如何将 JavaScript 变量传递给 PHP? [英] How do I pass JavaScript variables to PHP?
问题描述
我想使用表单中的隐藏输入将 JavaScript 变量传递给 PHP.
I want to pass JavaScript variables to PHP using a hidden input in a form.
但我无法将 $_POST['hidden1'] 的值转换为 $salarieid.有什么问题吗?
But I can't get the value of $_POST['hidden1'] into $salarieid. Is there something wrong?
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>
推荐答案
您不能将变量值从当前页面 JavaScript 代码传递到当前页面 PHP 代码... PHP 代码运行在服务器端，它不会对客户端发生的事情一无所知.
You cannot pass variable values from the current page JavaScript code to the current page PHP code... PHP code runs at the server side, and it doesn't know anything about what is going on on the client side.
您需要使用另一种机制将变量从 HTML 表单传递给 PHP 代码，例如使用 GET 或 POST 方法提交表单.
You need to pass variables to PHP code from the HTML form using another mechanism, such as submitting the form using the GET or POST methods.
<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>
这篇关于如何将 JavaScript 变量传递给 PHP?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
