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<?phpinclude_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>
1 回答
ITMISS
TA贡献1871条经验 获得超8个赞
您需要将用户 ID 放入表单中，并在插入时使用。
此外，表格需要在里面<body>，而不是在它之后。
<?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>
<?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'];
?>
<form action ="" method="post">
<button value = "<?php echo $ide; ?>" name="like">like</button>
</form>
<?php
}
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