将数组数据作为对象附加到json文件

我有一个简单的json文件,它具有直接的图像链接和每个json对象的文件夹名称:

[
        {
                "image": "http://placehold.it/350x150",
                "folder": "Small"
        },
        {
                "image": "http://placehold.it/450x250",
                "folder": "Medium"
        },
        {
                "image": "http://placehold.it/550x350",
                "folder": "Medium"
        }
]

我想从帖子中追加这两个值,但结果是一个不变的json文件。 这里是PHP代码w / comments:

$directLink = $_POST['txtDirectLink'];
$category = $_POST['selectCategoriesImages'];
$util->addToJsonImageList($directLink, $category);

//decodes the json image list, merges to the decoded array a new image, then encodes back to a json file
function addToJsonImageList($link, $category) {
    $file = file_get_contents('./images_list.json', true);
    $data = json_decode($file);
    unset($file);

    $newImage = array('image' => $link, 'folder' => $category);
    //$newImage['image'] = $link;
    //$newImage['folder'] = $category;
    $result = array_merge((array)$data, (array)$newImage);

    json_encode($result);
    file_put_contents('./images_list.json', $result);
    unset($result);
}

逻辑是,将文件json_decode解码为数组应该很简单,将新数组合并到该数组上,然后对结果进行编码并放入同一个文件中。 我一直无法捕捉任何类型的错误。


$data = json_decode($file, true);
//When TRUE, returned objects will be converted into associative arrays.
unset($file);

$newImage = array('image' => $link, 'folder' => $category);
//$newImage['image'] = $link;
//$newImage['folder'] = $category;
$result = array_merge($data, $newImage);
//don't need to cast

参考PHP json_decode手册

编辑下面的代码工程(测试)

function addToJsonImageList($link, $category) {
    $file = file_get_contents('./images_list.json', true);
    $data = json_decode($file,true);
    unset($file);
    //you need to add new data as next index of data.
    $data[] = array('image' => $link, 'folder' => $category);
    $result=json_encode($data);
    file_put_contents('./images_list.json', $result);
    unset($result);
}

编辑2增加了很多错误报告和调试。 请让我知道以下的输出。 下面的代码没有经过测试(只是在这里输入)。如果发现任何语法错误,请修复。 这里迟到,只能明天我的时间,但可以回复。

<?php
//display all errors and warnings
error_reporting(-1);
addToJsonImageList('test link', 'test category');

function addToJsonImageList($link, $category) {
    $file = file_get_contents('./images_list.json', true);
    if ($file===false)die('unable to read file');
    $data = json_decode($file,true);
    if ($data ===NULL)die('Unable to decode');
    unset($file);
    echo "data beforen";
    var_dump ($data);
    //you need to add new data as next index of data.
    $data[] = array('image' => $link, 'folder' => $category);
    echo "data aftern";
    var_dump ($data);
    $result=json_encode($data);
    if (file_put_contents('./images_list.json', $result) === false){
        die('unable to write file');
    }
    unset($result);
}
?>

比方说,你有这个名为(playerJson)的.json文件

{

"Players":
   [
     {"Name":"Arun","Arm":"Gun","num":"1"},
     {"Name":"sssv","Arm":"Arc","num":"2"},
     {"Name":"Surya","Arm":"Bomb","num":"3"},
     {"Name":"sssv","Arm":"Fire","num":"4"}

   ]
}

现在我们在php(myPhpFile.php)中必须做的是:(我想你是从表单加载你的数据)

<?php

$json = file_get_contents('playerJson.json');
$json_data = json_decode($json,true);

$newar = array(
           'Name'=>$_POST['nameField'] ,
           'Arm' =>$_POST['armField'],
           'Num' =>$_POST['numField']
);
//saving data in Players object...
array_push($json_data['Players'], $newar);

$json = json_encode($json_data);

file_put_contents('playerJson.json', $json);

?>

就这样 ! 你在“玩家”对象中有一个新行。 但是如果你想添加一个新对象,请避免array_push函数中的$ json_data之后的[Players]。


你可以做的是解码json然后合并它们。

$data = json_decode(file_get_contents(./images_list.json));

$result = array_merge((array)$data, (array)$newImage);

然后输出json_encode

file_put_contents('./images_list.json', json_encode($result, JSON_FORCE_OBJECT));
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