PHP Facebook JSON转换工具到csv
我试图将Facebook上的数据输入电子表格,我使用的代码如下,但我遇到了一些问题: PHP警告:为第27行/opt/lampp/htdocs/xampp/tratajson.PHP中的foreach()提供的参数无效 这是我的密码:PHP Facebook JSON转换工具到csv,php,json,facebook,facebook-graph-api,csv,Php,Json,Facebook,Facebook Graph Api,Csv,我试图将Facebook上的数据输入电子表格,我使用的代码如下,但我遇到了一些问题: PHP警告:为第27行/opt/lampp/htdocs/xampp/tratajson.PHP中的foreach()提供的参数无效 这是我的密码: <?php $arquivo = fopen('/filepath.csv','w'); $request_url ="/filepath.json"; $requests = file_get_contents($request_url);
<?php
$arquivo = fopen('/filepath.csv','w');
$request_url ="/filepath.json";
$requests = file_get_contents($request_url);
$fb_response = json_decode($requests);
foreach ($fb_response->feed->data as $item)
{
$mensagem=trim($item->message);
$mensagem = str_replace(array("\r\n", "\r", "\n","(",")","chr(13)","\t", "\0", "\x0B"), "", $mensagem);
$quempostou=$item->from->name;
$data=$item->created_time;
$ncomentarios=0;
$nlikes=0;
foreach ($item->comments as $comentarios)
{
$ncomentarios++;
foreach ($comentarios as $comentario)
{
echo "Quem comentou:";
echo $comentario->from->name;
$comentou=$comentario->from->name;
echo " Curtir:";
echo $comentario->like_count;
$curtidas=$comentario->likecount;
$nlikes=$nlikes+$curtidas;
echo " Data:";
$datacomentario = $comentario->created_time;
echo $datacomentario;
echo " Comentario:";
$comentario=$comentario->message;
$comentario = str_replace(array("\r\n", "\r", "\n","(",")"), "", $comentario);
echo $comentario;
$grava="COMENTARIO:".";".$comentario.";".$comentou.";".$datacomentario.";".$curtidas."\n";
fwrite($arquivo,$grava);
echo "<br>";
}
}
echo "Quem postou: $quempostou <br>";
echo "Mensagem: $mensagem <br>";
echo "Data: $data <br>";
echo "Comentarios: $ncomentarios <br>";
echo "Likes na thread: $nlikes <br>";
echo "<br><br>";
$grava="POST:".";".$mensagem.";".$quempostou.";".$data.";".$ncomentarios.";".$nlikes."\n\n";
fwrite($arquivo,$grava);
}
fclose($arquivo);
?>
$request_url中请求的文件具有在页面提要的FB Graph API浏览器中返回的JSON文件的结构:
有人能帮我吗?尝试更改此
$fb\u response=json\u decode($requests)
进入这个$fb\u response=json\u decode($requests,true)代码>你能发布示例JSON吗?你可以在这里看到@AmalMurali:通过这个更改,我得到以下结果:-PHP注意:在第10行的/opt/lampp/htdocs/xampp/tratajson.PHP中尝试获取非对象的属性-PHP注意:在第10行的/opt/lampp/htdocs/xampp/tratajson.PHP中尝试获取非对象的属性-PHP警告:为第10行/opt/lampp/htdocs/xampp/tratajson.php中的foreach()提供的参数无效,在我获取行之前,不返回电子表格中的任何行,但有一些错误,如制表和在“喜欢”和“评论”中计算错误