我遇到的问题是我的onclick事件正在处理img但不在标签上。 我在php和jquery中创建了一个follow和unfollow函数。它没有工作,所以我决定在点击跟随或取消关注按钮后测试警报。然后它也无法正常工作。 代码就是这个
<?php
if ($my_id != $id) {
$check_follow_not = mysqli_query($conn, "SELECT * FROM follower WHERE user_id='$my_id' AND following='$id'");
$check_follow_not_rows = mysqli_num_rows($check_follow_not);
if ($check_follow_not_rows == 0) {
$editfollowunfollow = "<span class='btn_f_uf'><a onclick=\"don();\">Follow</a></span>";
} else {
$editfollowunfollow = "<span class='btn_f_uf'><a onclick=\"don();\">UnFollow</a></span>";
}
} else {
$editfollowunfollow = "<span class='btn_f_uf'><a>Edit Profile</a></span>";
}
echo $editfollowunfollow;
?>
不要去PHP代码它工作正常它是检查登录用户是否跟随另一个用户以下。问题是我已经插入了无效的onclick功能。
don()函数就是这个
function don(){
alert();
}
此功能在同一文件中的图像上正常工作。但是没有在跟随和取消关注中工作。
提前致谢。我感谢所有答案
修改
btn_f_uf类的Css代码
.btn_f_uf{
border:0;
outline:0;
padding:10px;
background:#53a93f;
border-radius:5px;
color:white;
float:right;
margin-right:20px;
margin-top:25px;
display:inline-block;
}
编辑2
个人资料页面整码
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8"></meta>
<title>Profcee</title>
<link rel="stylesheet" href="style/style.css">
<script src="js/jquery-2.1.3.js"></script>
<script src="js/check.js"></script>
<script src="js/jquery-ui.js"></script>
<script src="js/jquery-ui.min.js"></script>
</head>
<?php
session_start();
if(isset($_GET['id'])){
include "function/checklogin.php";
include "config/connect.php";
$my_profile_image = $_SESSION['profile_image'];
$my_email = $_SESSION['email'];
$my_id = $_SESSION['id'] ;
$my_name = ucfirst($_SESSION['name']);
$my_background_image = $_SESSION['background_image'];
$id = $_GET['id'];
$check = mysqli_query($conn,"SELECT * FROM users WHERE id='$id'");
$check_num = mysqli_num_rows($check);
if($check_num == 0){
header("location:home.php");
}else{
$fetch = mysqli_fetch_assoc($check);
$name = ucfirst($fetch['name']);
$email = $fetch['email'];
$profile_image = $fetch['profile'];
$background_image = $fetch['background'];
if($my_id != $id){
$check_follow_not = mysqli_query($conn,"SELECT * FROM follower WHERE user_id='$my_id' AND following='$id'");
$check_follow_not_rows = mysqli_num_rows($check_follow_not);
if($check_follow_not_rows == 0){
$editfollowunfollow = "<span class='btn_f_uf'><a class='btns'>Follow</a></span>";
} else{
$editfollowunfollow = "<span class='btn_f_uf'><a class='btns'>UnFollow</a></span>";
}
}else{
$editfollowunfollow = "<span class='btn_f_uf'><a>Edit Profile</a></span>";
}
?>
<?php
include "header.php"; ?>
<div id="profile_full">
<div id="background_image">
<img src="<?php echo $background_image;?>" width="100%" height="400px" onclick="don();">
</div>
<div id="profile_image">
<img src="<?php echo $profile_image?>" width="250px" height="250px">
</div>
<div id="profile_header">
<h1><?php echo $name;?></h1>
<div id="profile_follower">
<?php
$follower = mysqli_query($conn,"SELECT * FROM follower WHERE following='$id'");
$follower_count = mysqli_num_rows($follower);
?>
<h2><?php echo $follower_count;?> Follower</h2>
</div>
<div id="profile_following">
<?php
$following = mysqli_query($conn,"SELECT * FROM follower WHERE user_id='$id'");
$following_count = mysqli_num_rows($following);
?>
<h2><?php echo $following_count;?> following </h2>
</div>
<?php echo $editfollowunfollow;?>
</div>
</div>
<?php
include "footer.php";
?>
<?php
}
}else{
header("location:home.php");
}
?>
check.js文件
$(document).on("click",".btns", function(e){
e.preventDefault();
alert();
});
答案 0 :(得分:0)
像其他人说的那样,给它类名btns
并删除该锚标记上的onclick
内联javascript:
.....
if($check_follow_not_rows == 0){
$editfollowunfollow = "<span class='btn_f_uf'><a class='btns'>Follow</a></span>";
} else{
$editfollowunfollow = "<span class='btn_f_uf'><a class='btns'>UnFollow</a></span>";
}
....
在js代码中:
$(document).on("click",".btns", function(e){
e.preventDefault();
alert();
});
如果你想坚持function don()
,请使用此(但必须确保删除onclick内联javascript ):
$(document).on("click",".btns", don );
在don
函数中:
function don(){
alert();
return false;
}
答案 1 :(得分:0)
是的,它已经完成了。问题在于div的定位。所以我从头开始