我可以使用普通的字符串函数执行此操作,但万一我想知道这件事是否可以用正则表达式方式完成。
$list = array("animal","human","bird");
$input1 = "Hello, I am an /animal/1451/ and /bird/4455";
$input2 = "Hello, I am an /human/4461451";
$input3 = "Hello, I am an /alien/4461451";
$output1 = ["type"=>"animal","number"=>1451],["type"=>"bird","number"=>4455]];
$output2 = [["type"=>"human","number"=>4461451]];
$output3 = [[]];
function doStuff($input,$list){
$input = explode(" ",$input);
foreach($input as $in){
foreach($list as $l){
if(strpos($in,"/".$l) === 0){
//do substr to get number and store in array
}
}
}
}
答案 0 :(得分:0)
正则表达式解决方案:
$regex = '~/(animal|human|bird)/(\d+)~';
$strs = [
"Hello, I am an /animal/1451/ and /bird/4455",
"Hello, I am an /human/4461451",
"Hello, I am an /alien/4461451",
];
$outs = [];
foreach ($strs as $s) {
$m = [];
preg_match_all($regex, $s, $m);
// check $m structure
echo'<pre>',print_r($m),'</pre>' . PHP_EOL;
if (sizeof($m[1])) {
$res = [];
foreach ($m[1] as $k => $v) {
$res[] = [
'type' => $v,
'number' => $m[2][$k],
];
}
$outs[] = $res;
}
}
echo'<pre>',print_r($outs),'</pre>';
答案 1 :(得分:0)
在JavaScript中你可以这样做
var list = ["animal","human","bird"];
var input1 = "Hello, I am an /animal/1451/ and /bird/4455";
var input2 = "Hello, I am an /human/4461451";
var input3 = "Hello, I am an /alien/4461451";
function get(input) {
var regex = new RegExp('(' + list.join('|') + ')\/(\\d+)', 'g');
var result = [];
var match;
while ((match = regex.exec(input))) {
result.push({ type: match[1], number: match[2] });
}
return result;
}
console.log(
get(input1),
get(input2),
get(input3)
);
答案 2 :(得分:0)
使用preg_match_all
和array_map
函数的简短解决方案:
$pattern = "/\/(?P<type>(". implode('|', $list)."))\/(?P<number>\d+)/";
$result = [];
foreach ([$input1, $input2, $input3] as $str) {
preg_match_all($pattern, $str, $matches, PREG_SET_ORDER);
$result[] = array_map(function($a){
return ['type'=> $a['type'], 'number' => $a['number']];
}, $matches);
}
print_r($result);