I'm having trouble finding a correct regex to achieve what I want.
I have a sentence like that :
Hi, my name is Stan, you are welcome, hello.
and I would like to transform it like that :
[hi|hello|welcome], my name is [stan|jack] you are [hi|hello|welcome] [hi|hello|welcome].
Right now my regex is half working, because somes words are not replaced, and those replaced are deleting some characters
Here is my test code
<?php
$test = 'Hi, my name is Stan, you are welcome, hello.';
$words = array(
array('hi', 'hello', 'welcome'),
array('stan', 'jack'),
);
$result = $test;
foreach ($words as $group) {
if (count($group) > 0) {
$replacement = '[' . implode('|', $group) . ']';
foreach ($group as $word) {
$result = preg_replace('#([^\[])' . $word . '([^\]])#i', $replacement, $result);
}
}
}
echo $test . '<br />' . $result;
Any help will be appreciated
The regex you are using is overcomplicated. You simply need to use a regex substitution using regular brackets ():
<?php
$test = 'Hi, my name is Stan, you are welcome, hello.';
$words = array(
array('hi', 'hello', 'welcome'),
array('stan', 'jack'),
);
$result = $test;
foreach ($words as $group) {
if (count($group) > 0) {
$imploded = implode('|', $group);
$replacement = "[$imploded]";
$search = "($imploded)";
$result = preg_replace("/$search/i", $replacement, $result);
}
}
echo $test . '<br />' . $result;
Your regular expression:
'#([^\[])' . $word . '([^\]])#i'
matches one character before and after $word as well. And as they do, they replace it. So your replacement string needs to reference these parts, too:
'$1' . $replacement . '$2'
Demo
preg_replace supports array as parameter. No need to iterate with a loop.
$s = array("/(hi|hello|welcome)/i", "/(stan|jack)/i");
$r = array("[hi|hello|welcome]", "[stan|jack]");
preg_replace($s, $r, $str);
or dynamically
$test = 'Hi, my name is Stan, you are welcome, hello.';
$s = array("hi|hello|welcome", "stan|jack");
$r = array_map(create_function('$a','return "[$a]";'), $s);
$s = array_map(create_function('$a','return "/($a)/i";'), $s);
echo preg_replace($s, $r, $str);
//[hi|hello|welcome], my name is [stan|jack], you are [hi|hello|welcome], [hi|hello|welcome].
Related
I am trying to check if word is occur in a string but not to be the first and last word, if its true then remove the space after and before of the word and replace with a underscore.
Input:
$str = 'This is a cool area";
Output:
$str = 'This is a_cool_area";
I want to check that the word 'cool' is inside the string but not a first and last word. if yes the remove the space & replace with '_'
You can use preg_replace to do this job, using this regex:
/(?<=\w)\s+(' . $word . ')\s+(?=\w)/i
which looks for the word, surrounded by at least one word character on either side (to prevent matching at the beginning or ending of the sentence). Usage in PHP:
$str = 'This is a cool area';
$word = 'cool';
$str = preg_replace('/(?<=\w)\s+(' . $word . ')\s+(?=\w)/i', '_$1_', $str);
echo $str . "\n";
$str = ' Cool areas are cool ';
$str = preg_replace('/(?<=\w)\s+(' . $word . ')\s+(?=\w)/i', '_$1_', $str);
echo $str . "\n";
Output:
This is a_cool_area
Cool areas are cool
Demo on 3v4l.org
function checkWord($str, $word)
{
$arr = explode(" ", $str);
$newArr = array_slice($arr, 1, -1);
$key = array_search($word, $newArr);
if($key !== false)
{
return implode('_',array_slice($arr, $key, 3));
}
else
{
return $str;
}
}
echo checkWord('This is a cool area', 'cool');
In my article, I want to automatically add links to keywords.
My keywords array:
$keywords = [
0=>['id'=>1,'slug'=>'getName','url'=>'https://example.com/1'],
1=>['id'=>2,'slug'=>'testName','url'=>'https://example.com/2'],
2=>['id'=>3,'slug'=>'ign','url'=>'https://example.com/3'],
];
This is my code:
private function keywords_replace(string $string, array $key_array)
{
$array_first = $key_array;
$array_last = [];
foreach ($array_first as $key=>$value)
{
$array_last[$key] = [$key, $value['slug'], '<a target="_blank" href="' . $value['url'] . '" title="' . $value['slug'] . '">' . $value['slug'] . '</a>'];
}
$count = count($array_last);
for ($i=0; $i<$count;$i++)
{
for ($j=$count-1;$j>$i;$j--)
{
if (strlen($array_last[$j][1]) > strlen($array_last[$j-1][1]))
{
$tmp = $array_last[$j];
$array_last[$j] = $array_last[$j-1];
$array_last[$j-1] = $tmp;
}
}
}
$keys = $array_last;
foreach ($keys as $key)
{
$string = str_ireplace($key[1],$key[0],$string);
}
foreach ($keys as $key)
{
$string = str_ireplace($key[0],$key[2],$string);
}
return $string;
}
result:
$str = "<p>Just a test: getName testName";
echo $this->keywords_replace($str,$keywords);
like this:Just a test: getName testName
very import: If the string has no spaces, it will not match.Because I will use other languages, sentences will not have spaces like English. Like Wordpress key words auto link
I think my code is not perfect，Is there a better algorithm to implement this function? Thanks!
You can use array_reduce and preg_replace to replace all occurrences of the slug words in your string with the corresponding url values:
$keywords = [
0=>['id'=>1,'slug'=>'getName','url'=>'https://www.getname.com'],
1=>['id'=>2,'slug'=>'testName','url'=>'https://www.testname.com'],
2=>['id'=>3,'slug'=>'ign','url'=>'https://www.ign.com'],
];
$str = "<p>Just a test: getName testName";
echo array_reduce($keywords, function ($c, $v) { return preg_replace('/\\b(' . $v['slug'] . ')\\b/', $v['url'], $c); }, $str);
Output:
<p>Just a test: https://www.getname.com https://www.testname.com
Demo on 3v4l.org
Update
To change the text into links, you need to use this:
echo array_reduce($keywords,
function ($c, $v) {
return preg_replace('/\\b(' . $v['slug'] . ')\\b/',
'$1', $c);
},
$str);
Output:
<p>Just a test: getName testName
Updated demo
Update 2
Because some of the links that are being substituted include words that are also values of slug, it's necessary to do all the replacements at once using the array format of strtr. We build an array of patterns and replacements using array_column, array_combine and array_map, then pass that to strtr:
$reps = array_combine(array_column($keywords, 'slug'),
array_map(function ($k) { return '' . $k['slug'] . ''; }, $keywords
));
$newstr = strtr($str, $reps);
New demo
First you need to change structure of array to key/value using loop that result stored in $newKeywords. Then using preg_replace_callback() select every word in string and check that it exist in key of array. If exist, wrap it in anchor tag.
$newKeywords = [];
foreach ($keywords as $keyword)
$newKeywords[$keyword['slug']] = $keyword['url'];
$newStr = preg_replace_callback("/(\w+)/", function($m) use($newKeywords){
return isset($newKeywords[$m[0]]) ? "<a href='{$newKeywords[$m[0]]}'>{$m[0]}</a>" : $m[0];
}, $str);
Output:
<p>Just a test: <a href='https://www.getname.com'>getName</a> <a href='https://www.testname.com'>testName</a></p>
Check result in demo
My answer uses preg_replace as does Nick's above.
It relies on the patterns and replacements being equally sized arrays, with corresponding patterns and replacements.
Word boundaries need to be respected, which I doubt you can do with a simple string replacement.
<?php
$keywords = [
0=>['id'=>1,'slug'=>'foo','url'=>'https://www.example.com/foo'],
1=>['id'=>2,'slug'=>'bar','url'=>'https://www.example.com/bar'],
2=>['id'=>3,'slug'=>'baz','url'=>'https://www.example.com/baz'],
];
foreach ($keywords as $item)
{
$patterns[] = '#\b(' . $item['slug'] . ')\b#i';
$replacements[] = '$1';
}
$html = "<p>I once knew a barbed man named <i>Foo</i>, he often visited the bar.</p>";
print preg_replace($patterns, $replacements, $html);
Output:
<p>I once knew a barbed man named <i>Foo</i>, he often visited the bar.</p>
This is my answer: thanks for #Nick
$content = array_reduce($keywords , function ($c, $v) {
return preg_replace('/(>[^<>]*?)(' . $v['slug'] . ')([^<>]*?<)/', '$1$2$3', $c);
}, $str);
I'm trying in PHP to get something like this:
$mail = "fakemail#le.mail.uk.test";
$rep = "le.mail";
I tried like this:
function test($mail) {
$pattern = '/^([\w-]+(?:\.[\w-]+)*)#((?:[\w-]+\.)*\w[\w-]{0,66})\.([a-z]{2,6}(?:\.[a-z]{2})?)$/';
preg_match($pattern, $mail, $matches);
echo $matches[2] . "\n";
}
test("fakemail#le.mail.uk");
// result = le.mail
but if i have another . in my mail it's broken
function test($mail) {
$pattern = '/^([\w-]+(?:\.[\w-]+)*)#((?:[\w-]+\.)*\w[\w-]{0,66})\.([a-z]{2,6}(?:\.[a-z]{2})?)$/';
preg_match($pattern, $mail, $matches);
echo $matches[2] . "\n";
}
test("fakemail#le.mail.uk.test.test");
// result = le.mail.uk.test.test.test
// whatIwant = le.mail
or I just want all character between # and until the next ..
I think I have to do a loop with an if but I'm not sure if it's possible
with only regex.
A php way without REGEX, using implode() and explode()
<?php
$sep = '.';
$str1 = 'email#test.com.robot';
$str2 = 'email#test.fr.uk.robot';
$get1 = explode($sep,explode('#',$str1)[1]);
$get2 = explode($sep,explode('#',$str2)[1]);
echo implode($sep,[$get1[0],$get1[1]]);
echo PHP_EOL;
echo implode($sep,[$get2[0],$get2[1]]);
?>
Output:
test.com
test.fr
DEMO: https://3v4l.org/hJu52
Using the following code:
$text = "أطلقت غوغل النسخة المخصصة للأجهزة الذكية العاملة بنظام أندرويد من الإصدار “25″ لمتصفحها الشهير كروم.ولم تحدث غوغل تطبيق كروم للأجهزة العاملة بأندرويد منذ شهر تشرين الثاني العام الماضي، وهو المتصفح الذي يستخدمه نسبة 2.02% من أصحاب الأجهزة الذكية حسب دراسة سابقة. ";
$tags = "غوغل, غوغل النسخة, كروم";
$tags = explode(",", $tags);
foreach($tags as $k=>$v) {
$text = preg_replace("/\b{$v}\b/u","$0",$text, 1);
}
echo $text;
Will give the following result:
I love PHP">love PHP</a>, but I am facing a problem
Note that my text is in Arabic.
The way is to do all in one pass. The idea is to build a pattern with an alternation of tags. To make this way work, you must before sort the tags because the regex engine will stop at the first alternative that succeeds (otherwise 'love' will always match even if it is followed by 'php' and 'love php' will never be matched).
To limit the replacement to the first occurence of each word you can remove tag from the array once it has been found and you test if it is always present in the array inside the replacement callback function:
$text = 'I love PHP, I love love but I am facing a problem';
$tagsCSV = 'love, love php, facing';
$tags = explode(', ', $tagsCSV);
rsort($tags);
$tags = array_map('preg_quote', $tags);
$pattern = '/\b(?:' . implode('|', $tags) . ')\b/iu';
$text = preg_replace_callback($pattern, function ($m) use (&$tags) {
$mLC = mb_strtolower($m[0], 'UTF-8');
if (false === $key = array_search($mLC, $tags))
return $m[0];
unset($tags[$key]);
return '<a href="index.php?s=news&tag=' . rawurlencode($mLC)
. '">' . $m[0] . '</a>';
}, $text);
Note: when you build an url you must encode special characters, this is the reason why I use preg_replace_callback instead of preg_replace to be able to use rawurlencode.
If you have to deal with an utf8 encoded string, you need to add the u modifier to the pattern and you need to replace strtolower with mb_strtolower)
the preg_split way
$tags = explode(', ', $tagsCSV);
rsort($tags);
$tags = array_map('preg_quote', $tags);
$pattern = '/\b(' . implode('|', $tags) . ')\b/iu';
$items = preg_split($pattern, $text, -1, PREG_SPLIT_DELIM_CAPTURE);
$itemsLength = count($items);
$i = 1;
while ($i<$itemsLength && count($tags)) {
if (false !== $key = array_search(mb_strtolower($items[$i], 'UTF-8'), $tags)) {
$items[$i] = '<a href="index.php?s=news&tag=' . rawurlencode($tags[$key])
. '">' . $items[$i] . '</a>';
unset($tags[$key]);
}
$i+=2;
}
$result = implode('', $items);
Instead of calling preg_replace multiple times, call it a single time with a regexp that matches any of the tags:
$tags = explode(",", tags);
$tags_re = '/\b(' . implode('|', $tags) . ')\b/u';
$text = preg_replace($tags_re, '$0', $text, 1);
This turns the list of tags into the regexp /\b(love|love php|facing)\b/u. x|y in a regexp means to match either x or y.
Could you tell how to replace string by preg-replace (need regular expression):
/user/{parent_id}/{action}/step/1
At the equivalent values of an array:
array('parent_id'=>32, 'action'=>'some');
To make:
/user/32/some/step/1
Addition
This is a typical problem, so I probably will not know what the names of variables come
You can use str_replace
For example:
str_replace(array("{parent_id}", "{action}"), array(32, 'some'), "/user/{parent_id}/{action}/step/1");
$arr = array('parent_id'=>32, 'action'=>'some');
$out = str_replace(array_keys($arr),array_values($arr),$in);
no need for regexps!
Say you have:
$arr = array('parent_id'=>32, 'action'=>'some');
$in = '/usr/{parent_id}/{action}/step/1';
This will replace the braces:
function bracelize($str) {
return '{' . $str . '}';
}
$search = array_map('bracelize', array_keys($arr));
$out = str_replace($search, $arr, $in);
Or if you are using PHP >= 5.3 you can use lambdas:
$search = array_map(function ($v) { return '{'.$v.'}';}, array_keys($arr));
$out = str_replace($search, $arr, $in);
$s = '/user/{parent_id}/{action}/step/1';
$replacement = array('parent_id'=>32, 'action'=>'some');
$res = preg_replace(array('/\\{parent_id\\}/', '/\\{action\\}/'), $replacement, $s);
Of course, you could just as well use str_replace (in fact, you ought to).
<?php
$string = '/usr/{parent_id}/{action}/step/1';
$pattern = array('#{parent_id}#', '#{action}#');
$values = array('32', 'some');
echo preg_replace($pattern, $values, $string);
?>
If your problem is not more complicated than this, i would recommend changing preg_replace to str_replace though.
EDIT: I see you don't know the variable names in advance. In which case you could do something like this.
<?php
function wrap_in_brackets(&$item)
{
$item = '{' . $item . '}';
return true;
}
$string = '/usr/{parent_id}/{action}/step/1';
$variables = array('parent_id' => 32, 'action' => 'some');
$keys = array_keys($variables);
array_walk($keys, 'wrap_in_brackets');
echo str_replace($keys, array_values($variables), $string);
?>
Expanding on mvds' answer:
$in = 'user/{parent_id}/{action}/step/1';
$arr = array('{parent_id}' => 32, '{action}' => 'some');
$out = str_replace(array_keys($arr), $arr, $in);
Or:
$in = 'user/{parent_id}/{action}/step/1';
$arr = array('parent_id' => 32, 'action' => 'some');
$arr[] = '';
$find = array_merge(array_keys($arr), array('{', '}'));
$out = str_replace($find, $arr, $in);