php取json子对象属性 php中输出json对象的值(实现方法)

实例如下所示：

$json = '{"report":{"date":"-04-10","content":"abcdefght"}}';

$arr = (array) json_decode($json,true);

echo '当前日期是：'. $arr['report']['date'];

echo "

";

echo '

';

print_r($arr);

echo '

';

foreach($arr as $arrvalue)

{

foreach($arrvalue as $key=>$value)

{

echo "[$key] = $value

";

}

}

echo "

";

?>

$file_exists = file_exists("json.txt");

if( $file_exists ){

$myfile = fopen("json.txt", "r") or die("");

$content = fread($myfile,filesize("json.txt"));

fclose($myfile);

echo $content;

echo "

";

$arr=(array)json_decode($content,true);

echo $arr['CSS'];

echo "

";

echo $arr['AJAX'];

}

?>

json.txt:

{

"AJAX": "Asynchronous JavaScript and XML",

"CSS": "Cascading Style Sheets"

}

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