我试图使用PHP与XAMPP(Apache和MYSQL)做一个简单的注册表单。我目前有一个密码字段的问题,目前显示它的输入为明文,而不是隐藏它从视图。回显密码输入字段显示为文本
?php
$regfields=array("remail"=>"Email","remail2"=>"Re-enter Email","rusername"=>"Username","rpassword"=>"Password","rpassword2"=>"Re-enter Password","rbiz"=>"BizName","rdesc"=>"Desc");
?>
<html>
<body>
<h3>Registration</h3>
<form action="Login.php" name="RegisterForm" method="post" >
<table border="0">
<?php
if(isset($errormsg))
{
echo"$errormsg";
}
foreach($regfields as $field=>$value)
{
if($field!=="rdesc")
{
echo"<tr>";
echo"<td>";
echo"<label for='$field'>$value: </label>";
echo"</td>";
echo"<td>";
echo"<input type='text' name='$field' size='40' maxlength='50' /> </td></tr>";
}
else if($field=="remail"||$field=="remail2")
{
echo"<tr>";
echo"<td>";
echo"<label for='$field'>$value: </label>";
echo"</td>";
echo"<td>";
echo"<input type='text' name='$field' size='40' maxlength='50' /></td></tr>";
}
else if($field=="rpassword"||$field=="rpassword2")
{
echo"<tr>";
echo"<td>";
echo"<label for='$field'>$value: </label>";
echo"</td>";
echo"<td>";
echo"<input type='password' name='$field' size='40' maxlength='50' /></td></tr>";
}
else
{
echo"<tr>";
echo"<td>";
echo"<label for=$field>$value: </label>";
echo"</td>";
echo"<td>";
echo"<input type='text' name='$field' size='300' maxlength='300' style='width:400px;height:100px; \n'/>";
echo"</td>";
echo"</tr>";
}
}
?>
</form>
</table>
<input type="submit" name="Button" value="Register" />
</body>
</html>
我怀疑此行
echo"<input type='password' name='$field' size='40' maxlength='50' /></td></tr>";
其中输入类型不是由于使用了单引号注册我的问题,茎,而不是双引号,导致输入型从字面上并没有解释被用作价值?
我将不胜感激任何输入。 谢谢!
移除
!
纠正你的代码是这样被显示在输入为纯文本或提交后..? –您在表格和提交之前关闭表单标记在表单之外。这将如何工作? – Rahul11
尝试使用'\“字符尝试转义此行'echo'