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php - PHP 喜欢系统随机

问题描述

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 lang="en">
<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>

标签： phprandom

解决方案

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 lang="en">
    <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
}

if(isset($_POST['like'])){
    $sql1= "INSERT INTO likes (id, likes)
        VALUES ('1', ?)";
    $stmt = $conn->prepare($sql1);
    $stmt->bind_param("s", $_POST['like']);
    if ($stmt->execute()) {
        echo "New record created successfully";
    } else {
        echo "Error: ".$sql1."<br>". $stmt->error;
    }
}
$conn->close();
?>
</body>
</html>

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