試圖讓我的外部php代碼顯示消息之前,但我有幾個問題。有人可以給我一個手或一些提示進展?PHP其他如果語句行錯誤的外部PHP
我想了解我哪裏去錯了,但任何幫助肯定會被讚賞。當我將此文件上傳到filezilla服務器時,出現以下錯誤。
Parse error: syntax error, unexpected 'echo' (T_ECHO) in/on line 13
HTML
<html xmlns="http://www.w3.org/1999/xhtml"><head>
<meta content="text/html; charset=utf-8" http-equiv="Content-Type">
<title>Enter your information in the form below</title>
</head>
<body>
<!-- Form1.html -->
<form action="Form1.php" method="post">
<fieldset><legend>Enter a number below </legend>
<b><p>Your number: </b> <input type="text" name="number" size="4" maxlength="4">
</p>
</fieldset>
<div align="center"><input type="submit" name="submit" value="Submit"></div>
</form>
</body>
</html>
PHP
<?php
if(isset($_POST['$number'])){
$number = $_POST['$number'];
if($number < 10){
echo "The number is smaller than 10";
} else ($number < 10 && $number > 100){
echo "The number is between 10 and 100";
} else ($number > 100){
echo "The number is larger than 100";
}
}
?>
s o我更新了代碼,當我運行我的html並提交時,它只會帶我到一個空白頁面,而不是回顯(打印)我的任何文本。不管我輸入什麼號碼?建議?我更新了php代碼,並且html代碼仍然與以前相同 –
<?php if(isset($ _ POST ['number'])){ $ number = $ _POST ['number']; $ number = 10;如果($ number <10){ echo「數字小於10」; }否則如果($ number <10 && $ number> 100){ echo「Number is between 10 and 100」; ($ number> 100){ echo「Number is large than 100」; } } ?> –
由於您的'$ number'值爲10,您希望它匹配哪種條件? –