Question

我从MySQL获取的地址如下：

$address="This Street, Nice Area, That Country";
$address="This Street, Superb Area, Which Country";
$address="This Street, Fine Area, A Country";

现在我要做的是找到AREA - 就像$area=array("Nice", "Good")。因此，如果在$area中找到$address中定义的值，则应列出或留空。

Answer 1

尝试类似

的内容

$area=array(1=>"Nice", "Good");
$address = 'This Street, Nice Area, That Country';
preg_match(', (\w+) Area,', $address, $res);
$is_area = array_search( $res[1], $area, true) ;
if($is_area) echo $address;

<强>解释

第一行设置区域数组以索引1而不是零开始（确保查找不返回零）

preg_match行从地址

中提取区域

数组搜索行查找区域数组中的区域

Answer 2

我认为你所要求的内容可以解释为：

“列出包含$area数组中找到的字符串实例的所有$ address字符串。”

在哪种情况下：

// Set up.
$address = array(
    "This Street, Nice Area, That Country",
    "This Street, Superb Area, Which Country",
    "This Street, Fine Area, A Country",
);
$area = array("Nice", "Good");

// The search.
$results = array();
foreach ($address as $haystack) {
    foreach ($area as $needle) {
        $regexPattern = "#{$needle}#";
        if (preg_match($regexPattern, $haystack)) 
            array_push($results, $haystack);
    }
}

// The results.
foreach ($results as $result) {
    echo "<p>{$result}</p>";
}

搜索字符串中数组的值

2 个答案: