[英]Post and Get value by jquery ajax
How can I add my values owdid and visited id after clicking below link by ajax? 单击ajax下面的链接后,如何添加我的值owdid和Visited ID?
<a href="index.php" onclick="insertvisit(<?php echo $interestid;?>)">member1</a>
Below is my insertvisit function. 以下是我的insertvisit函数。 I have defined owdid and interestid
我已经定义了owdid和interestid
function insertvisit(visitedid) {
$.ajax({
type: "POST",
url: 'insertvisit.php',
data:{'field1' : owdid, 'field2' : visitedid},
success:function(html) {
}
});
}
and below is insertvisit.php 下面是insertvisit.php
global $pdo;
$ownid = $_GET['field1'];
$interestid =$_GET['field2'];
$query = $pdo->prepare("UPDATE tablem SET field1= ? WHERE field2= ?");
$query -> bindValue(1, $ownid);
$query -> bindValue(2, $interestid);
$query -> execute();
Please help thanks. 请帮忙谢谢。
You need to pass both value in function with , separated and also you need to change your function call like bellow 您需要在函数中使用,分隔传递两个值,还需要像下面这样更改函数调用
<a href="index.php" onclick="insertvisit(<?php echo $interestid.','.$owdid;?>)">member1</a>
And your function : 和你的功能:
function insertvisit(visitedid,owdid) {
$.ajax({
type: "POST",
url: 'insertvisit.php',
data:{'field1' : owdid, 'field2' : visitedid},
success:function(html) {
}
});
}
and also you need to change your method $_GET to $_POST like below 并且您还需要将方法$ _GET更改为$ _POST,如下所示
$ownid = $_POST['field1'];
$interestid =$_POST['field2'];
$query = $pdo->prepare("UPDATE tablem SET field1= ? WHERE field2= ?");
$query -> bindValue(1, $ownid);
$query -> bindValue(2, $interestid);
$query -> execute();
i hope this will help you. 我希望这能帮到您。
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.