[英]Input and output values for php into the browser?
我剛開始在 php 上學習我的模塊,兩個月前完成了 Javascript 我熟悉很多元素,但有點生疏。 基本上,我用 html 設置了一個表格,並添加了一點 css 使它漂亮。 學校讓我們做一些通用的練習。 我現在如何使用 Javascript 來獲取值等
document.getElementById("textBox").value;
並通過我創建的 div 返回 output
document.getElementById("return").innerHTML =... ;
老實說,我不知道如何用 php 做到這一點,我相信這很容易,但我一生都無法在網上找到解決方案,而且學校的視頻也很模糊。
練習者是“創建一個向用戶詢問數字 (1-7) 並返回與輸入的數字相關聯的星期幾的應用程序。您必須驗證數字”
這是我到目前為止的代碼....
<!DOCTYPE html>
<html>
<head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.4.1/jquery.min.js"></script>
<title>PHP exercise 1</title>
<link rel="stylesheet" type="text/css" href="./styling1.css"/>
</head>
<body>
<!--E X C E R S I S E 3 !-->
<section id="allContent">
<section>
<button class="hide" id="clic"> <h4> E X E R C I C E 1 </h4></button>
</section>
<div id="container" class="hidden">
<div id="Barbox"><p class="pClass"> Create an application that ask the user for number (1-7) and return the day of the week associate with the number entered. You must validate the number</p>
</div>
<div id="background"></div>
<div class="ball"></div><div id="ball2"></div>
<div id="titleBox">
<p id="titlePRG">Enter a number!</p>
</div>
<form>
<input type="text" value="<?php echo $number = 1 ; ?>" id="textField">
<input type="button" value="Enter" class="button" onclick="return_day()">
</form>
<div id="valueReturned">
<p id="returnText"><p>
</div>
<a href="index4.html" id="jButton"><h3> Next Exercise </h3></a>
</div>
</section>
<?php
function return_day(){
$number = 1 ;
if ($number == 1){
echo "The day of the week is Monday";
}
else if ($number == 2){
echo "The day of the week is Tuesday";
}
else if ($number == 3){
echo "The day of the week is Wednesday";
}
else if ($number == 4){
echo "The day of the week is Thursday";
}
else if ($number == 5){
echo "The day of the week is Friday";
}
else if ($number == 6){
echo "The day of the week is Saturday";
}
else if ($number == 7){
echo "The day of the week is Sunday";
}
else{
echo ("Wrong input, try again!");
}
}//end of function
?>
如果它有任何區別,這就是瀏覽器的樣子,所以你理解我的問題
答案應該顯示在底部的白框中,即
下面的代碼應該可以幫助您理解。
<?php
$response = "";
if(isset($_POST) && isset($_POST["mytext"])): /// check if user submitted form
$num = $_POST["mytext"]; //put form mytext variable.. retrieved from html <input name="mytext"....
if(is_numeric($num)):
switch($num):
case "1":
$response = "Monday";
break;
case "2":
$response = "Tuesday";
break;
case "3":
$response = "Wednesday";
break;
case "4":
$response = "Thursday";
break;
case "5":
$response = "Friday";
break;
case "6":
$response = "Saturday";
break;
case "7":
$response = "Sunday";
break;
default:
$response = "Invalid number";
endswitch;
else:
$response = "Invalid number";
endif;
endif;
?>
<!DOCTYPE html>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<title>PhpFiddle Initial Code</title>
<!--
http://phpfiddle.org
-->
<script type="text/javascript">
</script>
<style type="text/css">
</style>
</head>
<body>
<div style="margin: 30px 10%;">
<h3>My form</h3>
<form id="myform" name="myform" action="" method="post">
<label>Text</label> <input type="text" value="" size="30" maxlength="100" name="mytext" id="" /><br /><br />
<br />
<button id="mysubmit" type="submit">Submit</button><br /><br />
<?php if (isset($_POST)): ?>
<div style="background-color:red;color:white;"><?php echo $response; ?></div>
<?php endif; ?>
</div>
</form>
</div>
<?php
?>
</body>
</html>
與 Javascript 不同,php 完全不同,Javascript 的功能可以在 HTML 中通過使用 ZC0BB2196426022E8AEDZFA 來訪問,加載。 等等,屬性。 但是php的功能卻大不相同,只能通過php代碼訪問。 首先,您需要使用一種方法創建一個舞會,然后發送到同一頁面(在您的情況下),然后運行您應用的條件,最后像我在其中所做的那樣回顯結果。
<!DOCTYPE html>
<html>
<head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.4.1/jquery.min.js">
</script>
<title>PHP exercise 1</title>
<link rel="stylesheet" type="text/css" href="./styling1.css" />
</head>
<body>
<!--E X C E R S I S E 3 !-->
<section id="allContent">
<section>
<button class="hide" id="clic">
<h4> E X E R C I C E 1 </h4>
</button>
</section>
<div id="container" class="hidden">
<div id="Barbox">
<p class="pClass"> Create an application that ask the user for number (1-7) and return the day of the week associate with the number entered. You must validate the number</p>
</div>
<div id="background"></div>
<div class="ball"></div>
<div id="ball2"></div>
<div id="titleBox">
<p id="titlePRG">Enter a number!</p>
</div>
<form method="get" action="<?php $_SERVER['PHP_SELF']; ?>">
<input type="text" value="" id="textField" name="number">
<input type="submit" value="Enter" class="button" name="submit">
</form>
<?php
if (isset($_GET['submit']))
$number = isset($_GET['number']) ? $_GET['number'] : '';
if ($number == 1) {
$returntext = "The day of the week is Monday";
} else if ($number == 2) {
$returntext = "The day of the week is Tuesday";
} else if ($number == 3) {
$returntext = "The day of the week is Wednesday";
} else if ($number == 4) {
$returntext = "The day of the week is Thursday";
} else if ($number == 5) {
$returntext = "The day of the week is Friday";
} else if ($number == 6) {
$returntext = "The day of the week is Saturday";
} else if ($number == 7) {
$returntext = "The day of the week is Sunday";
} else {
$returntext = ("Wrong input, try again!");
}
?>
<div id="valueReturned">
<p id="returnText"><?php echo $returntext; ?><p>
</div>
<a href="index4.html" id="jButton">
<h3> Next Exercise </h3>
</a>
</div></section>
希望您必須在合適的服務器上運行這些 php 代碼! 好走!
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