Question

我正在尝试从php发送一个数组（我从mysql表中发送到js）。虽然有很多例子，但我似乎无法使其中的任何一个工作。我到目前为止所获得的代码是：

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);
        });
     });
   });

这是我第一个使用此类内容的应用程序，请耐心等待。

Answer 1

现在您正在发送与您的json响应混合的完整页面标记，这当然不起作用。

例如，假设你有以下PHP脚本，它假设返回一个json响应：

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

此页面的响应不是json，因为它也会返回包装div元素。

您可以将您的PHP代码移动到页面顶部或只删除所有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;
?>

使用ajax和json将数组从php发送到js

1 个答案: