当我输入游戏的ID号时,我总是得到我的if else表达式的第3个语句.我哪里错了？!!
<html>
<body>
<p>Type in the ID and hit search:</p>
<button onclick="myFunction()">Search</button>
<p id="demo"></p>
<form>
<input id="textbox" type="text" />
</form>
<script>
var textboxValue = document.getElementById("textbox").value;
function myFunction() {
var message;
if (textboxValue == 1) {
message = "Fantasy World";
} else if (textboxValue == 2) {
message = "Sir Wags A Lot";
} else {
message = "Take a Path";
}
document.getElementById("demo").innerHTML = message;
}
</script>
</body>
</html>
我不确定为什么这个简单的PHP脚本不起作用,
我的浏览器不会加载该页面.我认为它是逻辑上的缺陷而不是语法,但也许这里的某个人会很友好地指出我出错的原因/原因.
<html>
<head>
<title>My Encryption Program</title>
</head>
<body>
<?PHP
$ConvertedLetter ="";
$SecretMessage= "Kiss My Shiny Metal...";
$MessageLength = strlen($SecretMessage);
$Counter = 0;
For($Counter;$MessageLength;$Counter++){
$LetterToEncrypt = substr($SecretMessage,$Counter,1);
$AsciiNumber = ord($LetterToEncrypt) + 3;
$ConvertedLetter = $ConvertedLetter + Chr($AsciiNumber);
}
echo $ConvertedLetter;
?>
</body>
</html>