PHP在數字邏輯中設置了最少4位數

Question

我是PHP新手，我試圖在變量中設置最小4位數字並創建成員ID。

$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); 

無論我做什麼，我總是得到ABC124的回報。

Answer 1

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

上面一行是做了以下內容：

1. 以數字串0123和增加小數1它，讓你124感謝type juggling
2. 然後將其前面加上「ABC」因爲你把它告訴

如果你想強制執行前導零使用str_pad()：

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

或者添加1您前面加上之前的零：

$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); 

或者使用sprintf()：

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