Question

我有一个div也在同一页面上我有一些数据显示在网页上我只想在div上显示数据我该怎么办？我是非常非常新的php。谢谢这里是php文件。

    <?php
    define('HOST','xxxxxxxxxxx');
    define('USER','xxxxxxxxxxxxx');
    define('PASS','xxxxxxxx');
    define('DB','androidapi2');

    $con = mysqli_connect(HOST,USER,PASS,DB);

    $sql = "select * from users WHERE status = 1";

    $res = mysqli_query($con,$sql);
    $result = array();

        while($row = mysqli_fetch_array($res)){
        array_push($result,
        array(
         //  'id'=>$row[0],
          'email'=>$row[3], 
    ));
    } 
    echo json_encode(array("result"=>$result));
    mysqli_close($con);

    ?>

    <html>
<head>

<title>DIC ChatBox Beta 1</title>
<style>
#usersOnLine { 
    font-family:tahoma;
    font-size:12px;
    color:black;
    border: 3px teal solid;
    height: 525px;
    width: 250px;
    float: right;
 overflow-y:scroll;
}

.container{
    width:970px;
    height:auto;
    margin:0 auto;
}
</style>
</head>
<body>
<div class="container"> 
</div>
<h2> all contacts</h2>
<div id="usersOnLine">
</div>
</div>
</body>
</html>

这里是div id usersonline，我想要显示所有数据。如何制作它。

Answer 1

无需使用json_encode(array("result"=>$result));，直接使用$result

<div id="usersOnLine">
<?php
foreach($result as $key => $val)
{
    echo $val['email'];
    echo "<br>";
}
 ?>
</div>

如果您想在其他网页上使用

在php文件中创建新会话

<?php
session_start(); ///at the top of this file 

///after your query and creating array
$_SESSION["result"]= $result;////set $result in session
?>

在另一页上开课

<?php
session_start();

if(isset($_SESSION["result"]))
{
   $result = $_SESSION["result"];
   ///your foreach loop to print data 
}
?>

<强> READ SESSION DOCUMENTAION

php从网上挑选数据并在div中显示

1 个答案: