让 file_get_contents 只返回我想要的

Question

我有这个代码。我希望在每个职位上都可以访问。但我得到的结果是字符串。这是一个例子：

"string(7652)"{"id":"34","nume":"STUDENT","bilet":null,"log":"2014-06-16 11:41:28\n2014- 06-16 11:41:30\n2014-06-16 11:41:41\n2014-06-16 11:41:41\n2014-06-16 11:41:43\n2014-06-16 11:42 :03\n2014-06-16 11:42:05\n2014-06-16 11:42:06\n2014-06-16 11:42:06\n2014-06-16 11:42:06\n2014-06 -16 11:42:09\n2014-06-16 11:42:11\n2014-06-16 11:42: "

我只需要日志......所以我可以用这些日志制作一个 heml 表。想法？

public $datte = array();
public function aduDpServer(){
    $data = array ('foo' => 'bar', 'id' => 34); 
    $data = http_build_query($data); 
    $context_options = array ( 
        /*'http' => array ( 
            'method' => 'POST', 
            'header'=> "Content-type: application/x-www-form-urlencoded\r\n" 
            . "Content-Length: " . strlen($data) . "\r\n", 
            'content' => $data 
        ) */
        'http'=> array(
            'method'=>'POST',
            'header'=>"Content-type: application/x-www-form-urlencoded\r\n",
            'content'=>$data
        )
    ); 
    $context = stream_context_create($context_options);
    $this->datte = file_get_contents('http://www.jajaja/x/updt.php', false, $context); 

    //var_dump($fp);

}

public function afiseaza(){
    //print_r($this->datte);
    echo"<table border=1 cellpading=0 cellspacing=0 >";
    echo"<th>Accesari</th>";
    $this->datte = is_array($this->datte) ? $this->datte: array($this->datte); //fortare vector
    foreach ($this->datte as $d) {
        echo"<tr>";
        echo"<td>".$d."</td>";echo"<br>";
        echo"</tr>";

    }

}

Answer 1

您的字符串是 JSON，因此您可以使用 json_decode 将您的数据转换为 php 数组。

$this->datte = json_decode(file_get_contents('http://www.jajaja/x/updt.php', false, $context), true);