显示不同的按钮或刷新

Question

我一直在尝试为论坛帖子制作一个喜欢/不喜欢的按钮，我已经取得了一些成功，下面的代码可以正常工作，但我无法刷新页面。

例如 - 当我单击提交按钮（喜欢）时，数据库已更新，但未显示不同的按钮。如果我刷新页面，现在有一个不同的按钮，那么现在有一种方法可以用 jquery 来做到这一点，所以一旦点击了类似的按钮，它就会显示按钮不像

/* Check Database */

$likes = DB::run("SELECT * FROM likes WHERE id=? AND lid=? AND uid=? AND pid=?", [1, 1, $CURUSER['id'], $row['id']])->fetch();

/* Check Result */

var_export($likes);

/* If No Result Like */

if (!$likes) {

echo

"<form action='' method='post'>

<input type='submit' name='like' value='Like' />

</form>";

if(isset($_POST['like']))

{

    DB::run("INSERT INTO likes (id, lid, uid, pid) VALUES (?, ?, ?, ?)", [1, 1, $CURUSER['id'], $row['id']]);

}

}

/* If Result UnLike */

if ($likes) {

echo

"<form action='' method='post'>

<input type='submit' name='unlike' value='Unlike' />

</form>";

if(isset($_POST['unlike']))

{

DB::run("DELETE FROM likes WHERE id=? AND lid=? AND uid=? AND pid=?", [1, 1, $CURUSER['id'], $row['id']]);

}

}

/* Show Results */

echo '</br>';

echo $likes['id'];

echo $likes['1id'];

echo $likes['uid'];

echo $likes['pid'];

Answer 1

表单显示后数据库被修改，因此新的“喜欢”状态不会反映在页面上。一种解决方案是在检查“喜欢”状态并显示表单之前修改数据库。下面，我使用 PHP 的header.&nbsp;刷新页面不是绝对必要的，但它确实有助于防止用户刷新时重复提交。这称为Post/Redirect/Get 模式。Post/Redirect/Get (PRG) 是一种 Web 开发设计模式，它允许重新加载、共享或添加表单提交后显示的页面而不会产生不良影响，例如再次提交表单。if(isset($_POST['like'])) {&nbsp; &nbsp; DB::run("INSERT INTO likes (id, lid, uid, pid) VALUES (?, ?, ?, ?)", [1, 1, $CURUSER['id'], $row['id']]);&nbsp; &nbsp; header('Location: '.$_SERVER['REQUEST_URI']);&nbsp; &nbsp; exit;} elseif (isset($_POST['unlike'])) {&nbsp; &nbsp; DB::run("DELETE FROM likes WHERE id=? AND lid=? AND uid=? AND pid=?", [1, 1, $CURUSER['id'], $row['id']]);&nbsp; &nbsp; header('Location: '.$_SERVER['REQUEST_URI']);&nbsp; &nbsp; exit;}/* Check Database */$likes = DB::run("SELECT * FROM likes WHERE id=? AND lid=? AND uid=? AND pid=?", [1, 1, $CURUSER['id'], $row['id']])->fetch();if (!$likes) {&nbsp; &nbsp; /* If No Result Like */&nbsp; &nbsp; echo "<form action='' method='post'>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; <input type='submit' name='like' value='Like' />&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; </form>";} else {&nbsp; &nbsp; /* If Result UnLike */&nbsp; &nbsp; echo "<form action='' method='post'>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; <input type='submit' name='unlike' value='Unlike' />&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; </form>";}

Answer 2

没有人可以帮助你，因为我们看不到你的$likes结果，这里是伪代码，它是如何工作的，在哪里$likes硬编码为 false 开始（不喜欢），当你点击喜欢时，会显示不同按钮。在你得到这个之后，你应该调试$likes每个时刻包含的内容以及为什么它没有进入你的 if 语句：<?php/* Check Database */$likes = false;/* Check Result */var_export($likes);/* If No Result Like */if (!$likes) {echo"<form action='' method='post'><input type='submit' name='like' value='Like' /></form>";if(isset($_POST['like'])){&nbsp; &nbsp; $likes = true;}}/* If Result UnLike */if ($likes) {echo"<form action='' method='post'><input type='submit' name='unlike' value='Unlike' /></form>";if(isset($_POST['unlike'])){&nbsp; &nbsp; $likes = false;}}/* Show Results */