如何在PHP中将JavaScript文字对象转换为Json对象

时间:2021-12-11 16:03:55

I have a JS literal object string such as {name:{first:"George",middle:"William"},surname:"Washington"} and I have to convert it in Json. How can I do it using PHP?

我有一个JS文字对象字符串,如{name:{first:“George”,middle:“William”},姓:“Washington”},我必须在Json中将其转换。我怎么能用PHP做到这一点?

2 个解决方案

#1


2  

JS:

JS:

// Pretend we're POSTing this
var foo = {foo:{first:"George",middle:"William"}};

PHP:

PHP:

$foo = $_POST['foo'];
$foo = json_decode( stripslashes( $foo ) );
echo $foo->first;

Credit where credit is due: https://www.youtube.com/watch?v=pORFYsgOXog

信用到期的信用:https://www.youtube.com/watch?v = pORFYsgOXog

#2


-2  

Not json_encode, use $var = json_decode($_POST['names'], true). You can then use it like echo $var['surname'] to echo "Washington".

不是json_encode,使用$ var = json_decode($ _ POST ['names'],true)。然后你可以像echo $ var ['surname']一样使用它来回应“华盛顿”。

#1


2  

JS:

JS:

// Pretend we're POSTing this
var foo = {foo:{first:"George",middle:"William"}};

PHP:

PHP:

$foo = $_POST['foo'];
$foo = json_decode( stripslashes( $foo ) );
echo $foo->first;

Credit where credit is due: https://www.youtube.com/watch?v=pORFYsgOXog

信用到期的信用:https://www.youtube.com/watch?v = pORFYsgOXog

#2


-2  

Not json_encode, use $var = json_decode($_POST['names'], true). You can then use it like echo $var['surname'] to echo "Washington".

不是json_encode,使用$ var = json_decode($ _ POST ['names'],true)。然后你可以像echo $ var ['surname']一样使用它来回应“华盛顿”。