Question

我是初学者，我想写这个功能。但是当我尝试运行它时，为什么视图没有定义，尽管我已经定义了它。任何人都可以给我解决这个问题的方法吗？这是完整的源代码

<?php
    include ("connection.php");
    session_start();
    $user_login = $_SESSION['user'];
?>
<html>
    <head>
    <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
    <!--Let browser know website is optimized for mobile-->
    <meta name="viewport" content="width=device-width, initial-scale=1.0"/>
    <title>Messenger</title>
    <!--Import Google Icon Font-->
    <link href="http://fonts.googleapis.com/icon?family=Material+Icons" rel="stylesheet">
    <link rel="stylesheet" href="font-awesome-4.7.0\css\font-awesome.css">
    <!--Import materialize.css-->
    <link rel="stylesheet" href="materialize/css/materialize.min.css"  media="screen,projection"/>
    <!--Import jQuery before materialize.js-->
    <script src='//cdnjs.cloudflare.com/ajax/libs/jquery/2.1.3/jquery.min.js'></script>
    <!-- Compiled and minified JavaScript -->
    <script src="https://cdnjs.cloudflare.com/ajax/libs/materialize/0.97.8/js/materialize.min.js"></script>
    <script type="text/javascript" src="script.js"></script>
    <script type="text/javascript">
    function viewProfile($name){
        var name ='test';
        alert(name);
    }
    </script>
</head>
<body class="blue">
<div id="chat-page">
    <div id="chatbox" class="row container">
    <?php
    $select = mysqli_query($conn, "SELECT * FROM user WHERE Username != '$user_login'");
    $row = mysqli_num_rows($select);
    //echo $row_1;
        if($row != 0){
            while($list= mysqli_fetch_assoc($select)){
            $name = $list['Username'];
            echo $name;
            $userFriend = mysqli_query($conn, "SELECT * FROM user WHERE Username = '$name'");
            $fetch = mysqli_fetch_assoc($userFriend);
            echo "<div class='friend'>";
            echo "<a href='#' onclick=\"viewProfile($name)\" style='text-decoration:none'>";
            if($fetch['Display Picture'])
                $pic = 'avatar.jpg';
            echo "<img src='".$pic."' alt='' class='circle responsive-img'>";
            echo "<p>";
            echo "<strong>".$fetch['Username']."</strong><br>";
            echo "<span>".$fetch['Email']."</span>";
            echo "</p>";
            echo "</div>";
            echo "</a>";
            }
        }else
        echo "username not found";
    ?>
    </div>
</div>
</body>
</html>

Answer 1

您需要将 view（）函数放在脚本标记中。还要将javascript放在页面底部。

尝试下面的代码：

<?php
echo "<div class='friend'>";
echo "<a href='#' onclick=\"view()\" style='text-decoration:none'>Click here </a>";
echo "</div>";
?>


<script>
    function view(){
        var name = 'Function Test';
        alert(name);
    }
    </script>

Answer 2

如果php文件中有函数，则添加<script> // js code </script>

像这样使用：

＆＃13;

function view(){
    var name = 'Test';
    alert(name);
}

＆＃13;

<a href='#' onclick="view()" style='text-decoration:none'>Click Here</a>

＆＃13;

Answer 3

在php代码之后将你的javascript放在底部，$name无法识别它的存在， PHP总是在javascript 之前评估，在你的视图函数中你有php变量$name

<script type="text/javascript">   
  function view(){
    var name = "<?php echo $name; ?>";
    alert(name);
  }
</script>

参考错误：未定义视图

3 个答案: