Question

我想在javascript中阅读网址的内容。该URL不在我的域中，因此我需要一个可以访问跨域的中间层。我尝试使用PHP函数来读取URL并使用jquery将结果返回给javascript，但它不起作用。

这是我的试用版：

我创建了一个名为“phpjs_test.php”的php文件

<?php
function get_data(){
  $url='http://asmary.dreameg.com/texttable.txt';
  $ch = curl_init($url);
  curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
  curl_setopt($ch, CURLOPT_BINARYTRANSFER, true);
  $content = curl_exec($ch);
  $content = htmlspecialchars($content);
  curl_close($ch);
  $content = nl2br($content);
  return $content;
}
?>

这是javascript代码：

<script>
        $(document).ready(function () {
            //httpQuery("http://asmary.dreameg.com/texttable.txt");
            getOutput();
        });

        function getRequest() {
            var req = false;
            try {
                // most browsers
                req = new XMLHttpRequest();
            } catch (e) {
                // IE
                try {
                    req = new ActiveXObject("Msxml2.XMLHTTP");
                } catch (e) {
                    // try an older version
                    try {
                        req = new ActiveXObject("Microsoft.XMLHTTP");
                    } catch (e) {
                        return false;
                    }
                }
            }
            return req;
        }

        function getOutput() {
            var ajax = getRequest();
            ajax.onreadystatechange = function() {
                if (ajax.readyState == 4) {
                    document.getElementById('output').innerHTML = ajax.responseText;
                }
            };
            ajax.open("GET", "phpjs_test.php", true);
            ajax.send(null);
        }

我是PHP的新手，所以我甚至不知道PHP函数是否正确。

Answer 1

你应该只使用jQuery ajax方法而不是创建XMLHTTPRequest，你不必费心为IE添加更多代码，而且你已经加载了jQuery库。此外，如果您在PHP文件中将标头设置为Allow-Origin-Access并指定您要求的其他域，那么您可以发出AJAX请求并获得响应，否则它将不返回任何内容或在您的开发工具网络选项卡中将显示403 - 禁止。

Access-Control-Allow-Origin syntax

Answer 2

将PHP文件更改为：

<?php
$url='http://asmary.dreameg.com/texttable.txt';
$ch = curl_init($url);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
$content = curl_exec($ch);
curl_close($ch);
$content = explode('~',$content);

foreach($content as $c)
{
    $records[] = explode('|',$c);
}
$content = json_encode($records);
echo $content;
?>

Javascript将收到像这样的json数组

[["1","name1","10","city1"],["2","name2","20","city2"],["3","name3","30","city3"],["4","name4","40","city4"],["5","name5","50","city5"],["6","name6","60","city6"],["7","name7","7","city7"],["8","name8","80","city8"],["9","name9","90","city9"],["10","name10","100","city10"],["11","name11","11","city11"],["12","name12","12","city12"],["13","name13","13","city13"],["14","name14","14","city14"],["15","name15","15","city15"],["16","name16","16","city16"],["17","name17","17","city17"],["18","name18","18","city18"],["19","name19","19","city19"],["20","name20","20","city20"],[""]]

使用php在javascript中读取URL内容

2 个答案: