2015-01-20 42 views
-2

我有一个应用程序显示一个网站列表和一些基于JSON文件的信息。我试图让应用程序的用户可以通过自己的网页添加新的网站列表。PHP - 试图让json_encode添加逗号和括号

我遇到的问题是,当我使用JSON_ENCODE时,它附加的JSON文件上没有开头或者右括号 - 并且没有逗号分隔每个对象。

我的应用程序不会读取没有添加这些字符的JSON文件。我做了大量的研究,并且在这一天里我的头撞墙了几天。我对PHP非常陌生,所以请原谅我,如果这已经得到了答复,但我已经搜索,没有运气。

这是我的PHP代码:

<?php 

if(isset($_POST['age']) && isset($_POST['id']) && isset($_POST['image']) && isset($_POST['name']) && isset($_POST['snippet0']) && isset($_POST['snippet1']) 
    && isset($_POST['snippet3']) && isset($_POST['snippet4']) && isset($_POST['snippet5']) && isset($_POST['snippet6'])) { 

    if(empty($_POST['age']) || empty($_POST['id']) || empty($_POST['image']) || empty($_POST['name']) || empty($_POST['snippet0']) || empty($_POST['snippet1']) 
    || empty($_POST['snippet2']) || empty($_POST['snippet3']) || empty($_POST['snippet4']) || empty($_POST['snippet5']) || empty($_POST['snippet6'])) { 
    echo 'All fields are required'; 
    } 
    else { 
    $postArray = array(
     "age" => $_POST['age'], 
     "id" => $_POST['id'], 
     "image" => $_POST['image'], 
     "name" => $_POST['name'], 
     "snippet0" => $_POST['snippet0'], 
     "snippet1" => $_POST['snippet1'], 
     "snippet2" => $_POST['snippet2'], 
     "snippet3" => $_POST['snippet3'], 
     "snippet4" => $_POST['snippet4'], 
     "snippet5" => $_POST['snippet5'], 
     "snippet6" => $_POST['snippet6'] 
    ); 


$jsondata = json_encode ($postArray, JSON_PRETTY_PRINT); 


$file = 'data/formdata.json'; 
if(file_put_contents($file, $jsondata, FILE_APPEND)) echo 'Data saved'; 
else echo 'Unable to save data'; 
    } 
} 
else echo 'Form fields not submitted'; 
?> 

下面是JSON输出的一个例子,我得到它:

{ 
    "age": "1", 
    "id": "bob", 
    "image": "bob.png", 
    "name": "Bob.com", 
    "snippet0": "sub.bob.com", 
    "snippet1": "sub.bob.com", 
    "snippet2": "sub.bob.com", 
    "snippet3": "sub.bob.com", 
    "snippet4": "sub.bob.com", 
    "snippet5": "sub.bob.com", 
    "snippet6": "sub.bob.com" 
}{ 
    "age": "1", 
    "id": "bob", 
    "image": "bob.png", 
    "name": "Bob.com", 
    "snippet0": "sub.bob.com", 
    "snippet1": "sub.bob.com", 
    "snippet2": "sub.bob.com", 
    "snippet3": "sub.bob.com", 
    "snippet4": "sub.bob.com", 
    "snippet5": "sub.bob.com", 
    "snippet6": "sub.bob.com" 
} 

最后,JSON输出的一个例子,我想:

[{ 
    "age": "1", 
    "id": "bob", 
    "image": "bob.png", 
    "name": "Bob.com", 
    "snippet0": "sub.bob.com", 
    "snippet1": "sub.bob.com", 
    "snippet2": "sub.bob.com", 
    "snippet3": "sub.bob.com", 
    "snippet4": "sub.bob.com", 
    "snippet5": "sub.bob.com", 
    "snippet6": "sub.bob.com" 
},{ 
    "age": "1", 
    "id": "bob", 
    "image": "bob.png", 
    "name": "Bob.com", 
    "snippet0": "sub.bob.com", 
    "snippet1": "sub.bob.com", 
    "snippet2": "sub.bob.com", 
    "snippet3": "sub.bob.com", 
    "snippet4": "sub.bob.com", 
    "snippet5": "sub.bob.com", 
    "snippet6": "sub.bob.com" 
}] 
+0

file_put_contents从文件中获取JSON之前,使用'json_decode',添加新值,使用'json_encode',并运行'file_put_contents'没有FILE_APPEND – 2015-01-20 19:19:19

+0

我需要什么值增加对json_decode? – blasted 2015-01-20 19:39:04

+0

新数据 - $ postArray – 2015-01-20 19:41:19

回答

0

首先,从文件中读取

$file_content = json_decode(file_get_contents('data/formdata.json')); 

然后添加数据

$file_content []= 
    array(
     "age" => $_POST['age'], 
     "id" => $_POST['id'], 
     "image" => $_POST['image'], 
     "name" => $_POST['name'], 
     "snippet0" => $_POST['snippet0'], 
     "snippet1" => $_POST['snippet1'], 
     "snippet2" => $_POST['snippet2'], 
     "snippet3" => $_POST['snippet3'], 
     "snippet4" => $_POST['snippet4'], 
     "snippet5" => $_POST['snippet5'], 
     "snippet6" => $_POST['snippet6'] 
    ); 

最终只是把内容回文件

$file_content = json_encode ($file_content, JSON_PRETTY_PRINT); 
file_put_contents('data/formdata.json', $file_content); 
0

这很容易,只要在第二个参数选项json_encode功能

JSON_UNESCAPED_SLASHES|JSON_PRETTY_PRINT|JSON_UNESCAPED_UNICODE 

例如:

json_encode($date,JSON_UNESCAPED_SLASHES|JSON_PRETTY_PRINT|JSON_UNESCAPED_UNICODE)