繁体   English   中英

使用ajax和json将数组从php发送到js

[英]Send array from php to js with ajax and json

我正在尝试从php发送一个数组(我已经从mysql表中获取了一个数组到js)。 尽管那里有很多示例,但我似乎无法使其中的任何一个正常工作。 到目前为止,我到达的代码是:

php_side.php

<!DOCTYPE html>
<html>
<body>

<?php
//$q = intval($_GET['q']);
header("Content-type: text/javascript");

$con = mysqli_connect("localhost","root","","Tileiatriki"); 
if (!$con) {
    die('Could not connect: ' . mysqli_error($con));
}

//mysqli_select_db($con,"users_in_calls");
$sql="SELECT * FROM users_in_calls";
$result = mysqli_query($con,$sql);


/*while($row = mysqli_fetch_array($result)) {
     echo $row['User1_number'];
     echo "<br/>";
     echo $row['User2_number'];
         echo "<br/>";
     echo $row['canvas_channel'];
         echo "<br/>";
}*/
echo json_encode($result);

    mysqli_close($con);
    ?>
    </body>
    </html>  

test_ajax.html

    $(document).ready(function(){
      $.getJSON('php_side.php', function(data) {
        $(data).each(function(key, value) {
            // Will alert 1, 2 and 3
            alert(value);
        });
     });
   });

这是我使用的第一个应用程序,因此,请耐心等待。

现在,您将发送完整的页面标记以及json响应,这当然是行不通的。

例如,假设您有以下php脚本,它们假定返回json响应:

<div><?php print json_encode(array('domain' => 'example.com')); ?></div>

此页面的响应将不是json,因为它还将返回包装的div元素。

您可以将您的php代码移到页面顶部,或者直接删除所有html:

<?php
 // uncomment the following two lines to get see any errors
 // ini_set('display_errors', 1);
 // error_reporting(E_ALL);

 // header can not be called after any output has been done
 // notice that you also should use 'application/json' in this case
 header("Content-type: application/json");

 $con = mysqli_connect("localhost","root","","Tileiatriki"); 
 if (!$con) {
   die('Could not connect: ' . mysqli_error($con));
 }

 $sql="SELECT * FROM users_in_calls";
 $result = mysqli_query($con,$sql);

 // fetch all rows from the result set
 $data = array();
 while($row = mysqli_fetch_array($result)) {
   $data[] = $row;
 }
 mysqli_close($con);

 echo json_encode($data);

 // terminate the script
 exit;
?>

暂无
暂无

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM