如何将 JavaScript 变量传递给 PHP？

php javascript variables

2022-08-30 02:03:06

我想使用表单中的隐藏输入将JavaScript变量传递给PHP。

但是我无法将的价值放入 .有什么问题吗？$_POST['hidden1']$salarieid

代码如下：

<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);
?>

<table>
   Code for displaying the query result.
</table>

答案 1

您无法将变量值从当前页面 JavaScript 代码传递到当前页面 PHP 代码...PHP代码在服务器端运行，它对客户端发生的事情一无所知。

您需要使用其他机制将变量从 HTML 表单传递到 PHP 代码，例如使用 GET 或 POST 方法提交表单。

<DOCTYPE html>
<html>
  <head>
    <title>My Test Form</title>
  </head>

  <body>
    <form method="POST">
      <p>Please, choose the salary id to proceed result:</p>
      <p>
        <label for="salarieids">SalarieID:</label>
        <?php
          $query = "SELECT * FROM salarie";
          $result = mysql_query($query);
          if ($result) :
        ?>
        <select id="salarieids" name="salarieid">
          <?php
            while ($row = mysql_fetch_assoc($result)) {
              echo '<option value="', $row['salaried'], '">', $row['salaried'], '</option>'; //between <option></option> tags you can output something more human-friendly (like $row['name'], if table "salaried" have one)
            }
          ?>
        </select>
        <?php endif ?>
      </p>
      <p>
        <input type="submit" value="Sumbit my choice"/>
      </p>
    </form>

    <?php if isset($_POST['salaried']) : ?>
      <?php
        $query = "SELECT * FROM salarie WHERE salarieid = " . $_POST['salarieid'];
        $result = mysql_query($query);
        if ($result) :
      ?>
        <table>
          <?php
            while ($row = mysql_fetch_assoc($result)) {
              echo '<tr>';
              echo '<td>', $row['salaried'], '</td><td>', $row['bla-bla-bla'], '</td>' ...; // and others
              echo '</tr>';
            }
          ?>
        </table>
      <?php endif?>
    <?php endif ?>
  </body>
</html>

答案 2

只需将其保存在饼干中：

$(document).ready(function () {
  createCookie("height", $(window).height(), "10");
});

function createCookie(name, value, days) {
  var expires;
  if (days) {
    var date = new Date();
    date.setTime(date.getTime() + (days * 24 * 60 * 60 * 1000));
    expires = "; expires=" + date.toGMTString();
  }
  else {
    expires = "";
  }
  document.cookie = escape(name) + "=" + escape(value) + expires + "; path=/";
}

然后用PHP阅读它：

<?PHP
   $_COOKIE["height"];
?>

这不是一个漂亮的解决方案，但它有效。