我正在学习更多的PHP，并且在PHP本身工作后，我似乎无法正确验证任何表单字段。我的目标是检查第一个名称字段是否为空，如果是的，它将以红色给出一条消息。红色作品中的消息，但这仅仅是因为echo脚本是通过表单提交来调用的，而不是因为它已经检测到了任何空字段，因为当我做出其他陈述时，如果不是空的，我就会说对于字段为空时，相同的消息。另外，是否可以像使用JavaScript一样一次检查多个输入字段？例如，如果输入1 ==''||input2 ==''等等。这是我的html：
<html>
<head>
<title>Welcome</title>
</head>
<body>
<form action="welcome.php" method="post">
<fieldset>
<legend>Personal Info</legend>
First name <input name="name" type="text">
Middle name <input name="middlename" type="text">
Surname <input name="lastname" type="text">
Age <input name="age" type="number">
Date of birth <input name="dob" type="date">
</fieldset>
<fieldset>
<legend>Regional & location info</legend>
Continent
<select>
<option value="europe">Europe</option>
<option value="americas">America</option>
<option value="africa">Africa</option>
<option value="asia">Asia</option>
<option value="australia">Australia</option>
<option value="eurasia">Eurasia</option>
</select>
Country <input name="country" type="text"> State <input type="text">
City <input name="city" type="text">
Street number <input name="streetno" type="number">
Street name <input name="streetname" type="text"> <br><br>
Suburb <input name="suburb" type="text"> Postcode <input name="postcode" type="number">
If none of these apply to your accommodations, enter a typed location here <input type="text">
</fieldset>
<fieldset>
<legend>Previous lifestyle accommodations</legend>
Previous &/or most recent job title <input name="job" type="text">
First time job seeker <input type="checkbox" name="check1" value="ftjb">
I'm a student <input type="checkbox" name="check2" value="ias">
Previous &/or most recent acedemic title <input name="school" type="text">
First time applying for a qualification <input type="checkbox" name="check3" value="ftafaq">
I have work experience <input type="checkbox" name="check4" value="ihwe">
</fieldset>
<fieldset>
<legend>Details of arrival</legend>
Reason for arrival <input name="reason" type="text">
Date of arrival <input name="arrival" type="date">
Amount of stay expectancy
<input type="checkbox" name="check3">Temporary
<input type="checkbox" name="check4">Longterm
</fieldset>
<fieldset>
<legend>Signiture</legend>
<input name='signiture' type="text">
</fieldset>
<input type="submit" name="submit" value="Submit">
</form>
</body>
</html>
这是我的php代码：
<?php
$firstname = $_POST['name'];
$lastname = $_POST['lastname'];
$age = $_POST['age'];
$dob = $_POST['dob'];
$country = $_POST['country'];
$city = $_POST['city'];
$suburb = $_POST['suburb'];
$postcode = $_POST['postcode'];
$streetno = $_POST['streetno'];
$streetname = $_POST['streetname'];
$suburb = $_POST['suburb'];
$job = $_POST['job'];
$school = $_POST['school'];
$reason = $_POST['reason'];
$arrival = $_POST['arrival'];
$signiture = $_POST['signiture'];
if (isset($_POST['submit'])) {
if (empty($_POST[$firstname])) {
echo '<p style="color: red; text-align: center">Your first name is required</p>';
} else {
echo "wassaup";
}
}
?>
在您的if语句中，您需要执行此操作：
if (empty($_POST['name'])) { //Or replace $_POST['name'] with $firstname
echo '<p style="color: red; text-align: center">Your first name is required</p>';
} else {
echo "wassaup";
}
您的语法if (empty($_POST[$firstname]))错误。您应该在firstname中使用"单个派法"，然后删除此标志$。
将此if (empty($_POST[$name])) {更改为if (empty($_POST['firstname'])) {，然后再次检查。您的语法是错误的，这就是为什么它不起作用。
在您的代码中更改
if (!isset($_POST['firstname'] || empty($_POST['firstname'])) {
echo '<p style="color: red; text-align: center">Your first name is required</p>';
} else {
echo "wassaup";
}
首先，您可以创建错误数组，将所有错误放在其中
$error = array();
然后检查多个字段
if (empty($_POST['firstname'])) $error[] = "First name can't be empty";
if (empty($_POST['lastname'])) $error[] = "Last name can't be empty";
// and so on
完成所有这些制作声明后，以检查错误数组是否为空，如果不显示错误
if (empty($error)) {
// do something
} else {
// error exists you want to display all errors
foreach ($error as $value) {
echo '<ul><li>'.$value.'</li></ul>';
}
}
在HTML表单中，您必须在HTML标签中添加所需的属性，而无需使用PHP处理
First name <input name="name" type="text" required="required">
，如果用户未输入他的名字，他将收到错误消息（需要名字）