评论系统 - 创建后,并将其显示在页面上-1 [英] Comment system - Creating post and showing it on the page -1
问题描述
有很多关于网络的评论系统脚本。但我觉得这很混乱。当我们想定制它,它就像空穴。
我要创造我自己的简单的评论框,用户可以发布和评论。
下面是我的空间,进入后: http://jsfiddle.net/karimkhan/FNcs8/
<吊牌>添加后< /吊牌>< BR>
< textarea的行数=4NAME =AddpostCOLS =50占位符=后添加> < / textarea的>
<输入类型=提交值=分享> < /输入>
现在就点击链接,我可以存储在数据库中,但是如何在用户输入后显示出下降的每点击按钮?使用Ajax是jQuery的它曾经高效和易于 我想说明近门柱用户图像。网址是从表中来了。
我从那里数据即将表是:
CREATE TABLE`user_record`(
`ID` VARCHAR(40)NOT NULL,
`name` VARCHAR(50)DEFAULT NULL,
`email`字段为varchar(50)DEFAULT NULL,
`picture` VARCHAR(50),默认为空// URL的图片
PRIMARY KEY(`ID`)
)ENGINE = InnoDB的默认字符集=拉丁文;
家伙,我会把这个整个系统在GitHub上,帮助其他也学习。需要帮助!
更新1: ajax.php - 数据库
< PHP
$ CON = mysqli_connect('127.0.0.1:3306','根','根','测试');
如果(mysqli_connect_errno()){
回声无法连接到MySQL:。 mysqli_connect_error();
}
$查询=SELECT * FROM user_record其中id = 1660546353;
$结果= mysqli_query($ CON,$查询);
//建立html格式,你需要它..
而($行= mysqli_fetch_array($结果)){
回声'< DIV CLASS =意见>'$行['名称']'< IMG SRC =。$行['图片']'/>< / DIV> ;
}
?>
如果您正在使用jquery然后用这个code
HTML
<标签>加入后< /标签>< BR>
< textarea的ID =消息行=4COLS =50占位符=后添加> < / textarea的>
<输入类型=提交ID =提交值=分享> < /输入>
< DIV ID =commentsholder>< / DIV>
Javascript的
<脚本类型=文/ JavaScript的>
$(文件)。就绪(函数(){
$('#提交')。在('点击',函数(){
。VAR commentdata = $(信息)VAL();
$阿贾克斯({
键入:POST,
数据:{
评论:commentdata
},
网址:ajax.php
成功:功能(数据,textStatus){
//警报(数据);
$(#commentsholder)追加(数据);
}
},HTML);
});
});
< / SCRIPT>
在ajax.php
//插入注释到数据库中。
//得到谁是张贴的用户内容。
< PHP
$ CON = mysqli_connect('127.0.0.1:3306','根','根','测试');
如果(mysqli_connect_errno()){
回声无法连接到MySQL:。 mysqli_connect_error();
}
$查询=SELECT * FROM user_record其中id = 1660546353;
$结果= mysqli_query($ CON,$查询);
//建立html格式,你需要它..
而($行= mysqli_fetch_array($结果)){
回声'< DIV CLASS =意见>'$行['名称']'< IMG SRC =。$行['图片']'/>< / DIV> ;
}
?>
这就是它..它应该工作
There are plenty for scripts on web for comment system. But I felt it very confusing. When we want to customize it, it's like as hole.
I want to create my own simple comment box where user can post and comment.
Here is my space to enter post: http://jsfiddle.net/karimkhan/FNcs8/
<lable>Add post </lable><br>
<textarea rows="4" name="Addpost" cols="50" placeholder="Add post"> </textarea>
<input type="submit" value="share"> </input>
Now on button click I can store in the database but how to show in downward each when user enters post and click button? using ajax is jquery which ever efficient and easy I want to show user image near post. Url is coming from table below.
My table from where data is coming is:
CREATE TABLE `user_record` (
`id` varchar(40) NOT NULL,
`name` varchar(50) DEFAULT NULL,
`email` varchar(50) DEFAULT NULL,
`picture` varchar(50) default NULL //url for picture
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
Guys i will put this entire system on github which help other also to learn. Needs help!
UPDATE 1: ajax.php - for database
<?php
$con = mysqli_connect('127.0.0.1:3306', 'root', 'root', 'test');
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$query= "select * from user_record where id=1660546353";
$result = mysqli_query($con,$query);
//build html format as you need it..
while($row = mysqli_fetch_array($result)){
echo '<div class="comment">'.$row ['name'].'<img src="'.$row ['picture'].'"/></div>';
}
?>
If you are using jquery then use this code
HTML
<label>Add post </label><br>
<textarea id="message" rows="4" cols="50" placeholder="Add post"> </textarea>
<input type="submit" id="submit" value="share"> </input>
<div id="commentsholder"></div>
Javascript
<script type="text/javascript">
$(document).ready(function(){
$('#submit').on('click',function(){
var commentdata=$("message").val();
$.ajax({
type: "POST",
data:{
comment: commentdata
},
url: "ajax.php",
success: function(data, textStatus){
//alert(data);
$("#commentsholder").append(data);
}
},'html');
});
});
</script>
In ajax.php
//insert comment into database.
//get the user content who are posting it.
<?php
$con = mysqli_connect('127.0.0.1:3306', 'root', 'root', 'test');
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$query= "select * from user_record where id=1660546353";
$result = mysqli_query($con,$query);
//build html format as you need it..
while($row = mysqli_fetch_array($result)){
echo '<div class="comment">'.$row ['name'].'<img src="'.$row ['picture'].'"/></div>';
}
?>
That's it.. it should work
这篇关于评论系统 - 创建后,并将其显示在页面上-1的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!