使用下面的代码,我无法在提交时显示表单响应。我尝试过混合使用_GET和_POST,但是我不知道什么时候使用,因为我对PHP比较陌生。代码如何在提交时显示表单响应?
<?php {
$fDogErr = $lDogErr = "";
// only show the information if the button named "subButton" has been pressed
if (!isset($_POST['submit'])) {
// set the variable with the submitted value
if (empty($fDog = $_POST['favourite dog'])) {
$fDogErr = "Need favourite dog";
} else {
$fDog = $_POST['favourite dog'];
}
if (empty ($lDog = $_POST['least favourite dog'])) {
$lDogErr = "Need least favourite dog";
} else {
$lDog = $_POST['least favourite dog'];
}
if (empty($password = $_POST['pawsword'])) {
$password = "";
} else {
$password = $_POST['password'];
}
if (empty($dogcac = $_POST['dogcac'])) {
$dogcac = "";
} else {
$dogcac = $_POST['dogcac'];
}
$secretdoggo = $_POST['secretdoggo'];
}
// display the user inputs to the screen
echo "<p>Your favourite dog is <b>" . $fDog . "</b>.</p>";
echo "<p>Your least favourite dog is <b>" . $lDog . "</b>.</p>";
echo "<p>Your pawsword is <b>" . $password . "</b>.</p>";
echo "<p> Did you know? <b>" . $secretdoggo . "</b>.</p>";
}
?>
发布于 2018-09-28 09:34:52
您需要先定义一个包含所有输入标记的HTML表单。表单标记有一个操作属性,该属性声明数据将发送到何处(您所链接的PHP ),以及一个方法 (POST、GET等)。以及一个类型为 submit的按钮,用于触发表单数据的发送。然后,当单击submit按钮时,PHP文件可以显示表单响应。
如下例所示:
表单HTML示例:
<form method="post" action="response.php">
<label>Password:</label> <input type="text" id="favourite_dog" name="favourite_dog" />
<label>Password:</label> <input type="password" id="password" name="password" />
more input tags . . .
<button type="submit">Submit</button>
</form>
显示响应的
PHP示例:
response.php
<?php
if (!empty($_POST)){
$favourite_dog = $_POST['favourite_dog'];
$password= $_POST['password'];
echo "<p>Your favourite dog is <b>" . $favourite_dog . "</b>.</p>";
}
希望这能有所帮助。
https://stackoverflow.com/questions/52546450
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