Alw*_*ver 3 php debugging validation
我正在使用PHP构建一个网站,我需要验证用户输入的名称是否正确.由于JavaScript是客户端的,我不能完全依赖它,所以这是我的服务器端函数来验证用户名:
function validate_name($name) {
$name = trim($name); // only for the purpose of debugging <---- edited comment
echo $name;
if (strlen($name) <= 1) {
return "small";
} else if (has_numbers($name)) {
return "numbers";
} else {
return true;
}
}
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在此之后,我相应地检查输入和显示结果:
function final_check() {
if (validate_name($_POST["first_name"]) == "small") {
echo "<span class='error'>Your first name cannot be empty</span>";
return false;
} else if (validate_name($_POST["first_name"]) == "numbers") {
echo "<span class='error'>Numbers are not allowed in your first name</span>";
return false;
}
return true;
}
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当我在first_name字段中输入任何内容时,我收到空错误消息; 当我输入数字时,我得到数字错误消息.但是,当我输入有效名称时,它会给我输入名称为空的错误消息.
这是帖子数据:
Array
(
[email] => ewf@gmail.com
[first_name] => qwe
[last_name] => wqe
[password] => qwe
[re_password] => qwe
[gender] => Male
)
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输出:
Your first name cannot be empty
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知道我做错了什么吗?在过去的一小时里,我一直在努力解决这个问题,但我一直无法找到解决方案.
Array
(
[email] => ewf@gmail.com
[first_name] => qwe
[last_name] => wqe
[password] => qwe
[re_password] => qwe
[gender] => Male
)
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// First_name长度= 3
function validate_name($name) {
$name = trim($name);
echo $name;
if (strlen($name) <= 1) {
return "small";
} else if (has_numbers($name)) {
return "numbers";
} else {
return true; // satisfy this case return true
}
}
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在这里,它变得像
function final_check() {
if (validate_name($_POST["first_name"]) == "small") { // if(1 == 'small')
echo "<span class='error'>Your first name cannot be empty</span>";
return false;
} else if (validate_name($_POST["first_name"]) == "numbers") {
echo "<span class='error'>Numbers are not allowed in your first name</span>";
return false;
}
return true;
}
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if(1 =='small')它与Boolean的字符串比较,它总是返回true.
请查看手册中的此页面以了解问题.