将php数组作为变量传递给javascript加载url，然后返回php数组? [英] Passing php array as variables to javascript load url and back to php array?

问题描述

我看到许多相关的示例，但我仍然感到困惑.我正在使用ajax(我不太了解)来获取每xxx秒更新一次文件的结果.如果仅传递一个变量，则可以正常工作，但是如果我需要通过php传递数组，那么最好的方法是什么?

I saw many examples related but I'm still confused. I'm using ajax (which I don't know much about) to get the results of a file updated every xxx seconds. It's working perfectly if I pass just one variable, but what is the best way if I need to pass an array from php through it?

结构简单:

show_results.php

<?php
include_once('modWhosonlineCustom.inc.php');
$document->addScript("http://code.jquery.com/jquery-latest.js");
$document->addScript("ajax.js");
$array_name = modWhosonlineCustom::getOnlineUserNames();//the array I need to pass to javascript variable
?>

<script>
var whosonline = '<?php echo "$array_name"; ?>';
</script>

<div id="results"></div>

Ajax代码将在URL加载中构建多个参数:

Ajax code than would have more than one param to build in the url load:

  ajax.js

  $(document).ready(function() {
    $("#results").load("response.php?array_name[param1]&array_name[param2]");
  var refreshId = setInterval(function() {
  $("#results").load("response.php?array_name[param1]&array_name[param2]&randval="+ Math.random());
  }, 10000);
  $.ajaxSetup({ cache: false });
  });

回到PHP响应页面，如何再次使用通过url传递的数组参数?

And back to PHP response page, how could I use again the array params passed through url?

  response.php

  <?php
  $names = $_GET['array_name'];
  foreach ($names as $name) {
  //do something

真的很感谢任何建议，谢谢！

Any suggestions is really appreciated, thanks!

编辑

谢谢大家，我想我现在是正确的方法，但是提出了通过JavaScript中的URL传递此数组的问题.或者，也许我没有以正确的方式在php结束回调文件中得到它.我将向您展示修改后的内容:

Thanks guys, I think I'm the right way now, but ramains the problem to pass this array through a url in the javascript. Or maybe I'm not getting it in the right way in the php end callback file. I'll show you what a modifyed:

   show_results.php

   ...
   <?php
   $names = modWhosonlineCustom::getOnlineUserNames();
   ?>

   <script>
   var whosonline = '<?php echo "json_encode($names)"; ?>';
   </script>


   ajax.js


   $(document).ready(function() {
   $("#atendentes").load("response.php?names=" + whosonline);
   var refreshId = setInterval(function() {
   $("#atendentes").load("response.php?names=" + whosonline + "&randval="+ Math.random());
   }, 10000);
   $.ajaxSetup({ cache: false });
   });


   response.php


   $users = $_GET['names'];
   $users = json_decode($users);
   echo "user: $users";

   $names = $users;
   foreach ($names as $name) {
   ...

在另一面，我得到了:警告:第33行的response.php中为foreach()提供了无效的参数，并且回显为空

Here in the other side I'm getting: Warning: Invalid argument supplied for foreach() in response.php on line 33, and the echo is empty

缺少什么?

推荐答案

您的代码无效.如果有

$arr = array('a', 'b', 'c');
echo $arr;

您实际上得到了

Array

作为输出.不是数组的内容.要将数组从PHP输出"到JS，您必须将其转换为本地Javascript，这是json_encode出现的地方:

as the output. Not the contents of the array. To "output" an array from PHP to JS, you have to convert it to native Javascript, which is where json_encode comes in:

<?php
$arr = array('a', 'b', 'c');
?>
var js_array = <?php echo json_encode($arr) ?>;

会产生

var js_array = ["a","b","c"];

作为一般规则，每当您使用PHP生成JavaScript代码并用PHP值填充Javascript变量时，都应使用json_encode来确保生成有效的Javascript.一旦客户端开始尝试执行它，任何语法错误和整个Javascript代码块都将淹没在水中.

As a general rule, anytime you are using PHP to generate javascript code and are filling in Javascript variables with PHP values, you should use json_encode to ensure that you're generating valid Javascript. Any syntax errors and the whole Javascript code block is dead in the water once the client starts trying to execute it.

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