PHP-解码JSON

我有以下脚本从API获取搜索结果,然后切片并转储它,我无法将JSON解码为数组,它返回Array(0){
}
这是一个 wordpress短代码

以下是从api获取的Json示例：

[
  {
    "barcode": "000015426950",
    "name": "Karen's cowboy",
    "author": "Ann",
    "author_last": "Martin",
    "publisher": "New York : Scholastic, c2000.",
    "year_pub": "",
    "edition": "",
    "genre": "",
    "checkout": "out",
    "series": "",
    "callnum": "MAR",
    "fiction": "true",
    "in_house": "false",
    "timestamp": "1355835387",
    "outto": "000008388615",
    "duedate": "1372005722",
    "ISBN": "059052528X",
    "media_type": "",
    "print": "false",
    "BoxID": "2147483647",
    "uid": "10",
    "printed": ""
  },
  {
    "barcode": "000015426949",
    "name": "Karen's yo-yo",
    "author": "Ann M",
    "author_last": "Martin",
    "publisher": "New York : Scholastic, c2000.",
    "year_pub": "",
    "edition": "",
    "genre": "",
    "checkout": "out",
    "series": "",
    "callnum": "MAR",
    "fiction": "true",
    "in_house": "false",
    "timestamp": "1355835343",
    "outto": "000008388615",
    "duedate": "1373216918",
    "ISBN": "0590525115",
    "media_type": "",
    "print": "false",
    "BoxID": "",
    "uid": "10",
    "printed": ""
  },
...
  }
]

以下是用于获取JSON并对其进行分页的代码：

function book_search_form() {
?>
<form method='get'><input type='text' name='searchvalue' value='<? if (isset($_GET['searchvalue'])) echo $_GET['searchvalue'];?>' />&nbsp;<input type='submit' value='Search' /><input type='hidden' name='pagenum' value='1' /></form>
<br>
<?PHP 
if(isset($_GET['pagenum']) && isset($_GET['searchvalue']))
{
$page=$_GET['pagenum'];
$searchvalue = $_GET['searchvalue'];   // get the value of the page from your url
$recordsPerPage=10; // number of records you want on your page
$api_url = get_option('api_url');
    $api_key = get_option('api_key');
    $data = file_get_contents("$api_url/book/index.PHP/name/$searchvalue?key=2513619352");
    $array = (array)json_decode($data);

$index=($page*$recordsPerPage)-1;
$recordsToBedisplayed = array_slice($array,$index,$recordsPerPage);// this array contains all the records you would want to display on a page;



    $total_pages=ceil(count($array)/$recordsPerPage);
    }
else { 

//use default values

}
?>
<html>...<body><div id="records">
<?
echo '<pre>';
echo $recordsToBedisplayed;
echo '</pre>';?><!-- use the array created above to display records -->
    </div>
<div id="pagination">
<?
for($j=1;$j<=$total_pages;$j++){
                    if($j==$page)
                   {?>

                   <li style="display: inline;"><a href="?searchvalue=&pagenum=<?=$j?>"><u><?=$j?></u></a></li>
                   <?}else{?>
                        <li style="display: inline;"><a href="?searchvalue=&pagenum=<?=$j?>"><?=$j?></a></li>


                   <?}}?>
</div>
<?PHP


}

解决方法:

试试json_decode

$array = json_decode($data, true);

然后你会得到一个数组而不是一个对象.

示例#1 json_decode()示例

<?PHP
$json = '{"a":1,"b":2,"c":3,"d":4,"e":5}';

var_dump(json_decode($json));
var_dump(json_decode($json, true));

?>

上面的例子将输出：

object(stdClass)#1 (5) {
    ["a"] => int(1)
    ["b"] => int(2)
    ["c"] => int(3)
    ["d"] => int(4)
    ["e"] => int(5)
}

array(5) {
    ["a"] => int(1)
    ["b"] => int(2)
    ["c"] => int(3)
    ["d"] => int(4)
    ["e"] => int(5)
}

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