试图让我的外部php代码显示消息之前,但我有几个问题。有人可以给我一个手或一些提示进展?PHP其他如果语句行错误的外部PHP
我想了解我哪里去错了,但任何帮助肯定会被赞赏。当我将此文件上传到filezilla服务器时,出现以下错误。
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我更新了代码,当我运行我的html并提交时,它只会带我到一个空白页面,而不是回显(打印)我的任何文本。不管我输入什么号码?建议?我更新了php代码,并且html代码仍然与以前相同 –
<?php if(isset($ _ POST ['number'])){ $ number = $ _POST ['number']; $ number = 10;如果($ number <10){ echo“数字小于10”; }否则如果($ number <10 && $ number> 100){ echo“Number is between 10 and 100”; ($ number> 100){ echo“Number is large than 100”; } } ?> –
由于您的'$ number'值为10,您希望它匹配哪种条件? –