php 如何将 JSON 字符串转换为数组

声明:本页面是StackOverFlow热门问题的中英对照翻译,遵循CC BY-SA 4.0协议,如果您需要使用它,必须同样遵循CC BY-SA许可,注明原文地址和作者信息,同时你必须将它归于原作者(不是我):StackOverFlow 原文地址: http://stackoverflow.com/questions/7511821/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me): StackOverFlow

提示:将鼠标放在中文语句上可以显示对应的英文。显示中英文
时间:2020-08-26 02:50:19  来源:igfitidea点击:

How to convert JSON string to array

phparraysjson

提问by XMen

What I want to do is the following:

我想要做的是以下内容:

  1. taking JSON as input from text area in php
  2. use this input and convert it to JSON and pass it to php curl to send request.
  1. 将 JSON 作为来自 php 文本区域的输入
  2. 使用此输入并将其转换为 JSON 并将其传递给 php curl 以发送请求。

this m getting at php from get of api this json string i want to pass to json but it is not converting to array

这我从 api 获取这个 json 字符串我想传递给 json 但它没有转换为数组

echo $str='{
        action : "create",
        record: {
            type: "n$product",
            fields: {
                n$name: "Bread",
                n$price: 2.11
            },
            namespaces: { "my.demo": "n" }
        }
    }';
    $json = json_decode($str, true);

the above code is not returning me array.

上面的代码没有返回我的数组。

回答by RickN

If you pass the JSON in your post to json_decode, it will fail. Valid JSON strings have quoted keys:

如果您将帖子中的 JSON 传递给json_decode,它将失败。有效的 JSON 字符串具有带引号的键:

json_decode('{foo:"bar"}');         // this fails
json_decode('{"foo":"bar"}', true); // returns array("foo" => "bar")
json_decode('{"foo":"bar"}');       // returns an object, not an array.

回答by sepidol

Try this:

尝试这个:

$data = json_decode($your_json_string, TRUE);

the second parameter will make decoded json string into an associative arrays.

第二个参数将解码的 json 字符串转换为关联数组。

回答by jbeauchamp

If you are getting the JSON string from the form using $_REQUEST, $_GET, or $_POSTthe you will need to use the function html_entity_decode(). I didn't realize this until I did a var_dumpof what was in the request vs. what I copied into and echostatement and noticed the request string was much larger.

如果你是从形式获取JSON字符串使用$_REQUEST$_GET或者$_POST在您需要使用的功能html_entity_decode()。我没有意识到这一点,直到我做了var_dump请求中的内容与我复制到和echo语句中的内容并注意到请求字符串大得多。

Correct Way:

正确方法:

$jsonText = $_REQUEST['myJSON'];
$decodedText = html_entity_decode($jsonText);
$myArray = json_decode($decodedText, true);

With Errors:

有错误:

$jsonText = $_REQUEST['myJSON'];
$myArray = json_decode($jsonText, true);
echo json_last_error(); //Returns 4 - Syntax error;

回答by Arslan Ramay

Use json_decode($json_string, TRUE)function to convert the JSON object to an array.

使用json_decode($json_string, TRUE)函数将 JSON 对象转换为数组。

Example:

例子:

$json_string   = '{"a":1,"b":2,"c":3,"d":4,"e":5}';

$my_array_data = json_decode($json_string, TRUE);

NOTE: The second parameter will convert decoded JSON string into an associative array.

注意:第二个参数将解码的 JSON 字符串转换为关联数组。

===========

============

Output:

输出:

var_dump($my_array_data);

array(5) {

    ["a"] => int(1)
    ["b"] => int(2)
    ["c"] => int(3)
    ["d"] => int(4)
    ["e"] => int(5)
}

回答by Dinanath Thakur

your string should be in the following format:

您的字符串应采用以下格式:

$str = '{"action": "create","record": {"type": "n$product","fields": {"n$name": "Bread","n$price": 2.11},"namespaces": { "my.demo": "n" }}}';
$array = json_decode($str, true);

echo "<pre>";
print_r($array);

Output:

输出:

Array
 (
    [action] => create
    [record] => Array
        (
            [type] => n$product
            [fields] => Array
                (
                    [n$name] => Bread
                    [n$price] => 2.11
                )

            [namespaces] => Array
                (
                    [my.demo] => n
                )

        )

)

回答by piyush

If you are getting json string from URL using file_get_contents, then follow the steps:

如果您使用从 URL 获取 json 字符串file_get_contents,请按照以下步骤操作:

$url = "http://localhost/rest/users";  //The url from where you are getting the contents
$response = (file_get_contents($url)); //Converting in json string
 $n = strpos($response, "[");
$response = substr_replace($response,"",0,$n+1);
$response = substr_replace($response, "" , -1,1);
print_r(json_decode($response,true));

回答by Manav Akela

You can convert json Object into Array & String.

您可以将 json 对象转换为数组和字符串。

$data='{"resultList":[{"id":"1839","displayName":"Analytics","subLine":""},{"id":"1015","displayName":"Automation","subLine":""},{"id":"1084","displayName":"Aviation","subLine":""},{"id":"554","displayName":"Apparel","subLine":""},{"id":"875","displayName":"Aerospace","subLine":""},{"id":"1990","displayName":"Account Reconciliation","subLine":""},{"id":"3657","displayName":"Android","subLine":""},{"id":"1262","displayName":"Apache","subLine":""},{"id":"1440","displayName":"Acting","subLine":""},{"id":"710","displayName":"Aircraft","subLine":""},{"id":"12187","displayName":"AAC","subLine":""}, {"id":"20365","displayName":"AAT","subLine":""}, {"id":"7849","displayName":"AAP","subLine":""}, {"id":"20511","displayName":"AACR2","subLine":""}, {"id":"28585","displayName":"AASHTO","subLine":""}, {"id":"45191","displayName":"AAMS","subLine":""}]}';

$b=json_decode($data);

$i=0;
while($b->{'resultList'}[$i])
{
    print_r($b->{'resultList'}[$i]->{'displayName'});
    echo "<br />";
    $i++;
}

回答by Shahrukh Anwar

You can change a string to JSON as follows and you can also trim, strip on string if wanted,

您可以按如下方式将字符串更改为 JSON,如果需要,您还可以修剪、去除字符串,

$str     = '[{"id":1, "value":"Comfort Stretch"}]';
//here is JSON object
$filters = json_decode($str);

foreach($filters as $obj){
   $filter_id[] = $obj->id;
}

//here is your array from that JSON
$filter_id;

回答by Pradeep Dhawan

<?php
$str='{
    "action" : "create",
    "record" : {
                "type": "$product",
                "fields": {
                           "name": "Bread",
                           "price": "2.11"
                           },
                "namespaces": { "my.demo": "n" }
                }
    }';
echo $str;
echo "<br>";
$jsonstr = json_decode($str, true);
print_r($jsonstr);

?>

i think this should Work, its just that the Keys should also be in double quotes if they are not numerals.

我认为这应该有效,只是如果键不是数字,它们也应该用双引号引起来。

回答by Mourad MAMASSI

this my solution: json string $columns_validation = string(1736) "[{"colId":"N_ni","hide":true,"aggFunc":null,"width":136,"pivotIndex":null,"pinned":null,"rowGroupIndex":null},{"colId":"J_2_fait","hide":true,"aggFunc":null,"width":67,"pivotIndex":null,"pinned":null,"rowGroupIndex":null}]"

这是我的解决方案:json 字符串 $columns_validation = string(1736) "[{"colId":"N_ni","hide":true,"aggFunc":null,"width":136,"pivotIndex":null,"pinned":null,"rowGroupIndex":null},{"colId":"J_2_fait","hide":true,"aggFunc":null,"width":67,"pivotIndex":null,"pinned":null,"rowGroupIndex":null}]"

so i use json_decode twice like that :

所以我这样使用 json_decode 两次:

$js_column_validation = json_decode($columns_validation);
$js_column_validation = json_decode($js_column_validation); 

var_dump($js_column_validation);

and the result is :

结果是:

 array(15) { [0]=> object(stdClass)#23 (7) { ["colId"]=> string(4) "N_ni" ["hide"]=> bool(true) ["aggFunc"]=> NULL ["width"]=> int(136) ["pivotIndex"]=> NULL ["pinned"]=> NULL ["rowGroupIndex"]=> NULL } [1]=> object(stdClass)#2130 (7) { ["colId"]=> string(8) "J_2_fait" ["hide"]=> bool(true) ["aggFunc"]=> NULL ["width"]=> int(67) ["pivotIndex"]=> NULL ["pinned"]=> NULL ["rowGroupIndex"]=> NULL }