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PHP数组上的json_encode返回字符串

[英]json_encode on a PHP array returns a string

I've searched through the site and can't see a question quite like mine so I hope this isn't a copy. 我已经搜索了该网站,却找不到像我这样的问题,所以我希望这不是一个副本。

So I've got a PHP script which is supposed to return a JSON array to my AJAX, and I want to use the array to generate a URL. 因此,我有一个PHP脚本,该脚本应该将JSON数组返回到我的AJAX,并且我想使用该数组来生成URL。

However, even though I'm pretty sure I've got an array on the PHP end, when I json_encode I'm getting a simple string out on the other end. 但是,即使我非常确定我在PHP端都有一个数组,但是当我json_encode时,我在另一端得到了一个简单的字符串。

PHP Code: PHP代码:

    $n = 10;
    $all_titles = array();

    while($row = mysqli_fetch_array($result)) {
        $title = trim($row['Job Title']);
        if(array_key_exists($title, $all_titles)) {
            ++$all_titles[$title];
        } else {
            $all_titles[$title] = 1;
        }
    }

    arsort($all_titles);
    $top_titles = array_slice($all_titles, 0, $n);
    $title_list = array();

    foreach (array_values($top_titles) as $key => $val) {
        array_push($title_list, array_keys($top_titles)[$key]);
    }
    echo json_encode($title_list);

These array operations seem to be working so far, based on other tests I've done, so I'm pretty sure that $title_list is an array. 根据我所做的其他测试,到目前为止,这些数组操作似乎可以正常运行,因此,我很确定$ title_list是一个数组。

Here is my JS: 这是我的JS:

    xmlhttp.onreadystatechange=function() {
      if (xmlhttp.readyState==4 && xmlhttp.status==200) {
        alert("Generated URL: " + URL_gen(xmlhttp.responseText));
      }
    }

And finally where the problem arises: 最后出现问题的地方:

function URL_gen(list) {
    var url = list[2];
    return url;
}

I have varied the number in list[#] (list[0], list[1], etc.) and each one is a character, meaning list (which is passed in from onreadystatechange as the responsetext from the PHP function above) is a string, not a JSON array. 我已经更改了list [#]中的数字(list [0],list [1]等),每个数字都是一个字符,这意味着列表(从onreadystatechange传递为上述PHP函数的响应文本)是字符串,而不是JSON数组。

Any suggestions? 有什么建议么?

That's what it does. 这就是它的作用。 Returns a string. 返回一个字符串。 You need to parse it on the client side. 您需要在客户端解析它。

alert("Generated URL: " + URL_gen(JSON.parse(xmlhttp.responseText)));

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