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.
但是我无法将$_POST['hidden1']的值转换为$salarieid.有什么问题吗?
But I can't get the value of $_POST['hidden1'] into $salarieid. Is there something wrong?
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);
?>
<table>
Code for displaying the query result.
</table>
推荐答案
您不能将变量值从当前页面的JavaScript代码传递到当前页面的PHP代码... PHP代码在服务器端运行，但不会在服务器端运行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.
您需要使用其他机制(例如，使用GET或POST方法提交表单)将变量从HTML表单传递给PHP代码.
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>
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