我在 PHP 中使用以下函数来检测包含“near”的字符串中的实体和位置preg_match();。有没有更优化的方法来为此编写代码?我使用了很多 if 语句,似乎可以改进,但我不确定如何改进。
// Test cases
$q = "red robin near seattle";
//$q = "red robin near me";
//$q = "red robin nearby";
//$q = "red robin near my location";
function getEntityAndLocation($q){
$entityAndLocation = array("entity" => null, "location" => null);
if(preg_match('(nearby)', $q) === 1) {
$breakdown = explode("nearby", $q);
$entityAndLocation["entity"] = $breakdown[0];
$entityAndLocation["location"] = $breakdown[1];
return $entityAndLocation;
}
if(preg_match('(near my location)', $q) === 1) {
$breakdown = explode("near my location", $q);
$entityAndLocation["entity"] = $breakdown[0];
$entityAndLocation["location"] = $breakdown[1];
return $entityAndLocation;
}
if(preg_match('(near me)', $q) === 1) {
$breakdown = explode("near me", $q);
$entityAndLocation["entity"] = $breakdown[0];
$entityAndLocation["location"] = $breakdown[1];
return $entityAndLocation;
}
if(preg_match('(near)', $q) === 1) {
$breakdown = explode("near", $q);
$entityAndLocation["entity"] = $breakdown[0];
$entityAndLocation["location"] = $breakdown[1];
return $entityAndLocation;
}
}
if(preg_match('(near)', $q) === 1) {
$entityAndLocation = getEntityAndLocation($q);
print_r($entityAndLocation);
}
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