你好,例如我有json格式。我需要生成一个php数组来从那个
构建这个json{
"posts":
[
{
"detial" : {
"titel": "web design",
"date-publish" : "2015-02-08",
"pic-thumbnail" : "small-web-desing-jpg",
"pic-medium" : "medium-web-desing-jpg"
},
"content" : {
"description" : "description web design post",
"content" : "web design is modern technology in world"
}
}
]
}
请显示PHP数组构建此json的示例
答案 0 :(得分:1)
很容易!
只需使用PHP的json_decode()
函数即可实现。
最后,将json_encode()返回数组保存到变量中。
<强>输出:强>
Array
(
[posts] => Array
(
[0] => Array
(
[detial] => Array
(
[titel] => web design
[date-publish] => 2015-02-08
[pic-thumbnail] => small-web-desing-jpg
[pic-medium] => medium-web-desing-jpg
)
[content] => Array
(
[description] => description web design post
[content] => web design is modern technology in world
)
)
)
)
PHP代码:
$results = json_decode('{
"posts": [{
"detial" : {
"titel": "web design",
"date-publish" : "2015-02-08",
"pic-thumbnail" : "small-web-desing-jpg",
"pic-medium" : "medium-web-desing-jpg"
},
"content" : {
"description" : "description web design post",
"content" : "web design is modern technology in world"
}
}]
}', true);
echo "<pre>".print_r($results, true)."</pre>";
答案 1 :(得分:0)
使用PHP函数json_decode()生成php数组。
$jsonData = json_decode('{
"posts": [{
"detial" : {
"titel": "web design",
"date-publish" : "2015-02-08",
"pic-thumbnail" : "small-web-desing-jpg",
"pic-medium" : "medium-web-desing-jpg"
},
"content" : {
"description" : "description web design post",
"content" : "web design is modern technology in world"
}
}]
}',true);
echo "<pre>";
print_r($jsonData);