如何将JavaScript变量传递给PHP？
如何将JavaScript变量传递给PHP？
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 which the user has chosen
function func_load3(name){
var oForm = document.forms["myform"];
var oSelectBox = oForm.select3;
var iChoice = oSelectBox.selectedIndex;
//alert("you have choosen: " + 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 display the query result.
</table>
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3回答
-
叮当猫咪
您无法将当前页面javascript中的变量值传递给当前页面PHP代码... PHP代码在服务器端运行，并且它不知道客户端上发生了什么。您需要使用其他机制从html-form将变量传递给PHP代码，例如在GET或POST方法上提交表单。<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>', $row['bla-bla-bla'], '</td>' ...; // and others echo '</tr>'; } ?> </table> <?php endif?> <?php endif ?> </body></html> -
鸿蒙传说
只需将其保存在cookie中：$(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读取它<?PHP $_COOKIE["height"];?>它不是一个漂亮的解决方案，但它确实有效。干杯。 -
呼如林
有几种方法可以将变量从javascript传递到php（当然不是当前页面）你可以：以此处所述的形式发送信息，（将导致页面刷新）在ajax中传递它（这里有几个帖子）（没有页面刷新）通过XMLHttpRequest请求（没有页面刷新）发出http请求，如下所示： if (window.XMLHttpRequest){ xmlhttp=new XMLHttpRequest(); }else{ xmlhttp=new ActiveXObject("Microsoft.XMLHTTP"); } var PageToSendTo = "nowitworks.php?"; var MyVariable = "variableData"; var VariablePlaceholder = "variableName="; var UrlToSend = PageToSendTo + VariablePlaceholder + MyVariable; xmlhttp.open("GET", UrlToSend, false); xmlhttp.send();我确信这可以看起来更漂亮并循环遍历所有变量和诸如此类的东西 - 但我保持它基本，以便让新手更容易理解。
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