相关疑难解决方法(0)

我想使用表单中的隐藏输入将JavaScript变量传递给PHP.

但我不能得到的价值$_POST['hidden1']为$salarieid.有什么不对？

这是代码:

<script type="text/javascript">
    // View what the user has chosen
    function func_load3(name) {
        var oForm = document.forms["myform"];
        var oSelectBox = oForm.select3;
        var iChoice = oSelectBox.selectedIndex;
        //alert("You have chosen: " + oSelectBox.options[iChoice].text);
        //document.write(oSelectBox.options[iChoice].text);
        var sa = oSelectBox.options[iChoice].text;
        document.getElementById("hidden1").value = sa;
    }
</script>

<form name="myform" action="<?php echo $_SERVER['$PHP_SELF']; ?>" method="POST">
    <input type="hidden" name="hidden1" id="hidden1" />
</form>

<?php
   $salarieid = $_POST['hidden1'];
   $query = "select * from salarie where salarieid = ".$salarieid;
   echo $query;
   $result = mysql_query($query);
?> …

可能重复:
如何将JavaScript变量传递给PHP？

如何在这个php变量上分配asign javasript值

<script type="text/javascript">

var var_example = "2222";

</script>

<?php 
  echo $var_example_php =   ?        

  /// How to assign on this php variable asign javasript value 
?>