doupa2871 2015-06-28 06:17 采纳率: 100%
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使用ajax和json将数组从php发送到js

I am trying to send an array from php (that I have taken from a mysql table to js). Although there a lot of examples out there I can't seem to make any of them work. The code that I have reached so far is:

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);
        });
     });
   });

This is my first app that I use something like this, so please be a little patient.

  • 写回答

1条回答 默认 最新

  • doulu4413 2015-06-28 06:26
    关注

    Right now you're sending the complete page markup mixed with your json response, which of course will not work.

    For example imagine that you have the following php script which suppose to return a json response:

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

    The response from this page would not be json since it also will return the wrapping div element.

    You can move your php code to the top of the page or simply remove all the 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;
?>
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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