Question

我试图onclick Jquery并发布到msg.php。我的代码有什么问题吗？请让我知道问题所在。谢谢。

HTML

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.2/jquery.min.js"></script>
<script>
function markAsRead(msgID) {
    var url = "msg.php";
    var ownerid = "7";
    var messageid = "msgID";
    $.post(url,{messageid: messageid, ownerid: ownerid });
}
</script>
<a onclick="markAsRead(<?php echo $row['id'];?>)">how are you?</a>

`

msg.php

if (isset($_POST['messageid'])) {
    $messageid = preg_replace('#[^0-9]#i', '', $_POST['messageid']); 
    $ownerid = preg_replace('#[^0-9]#i', '', $_POST['ownerid']);
    mysqli_query("UPDATE private_messages SET opened='1' WHERE id='$messageid'
LIMIT 1");
}
}
else
    echo "There is no post value";

Answer 1

您应该设置$row['id']

的值

<?php
    $row['id'] = 1; // the id is 1 for example
?>

<a onclick="markAsRead(<?php echo $row['id'] ?>)">how are you?</a>

<script>
    function markAsRead(msgID) {
        var url = "msg.php";
        var ownerid = 7; // this should be unquoted
        var messageid = msgID; // this should be unquoted
        $.post(url, {
            messageid: messageid,
            ownerid: ownerid
        }, function () {
            alert('Marked!'); // added this as a callback
        }).error(function(){
            alert('Failed to mark.'); // as an error callback
        });
    }
</script>

PHP代码

<?php
    if (isset($_POST['messageid']))
    {
        $messageid = intval($_POST['messageid']); // seems your replacing numbers
        $ownerid = intval($_POST['ownerid']);  // seems your replacing numbers

        mysqli_query("UPDATE private_messages SET opened = 1 WHERE id = $messageid LIMIT 1");
    }
    else
    {
        echo "There is no post value";
    }
?>

Jquery帖子和onclick

1 个答案: