Question

目前我使用此类数组进行Mysql数据库的版本控制：

$pages_table = array (
    "GUID" => array (
        "type" => "CHAR(13)",
        "length" => 13,
    )
    "Number" => array (
        "type" => "TINYINT(4)",
        "length" => 4,
    )
    "Pagename" => array (
        "type" => "VARCHAR(30)",
        "length" => 30,
    )

它有效，但我想让它更干净，如：

$pages_table = array (
    "GUID" => "CHAR(13)",
    "Number" => "TINYINT(4)",
    "Pagename" => "VARCHAR(30)",
);

然后，如果我遍历数组，我想将$ new_length（INT）设置为$ new_type字符串括号之间的数字：

while ($column = key($pages_table)) {
    $new_type = current($pages_table);
    $new_length = //Get this value from $new_type;
    if ($existing_table[$column]['length'] < $new_length) {
        $modify[$column] = $new_type;
    }
    next($pages_table);
}

Answer 1

$new_length = (int) preg_replace('/\D/', '', $new_type);

Answer 2

使用正则表达式：

preg_match('/\(\d+\)/', $subject, $matches);
$new_length = $matches[0];

如果确保字符串中没有其他数字，则可以缩短模式：

preg_match('/\d+/', $subject, $matches);
$new_length = $matches[0];

while ($column = key($pages_table)) {
    $new_type = current($pages_table);
    $hasLength = (preg_match('/\(\d+\)/', $new_type, $matches) == 1);

    $new_length = intval($matches[0]);
    if ($hasLength && $existing_table[$column]['length'] < $new_length) {
        $modify[$column] => $new_type;
    }
    next($pages_table);
}

如何从PHP中的字符串括号中获取数字

2 个答案: