php中输出json对象的值(实现方法)
时间:2021-03-20 10:02:15|栏目:PHP代码|点击: 次
实例如下所示:
<!DOCTYPE html>
<html>
<body>
<?php
$json = '{"report":{"date":"2012-04-10","content":"abcdefght"}}';
$arr = (array) json_decode($json,true);
echo '当前日期是:'. $arr['report']['date'];
echo "<br/>";
echo '<pre>';
print_r($arr);
echo '</pre>';
foreach($arr as $arrvalue)
{
foreach($arrvalue as $key=>$value)
{
echo "[$key] = $value <br />";
}
}
echo "<br/>";
?>
</body>
</html>

<!DOCTYPE html>
<html>
<body>
<?php
$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 "<br/>";
$arr=(array)json_decode($content,true);
echo $arr['CSS'];
echo "<br/>";
echo $arr['AJAX'];
}
?>
</body>
</html>

json.txt:
{
"AJAX": "Asynchronous JavaScript and XML",
"CSS": "Cascading Style Sheets"
}
上一篇:cakephp常见知识点汇总
栏 目:PHP代码
下一篇:PHP类的特性实例分析
本文标题:php中输出json对象的值(实现方法)
本文地址:http://www.codeinn.net/misctech/84528.html






