我想:他有XXX得到它。或者:他必须拥有XXX。
$string = "He had had to have had it.";
echo preg_replace('/had/', 'XXX', $string, 1);
输出:
他XXX必须拥有它。
在案例中,'have'被替换为第一个。
我想使用第二个和第三个。不是从右边或左边读,“preg_replace”可以做什么?
答案 0 :(得分:2)
试试这个:
<?php
function my_replace($srch, $replace, $subject, $skip=1){
$subject = explode($srch, $subject.' ', $skip+1);
$subject[$skip] = str_replace($srch, $replace, $subject[$skip]);
while (($tmp = array_pop($subject)) == '');
$subject[]=$tmp;
return implode($srch, $subject);
}
$test ="He had had to have had it.";;
echo my_replace('had', 'xxx', $test);
echo "<br />\n";
echo my_replace('had', 'xxx', $test, 2);
?>
答案 1 :(得分:2)
$string = "He had had to have had it.";
$replace = 'XXX';
$counter = 0; // Initialise counter
$entry = 2; // The "found" occurrence to replace (starting from 1)
echo preg_replace_callback(
'/had/',
function ($matches) use ($replace, &$counter, $entry) {
return (++$counter == $entry) ? $replace : $matches[0];
},
$string
);
答案 2 :(得分:0)
试试这个
的解决方案的
function generate_patterns($string, $find, $replace) {
// Make single statement
// Replace whitespace characters with a single space
$string = preg_replace('/\s+/', ' ', $string);
// Count no of patterns
$count = substr_count($string, $find);
// Array of result patterns
$solutionArray = array();
// Require for substr_replace
$findLength = strlen($find);
// Hold index for next replacement
$lastIndex = -1;
// Generate all patterns
for ( $i = 0; $i < $count ; $i++ ) {
// Find next word index
$lastIndex = strpos($string, $find, $lastIndex+1);
array_push( $solutionArray , substr_replace($string, $replace, $lastIndex, $findLength));
}
return $solutionArray;
}
$string = "He had had to have had it.";
$find = "had";
$replace = "yz";
$solutionArray = generate_patterns($string, $find, $replace);
print_r ($solutionArray);
输出
Array
(
[0] => He yz had to have had it.
[1] => He had yz to have had it.
[2] => He had had to have yz it.
)
我管理此代码尝试优化它。
答案 3 :(得分:0)
可能不会用这个赢得任何优雅,但很短:
$string = "He had had to have had it.";
echo strrev(preg_replace('/dah/', 'XXX', strrev($string), 1));