Php 处理html下拉表单
我有以下html下拉表单:Php 处理html下拉表单,php,html,Php,Html,我有以下html下拉表单: <form method='post' action='signup.php'>$error <span class='fieldname'>Sex</span> <select name ="sex"> <option value="female">Female</option> <option value="male">Male</option> </s
<form method='post' action='signup.php'>$error
<span class='fieldname'>Sex</span>
<select name ="sex">
<option value="female">Female</option>
<option value="male">Male</option>
</select> <br />
_END;
给我们更多的脚本。可能是一些错误。我使用了下面的代码,它工作得很好
<?php
if(isset($_POST['sex']))
{
$sex = $_POST['sex'];
echo 'Result: '.$sex;
}
?>
<form method='post' action=''>
<span class='fieldname'>Sex</span>
<select name ="sex">
<option value="female">Female</option>
<option value="male">Male</option>
</select><input type="submit"> <br />
</form>
选项值,因此必须使用selected=selected任意一个。然后尝试下面的代码
<form method='post' action=''>
<span class='fieldname'>Sex</span>
<select name ="sex">
<option value="female" selected="selected">Female</option>
<option value="male">Male</option>
</select><input type="submit"> <br />
</form>
<?php
$sex = isset($_POST['sex']) ? mysql_real_escape_string($_POST['sex']) : '';
echo 'Gender: '. $sex;
?>
您不应该在适当的环境中包含这一点并参考PHP脚本吗?没有足够的信息。您没有显示任何与处理表单的PHP相关的内容,这就是错误所在;您在php代码中不回显$error,所以我不确定您是否知道php脚本是如何工作的,但您是否尝试使用include标记来结束html表单
<form method='post' action=''>
<span class='fieldname'>Sex</span>
<select name ="sex">
<option value="female" selected="selected">Female</option>
<option value="male">Male</option>
</select><input type="submit"> <br />
</form>
<?php
$sex = isset($_POST['sex']) ? mysql_real_escape_string($_POST['sex']) : '';
echo 'Gender: '. $sex;
?>