php - 将字符数限制在两个限制之间

Question

我有一个验证if语句，该语句主要检查两件事。首先:查看名为“why”的下拉框是否为空，如果为空，则会向前端发出错误消息，指出我们需要您提供适当的原因。很好现在，我有第二个条件，即如果出于任何原因，“为什么”下拉框中的值为“其他”，并且如果注释框为空，则将拍摄另一个错误。会说“请在评论框中说明您访问的性质!”

我的问题是我尝试使用它，以便注释框的长度在15到45个字符之间。我一直在努力进行signinpage.php验证大约20个小时，以使其完全符合我的要求，而我却无法到达所需的位置。任何帮助将是可爱的!

if(empty($why)) {
            $errors[] = 'Please make sure to select the proper reasoning for your vistit today!';
    } 
        elseif ($why ==='Other' && empty($comments)) {
            $errors[] = 'Please explain the nature of your visit in the comments box!';

        if (strlen($comments) < 15) {
            $errors[] = 'Your explaination is short, please revise!';
    }
        if(strlen($comments) > 45) {
            $errors[] = 'Your explaintion is to long, please revise!';
    }
    }

缩进:

if(empty($why)) {
    $errors[] = 'Please make sure to select the proper reasoning for your vistit today!';
} 
elseif ($why ==='Other' && empty($comments)) {
    $errors[] = 'Please explain the nature of your visit in the comments box!';

    if (strlen($comments) < 15) {
        $errors[] = 'Your explaination is short, please revise!';
    }
    if(strlen($comments) > 45) {
        $errors[] = 'Your explaintion is to long, please revise!';
    }
}

Answer 1

如果您正确地复制并粘贴了此内容，则说明您的括号弄乱了。将所有注释验证放到一个单独的部分中，只有$why==="Other"才被检查。

if(empty($why))
    $errors[] = 'Please make sure to select the proper reasoning for your vistit today!';
elseif ($why ==='Other'){
    if(empty($comments))
        $errors[] = 'Please explain the nature of your visit in the comments box!';
    else{
        if (strlen($comments) < 15)
            $errors[] = 'Your explaination is short, please revise!';
        if(strlen($comments) > 45)
            $errors[] = 'Your explaintion is to long, please revise!';
    }
}