Question

我是PHP的新手，我正在尝试在变量中设置最小4位数字并创建一个memberid。

$memberidVal = ABC0123;
$numMemberVal = preg_replace("/[^0-9]/", "", $memberidVal);
if (strlen($numMemberVal) == 3) {
    $numMemberVal = 0 . $numMemberVal;
} elseif (strlen($numMemberVal) == 2) {
    $numMemberVal = 00 . $numMemberVal;
} elseif (strlen($numMemberVal) == 1) {
    $numMemberVal = 000 . $numMemberVal;
}
$newMemberId = "ABC" . ($numMemberVal + 1);
echo ($newMemberId);

无论我做什么，我总是得到ABC124作为回报。

Answer 1

$newMemberId = "ABC" . ($numMemberVal + 1);

以上一行正在执行以下操作：

使用数字字符串0123并将小数点1添加到124，感谢type juggling
然后它是在“ABC”前面，因为你告诉它

如果您想强制使用前导零使用str_pad()：

$memberidVal = ABC0123;
$numMemberVal = preg_replace("/[^0-9]/", "", $memberidVal);
$newMemberId = "ABC" . str_pad(($numMemberVal + 1), 4, '0', STR_PAD_LEFT);

或者在添加{0}之前添加1：

$memberidVal = ABC0123;
$numMemberVal = preg_replace("/[^0-9]/", "", $memberidVal);
$numMemberVal++;
if (strlen($numMemberVal) == 3) {
    $numMemberVal = 0 . $numMemberVal;
} elseif (strlen($numMemberVal) == 2) {
    $numMemberVal = 00 . $numMemberVal;
} elseif (strlen($numMemberVal) == 1) {
    $numMemberVal = 000 . $numMemberVal;
}
$newMemberId = "ABC" . $numMemberVal;
echo ($newMemberId);

或使用sprintf()：

$memberidVal = ABC0123;
$numMemberVal = preg_replace("/[^0-9]/", "", $memberidVal);
$newMemberId = sprintf('%s%04d', 'ABC', ($numMemberVal + 1));

PHP在数字逻辑中设置最小4位数

1 个答案: