HTML 文件
<form method="post" name="mysearch" action="/abc/search.php" onSubmit="return ch_search()">
<div class="statea">Search</div>
<div class="select_a">
<label for="text2"></label>
<input type="text" name="skey" id="text2" onblur="if(this.value=='') this.value = 'Search';" onfocus="if(this.value=='Search') this.value = '';" value="">
</div>
<div class="select_a">
<label for="text2"></label>
<select name="skid[]" id="sele3" multiple="multiple">
<? echo $kalist;?>
</select>
</div>
<div class="select_a">
<label for="text2"></label>
<input type="text" name="scity" id="text2" onblur="if(this.value=='') this.value = 'City';" onfocus="if(this.value=='City') this.value = '';" /><span class="hnt">(Comma Seperated)</span>
</div>
<div class="select_a">
<select name="s_state[]" id="sel3" multiple="multiple">
<? echo $states;?>
</select>
</div>
<div class="select_a">
<label for="text2"></label>
<input type="submit" name="ssubmit" id="text2" value="Search"/>
</div>
</form>
PHP 文件 -
if(isset($_POST['ssubmit']))
{
my code here
}
我试过 print_r($_POST); var_dump($_POST); 一切都将归零。它在我的本地机器上运行良好,但在主服务器上上传时它在 $_POST 中没有给出任何值。我的这段代码有什么问题?
编辑:当我在 firefox firebug 面板中检查标题时,它显示已发布的数据,那么为什么我不能通过 php 脚本看到它????