下面的代码是postar.php
include "conexao.php";
$consulta = mysqli_query($conexao, "SELECT titus.titus, titus.des, titus.link from titus");
while($postagem = mysqli_fetch_object($consulta));
echo "<div> $postagem-></div>
<div>
$postagem->
$postagem->$postagem-></div>"下面这个是podcast.php
<!-- Button -->
<form method="post" action="postar.php">
<button type="submit" class="pos" >
publish</button>
</form>我需要单击publish按钮,以便它包含来自mysql数据库的最新信息,并在帖子中返回它。
在postar.php中有一个很小的php代码,它的目的是替换掉那些提图斯、des和link的信息。
示例:
titus is sun
des is moon
link is land(数据库中的数据)
所以它应该看起来像这样:
sun
moon
land但是当我运行代码时,只有箭头出现。我想知道如何解决这个问题,并将数据显示为帖子。
下面包含数据库连接代码,即conexao.php
define('HOST', '127.0.0.1');
define('USUARIO', 'root');
define('SENHA', '');
define('DB', 'login');
$conexao = mysqli_connect(HOST, USUARIO, SENHA, DB) or die ('Could not connect');这是数据库表:
CREATE TABLE titus(
idpod INT NOT NULL AUTO_INCREMENT,
titus VARCHAR(200) NOT NULL,
des VARCHAR(400) NOT NULL,
link VARCHAR (300) NOT NULL,
PRIMARY KEY (`idpod`)
);这段代码的目的是发布,当在数据库中添加新记录时,我执行这个发布过程,我知道使用触发器可以做到这一点,我不是一个专业的程序员,但我想理解这个问题,谢谢大家:)
发布于 2020-11-05 03:32:50
以下代码起作用了:
$consulta = mysqli_query($conexao, "SELECT * from titus
where idpod = (select max(idpod) from titus)");
while($postagem = mysqli_fetch_object($consulta)){
echo $postagem = $_POST['titus'] .'<br>';
echo $postagem = $_POST['des'] .'<br>';
echo $postagem = $_POST['link'] .'<br>';发布于 2020-11-04 07:04:47
您应该更改您的代码,如下所示:
$consulta = mysqli_query($conexao, "SELECT titus, des, link from titus");
while($postagem = mysqli_fetch_object($consulta));
if ($postagem) {
echo "<div>titus is $postagem->titus</div>
<div>des is $postagem->des</div>
<div>link is $postagem->link</div>";
} else {
echo "data is empty";
}https://stackoverflow.com/questions/64671784
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