嘿伙计们,我发现了我的问题,现在我需要帮助如何解决它。我通过 rand 从数据库中获取一个人,单击“喜欢”按钮后,它应该喜欢它显示的当前数字,但事实并非如此。它喜欢单击按钮后显示的下一个数字,我不知道如何更好地显示它。这是我的代码
<?php
include_once 'session.php';
include_once 'dbh.php';
?>
<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>The HTML5 Herald</title>
<meta name="description" content="The HTML5 Herald">
<meta name="author" content="SitePoint">
<link rel="stylesheet" href="cs.css">
</head>
<body>
</body>
</html>
<?php
$sql = "SELECT * FROM users ORDER BY RAND () LIMIT 1 ; ";
$result = mysqli_query($conn, $sql);
$resultCheck = mysqli_num_rows($result);
if ($resultCheck> 0){
$row = mysqli_fetch_assoc($result);
$ide = $row['idUsers'];
echo $ide;
if(isset($_POST['like'])){
$sql1= "INSERT INTO likes (id, likes)
VALUES ('1', '$ide')";
if ($conn->query($sql1) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: ".$sql1."<br>". $conn->error;
}
$conn->close();
}
}
?>
<form action ="" method="post">
<button value = "like" name="like">like</button>
</form>
ITMISS