json在php中解析响应

Question

我的所有JSON数据都是NULL 这是我的代码

$jsons1 = file_get_contents($api_url); // returns JSON
        $jsons=json_decode($jsons1);

        foreach ($jsons as $json)
        {
        echo "\n";
        echo "<ul>";
        echo "<li>";
        echo "\nTitle : ".$json->title."\n";
        echo "\nContent: ".$json->content."\n";
        echo "</li>";
        echo "</ul>";
        }

我无法像$ json-＆gt;标题那样获取它，我错过了什么？

输出

 - Title : Content: 

请在这方面提供帮助

这里是$jsons1变量

的输出

{"results":[{"title": "Temp, Russia - Wikipedia, the free encyclopedia","kwic": "Coordinates : 52°03′N 39°44′E  /  52.05°N 39.733°E  / 52.05; 39.733  tenmp ( Russian : УÑмань ) is a town and the administrative center of ...","content": "","url": "http://en.wikipedia.org/wiki/sdfds,_Russia","iurl": "","domain": "en.wikipedia.org","author": "","news": false,"votes": "1","date": 1357744155518,"related":[]}],"query": "tenp","suggestions":[],"count":96,"start":1,"length":10,"time": "412"}

Answer 1

尝试

$j = '{"results":[{"title": "Temp, Russia - Wikipedia, the free encyclopedia","kwic": "Coordinates : 52°03′N 39°44′E  /  52.05°N 39.733°E  / 52.05; 39.733  tenmp ( Russian : УÑмань ) is a town and the administrative center of ...","content": "","url": "http://en.wikipedia.org/wiki/sdfds,_Russia","iurl": "","domain": "en.wikipedia.org","author": "","news": false,"votes": "1","date": 1357744155518,"related":[]}],"query": "tenp","suggestions":[],"count":96,"start":1,"length":10,"time": "412"}';

$data = json_decode($j,true);

foreach ($data["results"] as $key=>$val){
    echo "\n";
    echo "<ul>";
    echo "<li>";
    echo "\nTitle : ".$val["title"]."\n";
    echo "\nContent: ".$val["content"]."\n";
    echo "</li>";
    echo "</ul>";
}

目前结果元素中只有一个数据，因此如果有更多内容，则循环更好。

Answer 2

您有包含数据的对象结果..

所以你只需要修复你的Foreach声明..＆gt;

    foreach ($jsons->results as $json) { ... }