php字典数组

时间:2014-05-28 11:25:22

标签: php json

我的php中有多个函数,看起来像这样

function linkExtractor($html){ 
    $doc = new DOMDocument(); 
    $last = libxml_use_internal_errors(TRUE); 
    $doc->loadHTML($html); 
    libxml_use_internal_errors($last); 
    $xp = new DOMXPath($doc); 

    $result = array(); 
    foreach ($xp->query("//*[contains(concat(' ', normalize-space(@class), ' '), ' infaa ')]") as $node) 
        $result[] = trim($node->textContent); 
    return $result;     
} 

我正在使用它将它们变成json:

echo json_encode(array("info" => linkExtractor($html),
"dates" => linkExtractor2($html),
"names" => linkExtractor3($html),
"images" => linkExtractor4($html),
"genres" => linkExtractor5($html)
));

但这会像这样返回json:

{
"name":["melter",...],
"date":["05/24/14",...],
"image":["pictu.jpg",...],
"genre":["art",...],
"info":["Lorem ipsum",...]
}

我希望对它们进行批处理,以便将每个结果的第一个放入如下的大括号中:

[ 
{ 
"name": "melter", 
"date": "05/24/14", 
"image": "pictu.jpg",
"genre": "art",
"info": "Lorem ipsum"
},
...
]

我该怎么做?


HTML片段:

<table width="703" border="0" align="center">
<tr>
<th width="697" scope="col">
<div id='gopro-hero-3'>
<a  href="Bits&Bobs/gopro-hero-3.html"><img src="pictu.jpg" alt="" width="700" height="525" class="images  gopro-hero-31" /></a>
</div>
</th>
</tr>
<tr>
<td valign="top" class="type" align="centre">
<table width="100%" border="0" align="right">
<tr>
<th width="48%" class="type genre" scope="col">art</th>
<th width="3%" class="" scope="col"> </th>
<th width="49%" class="date" scope="col">05/24/14</th>
</tr>
</table>
</td>
</tr>
<tr>
<td align="left">
<table width="100%" border="0">
<tr>
<th class="name" align="left" scope="col"><a  class="gopro-hero-31" href="Bits&Bobs/gopro-hero-3.html">melter</a>
</th>
</tr>
<tr>
  <th class="infaa" align="left" scope="col">Lorem ipsum</th>
</tr>
</table>

2 个答案:

答案 0 :(得分:1)

我想linkextractor函数返回的数组长度相同。

$arr = array();

$info = linkExtractor($html);
$dates = linkExtractor2($html);
$names = linkExtractor3($html);
$images = linkExtractor4($html);
$genres = linkExtractor5($html);

for ($i=0; $i<count($info); $i++) {
    $arr[] = array("info" => $info[$i], "date" => $dates[$i], "name" => $names[$i], "image" => $images[$i], "genre" => $genres[$i]);
}

echo json_encode($arr);

答案 1 :(得分:0)

假设每个数组具有相同数量的元素并假设您以相同的顺序提取它们,您可以像这样循环:

$count = count($names);
$result = array();
for ($i=0; $i<$countà; ++$i) {
    $item = new stdClass;
    $item->name = $names[$i];
    $item->date = $dates[$i];
    // ...
    $result[] = $item;
}
echo json_encode($result);

但是,您所做的工作非常低效,您应该重新考虑解析策略,以便一次性提取所有信息。