
$text = str_replace('%b%','million platnum coins',$text);
$text = str_replace('%m%','million gold coins',$text);
$text = str_replace('%%','gold coins',$text);
return $text;
}
I would greatly appreciate it if someone could explain to me what I am missing or getting wrong and guide me to the right answer. You may safely assume I know very little about regular expressions. I would like to understand why the solution works too if I can.
Using preg_replace_callback, you can do this in a single function call:
define ("re", '/[€$£]*(\.\d+|\d+(?:[.,]\d+)?)([mb]|)/i');
function funnymoney($text) {
return preg_replace_callback(re, function($m) {
return $m[1] .
($m[2] != "" ? " million" : "") . ($m[2] == "b" ? " platinum" : " gold") .
" coins";
}, $text);
}
// not call this function
echo funnymoney('$12.6m');
//=> "12.6 million gold coins"
echo funnymoney('£2b');
//=> "2 million platinum coins"
echo funnymoney('€99');
//=> "99 gold coins"
I am not sure how you intend to handle decimal places and thousands separators, so that part of my pattern may require adjustment. Beyond that, match the leading currency symbol (so that it is consumed/removed, then capture the numeric substring, then capture the optional trailing units (b or m).
Use a lookup array to translate the units to English. When the unit character is missing, apply the fallback value from the lookup array.
A lookup array will make your task easier to read and maintain.
Code: (Demo)
$str = '$1.1m
Foo
£2,2b
Bar
€99.9';
$lookup = [
'b' => 'million platinum coins',
'm' => 'million gold coins',
'' => 'gold coins',
];
echo preg_replace_callback(
'~[$£€](\d+(?:[.,]\d+)?)([bm]?)~iu',
function($m) use ($lookup) {
return "$m[1] " . $lookup[strtolower($m[2])];
},
$str
);
Output:
1.1 million gold coins
Foo
2,2 million platinum coins
Bar
99.9 gold coins
Your regex has a first full match on the string, and it goes on index 0 of the returning array, but it seems you just need the capturing groups.
$text = preg_replace('/[€$£]*([0-9\.,]+)([mb])/i','\1 %\2%',$text);
Funny question, btw!
Is this what you want?
<?php
/**
$12.6m ==> 12.6 million gold coins
£2b ==> 2 million platinum coins
€99 ==> 99 gold coins
*/
$str = <<<EOD
$12.6m
£2b
€99
EOD;
preg_match('/\$(.*?)m/', $str, $string1);
echo $string1[1] . " million gold coins \n";
preg_match('/\£(.*?)b/', $str, $string2);
echo $string2[1] . " million platinum coins \n";
preg_match('/\€([0-9])/', $str, $string3);
echo $string3[1] . " gold coins \n";
// output:
// 12.6 million gold coins
// 2 million platinum coins
// 9 gold coins
Having difficuties matching a date in the given string. Tried a myriad of regex suggestions. Keep on getting "No date found", while the date is obviously there: 07/02/2016.
What am I missing?
function matchDate($str) {
if (preg_match('/\b(0?[1-9]|1[012])[- /.](0?[1-9]|[12][0-9]|3[01])[- /.](19|20)?[0-9]{2}\b/', $str, $mDdate)) {
return $mDdate[0];
} else {
return "No date found.";
}
}
$str = "FISH HOUSE KINGS FISH HOUSE 100 W Broadway Long Beach, Ca. 90802 562-432-7463 Server: Ezbeth 07/02/2016 Table 44/1 8:38 PM 60018 10.00 18.75 enon Drop Fried atiish D I obster Crunchy Roll callap and Shrimp Char D 13.50 22.50 6.95 101.60 Cheeseburger 1/2lb D (2 14.95) 29.90 Caesar Salad Subtotal Tax 9.14 110.74 110.74 Total Ba 1ance Due KING'S FISH HOUSE Welcome To The House That Seafood Built Find Us Online #KingsFishHouse ";
echo matchDate($str);
For your given example, this is
\b\d{2}/\d{2}/\d{4}\b
See a demo on regex101.com.
The problem with your example regex is that you need to escape the / character in the pattern.
If you don't escape the / character, regex will understand it as the end of the regex pattern.
Based on your example, the solution should be:
\b(0?[1-9]|1[012])\/(0?[1-9]|[12][0-9]|3[01])\/(19|20)?[0-9]{2}\b
You can see it in action in this demo
I have the following output:
Item
Length : 130
Depth : 25
Total Area (sq cm): 3250
Wood Finish: Beech
Etc: etc
I want to remove the Total Area (sq cm): and the 4 digits after it from the string, currently I am trying to use str_replace like so:
$tidy_str = str_replace( $totalarea, "", $tidy_str);
Is this the correct function to use and if so how can I include the 4 random digits after this text? Please also note that this is not a set output so the string will change position within this.
You can practice php regex at http://www.phpliveregex.com/
<?php
$str = '
Item
Length : 130
Depth : 25
Total Area (sq cm): 3250
Wood Finish: Beech
Etc: etc
';
echo preg_replace("/Total Area \(sq cm\): [0-9]*\\n/", "", $str);
Item
Length : 130
Depth : 25
Wood Finish: Beech
Etc: etc
This will do it.
$exp = '/\(sq cm\): \d+/';
echo preg_replace($exp, '', $array);
Try with this:
preg_replace('/(Total Area \(sq cm\): )([0-9\.,]*)/' , '', $tidy_str);
You are looking for substr_replace:
$strToSearch = "Total Area (sq cm):";
$totalAreaIndex = strpos($tidy_str, $strToSearch);
echo substr_replace($tidy_str, '', $totalAreaIndex, strlen($strToSearch) + 5); // 5 = space plus 4 numbers.
If you want to remove the newline too, you should check if it's \n or \r\n. \n add one, \r\n add two to offset. Ie. strlen($strToSearch) + 7
So, my text I want to post on Twitter is sometimes more than 140 character, so, I need to check the lenght and then go without changes if less than 140 or slive the text into two pieces (the text and the link) and grab the text part and make it e.g. 100 characters long - chop the rest.
Then grab the - now 100 characters long part - and put it otgether with the url.
How to do that?
my code so far:
if (strlen($status) < 140) {
// continue
} else {
// 1. slice the $status into $text and $url (every message has url so
// checking is not important right now
// 2. shorten the text to 100 char
// something like $text = substr($text, 0, 100); ?
// 3. put them back together
$status = $text . ' ' . $url;
}
How should I change my code? I have biggest problem with the first part when getting the url and text part.
Btw. in each $status is only 1 url, so checking for mulitple urls is not necessary
Example of a text that is longer than it should be:
What is now Ecuador was home to a variety of indigenous groups that were gradually incorporated into the Inca Empire during the fifteenth century. The territory was colonized by Spain during the sixteenth century, achieving independence in 1820 as part of Gran Colombia, from which it emerged as its own sovereign state in 1830. The legacy of both empires is reflected in Ecuador's ethnically diverse population, with most of its 15.2 million people being mestizos, followed by large minorities of European, Amerindian, and African descendant. https://en.wikipedia.org/wiki/Ecuador
should become in the end this:
What is now Ecuador was home to a variety of indigenous groups that were gradually incorporated int https://en.wikipedia.org/wiki/Ecuador
If you can be sure that the URL does not contain any spaces (no well-formed URL should) and that it is always present, try it like that:
preg_match('/^(.*)(\\S+)$/', $status, $matches);
$text = $matches[1];
$url = $matches[2];
$text = substr($text, 0, 100);
But possibly the length of the text should be adapted to the length of the url, so you would use
$text = substr($text, 0, 140-strlen($url)-1);
$reg = '/\b(?:(?:https?|ftp|file):\/\/|www\.|ftp\.)[-A-Z0-9+&##\/%=~_|$?!:,.]*[A-Z0-9+&##\/%=~_|$]/i';
$string = "What is now Ecuador was home to a variety of indigenous groups that were gradually incorporated into the Inca Empire during the fifteenth century. The territory was colonized by Spain during the sixteenth century, achieving independence in 1820 as part of Gran Colombia, from which it emerged as its own sovereign state in 1830. The legacy of both empires is reflected in Ecuador's ethnically diverse population, with most of its 15.2 million people being mestizos, followed by large minorities of European, Amerindian, and African descendant. https://en.wikipedia.org/wiki/Ecuador";
preg_match_all($reg, $string, $matches, PREG_PATTERN_ORDER);
$cut_string = substr($string, 0, (140-strlen($matches[0][0])-1));
$your_twitt = $cut_string . " " . $matches[0][0];
echo $your_twitt;
// ouputs : "What is now Ecuador was home to a variety of indigenous groups that were gradually incorporated into t https://en.wikipedia.org/wiki/Ecuador"
This might be what you want :
$status = 'What is now Ecuador was home to a variety of indigenous groups that were gradually incorporated into the Inca Empire during the fifteenth century. The territory was colonized by Spain during the sixteenth century, achieving independence in 1820 as part of Gran Colombia, from which it emerged as its own sovereign state in 1830. The legacy of both empires is reflected in Ecuador\'s ethnically diverse population, with most of its 15.2 million people being mestizos, followed by large minorities of European, Amerindian, and African descendant. https://en.wikipedia.org/wiki/Ecuador';
if (strlen($status) < 140) {
echo 'Lenght ok';
} else {
$totalPart = round(strlen($status)/100);
$fulltweet = array();
for ($i=0; $i < $totalPart; $i++) {
if($i==0)
{
$fulltweet[$i] = substr($status, 0,100);
}else{
$fulltweet[$i] = substr($status, $i * 100);
}
}
}
If the string is longer than 140 chars then it'll explode it in an array of 100 char for each row
I explain my problem : I'm working on different kind of address
" 25 Down Street 15000 London "
" 25 B Down Street 15000 London "
" Building A 25 Down Street 15000 London "
I found a way to determine which is the number of the street on all case with this regex :
`^([1-9][0-9]{0,2}(?:\s*[A-Z])?)\b`
But now i got a problem that i can't solve, i need when the case is real to determine characters which are before the street's number .
Example : " Building 2 25 Down Street 15000 London " i need here to find only "Building 2"
I understand that i have to find characters before the first number of this string.
Keep searching on my own but will be great if someone got a solution for me .
Thank you .
Edit my code now is :
preg_match('/^(.*?)\d+\s+\D+/', $cleanAdressNode, $result, PREG_OFFSET_CAPTURE,0);
print $result[0][0];
return $result[0][0];
and the result now is : Résidence Les Thermes 1 15 boulevard Jean Jaurès instead of only : Résidence Les Thermes 1
How about:
preg_match('/^(\D*)/', $str, $match);
You will find in $match[1] everything that is not a digit at the begining of the string.
According to your example:
preg_match('/^(.*?)\d+\s+\D+/', $str, $match);
If you only want to match the first non-numeric characters, ^([^0-9]*) should do the trick. It uses class negation to grab every non-numeric characters at the start of the string.