Question

下午好，

我目前在使用PHP从json文件获取数据时遇到问题。

我现在得到的是

$jsondecoded = json_decode('[{"chat":{"username":"RobloxProtectorKing","message":":slender me","time":"2018-03-20 01:56:12"}}', true);

echo ($jsondecoded[0]->chat);

如你所见。我正试图获取聊天信息。但是，当我试图这样做时，它不会回应任何东西。我一直试图弄清楚为什么会这样，但我遗憾地找不到它。我必须遗漏一些东西，但我无法弄清楚我错过了什么。

先谢谢。

Answer 1

三个错误：

1）在json结束时丢失了。

2）您正在使用json_decode的“array assoc”选项，因此它不会返回对象。

3）您无法回显聊天对象。

试试这个：

$jsondecoded = json_decode('[{"chat":{"username":"RobloxProtectorKing","message":":slender me","time":"2018-03-20 01:56:12"}}]');

echo ($jsondecoded[0]->chat->username);

Answer 2

$jsondecoded = json_decode('[{"chat": {"username":"RobloxProtectorKing","message":":slender me","time":"2018-03-20 01:56:12"}}]', true);

此外，请记住，您已使用true作为json_decode的第二个参数，这意味着您要将结果转换为关联数组，要么删除true，要么更改访问：

var_dump($jsondecoded[0]['chat']);

Answer 3

你的json在json字符串的结尾添加']'结尾时无效，而php中的print数组使用print_r函数。 '[0]'用于获取第一个元素，['chat']用于获取聊天键的值

$jsondecoded = json_decode('[{"chat":{"username":"RobloxProtectorKing","message":":slender me","time":"2018-03-20 01:56:12"}}]', true);

print_r ($jsondecoded[0]['chat']);

输出将

Array
(
    [username] => RobloxProtectorKing
    [message] => :slender me
    [time] => 2018-03-20 01:56:12
)

Answer 4

以下是我所拥有的模板的示例。希望它有所帮助。

<?php
/* Status Codes

return 0 = Nothing to Update (n/a)
return 1 = Successful Insert Query
return 2 = Database Connection refused
return 3 = MySQL Query Error OR Wrong URL Parameters */

/* Disable Warnings so that we can return ONLY what we want through echo. */
mysqli_report(MYSQLI_REPORT_STRICT);

// First get raw POST input
$raw_post     = file_get_contents('php://input');
// Run through url_decode..
$url_decoded  = urldecode($raw_post);
// Run through json_decode...
// false to allow for reference to oject. eg. $column->name instead of $column["name"] in the foreach.
$json_decoded = json_decode($url_decoded, false);

$pk_line_item_id  = (strlen($json_decoded[0]->value)  > 0 ? $json_decoded[0]->value : null);

// INCLUDE DB CONNECTION STRING
include 'php_pdo_mysql_connect.php';

// SQL INSERT query...
$stmt = $link->prepare(" SELECT * FROM tbl_xyz ");

//$stmt->bindParam(':pk_line_item_id', $pk_line_item_id, PDO::PARAM_INT);   // INT

// Execute this SQL statement.
$stmt->execute();

// Fetch & Populate the single returned in var $resultSet.
$resultSet = $stmt->fetchAll(); 
// Returns an array indexed by column number as returned in your result set, starting at column 0. 
// https://www.ibm.com/support/knowledgecenter/en/SSEPGG_9.7.0/com.ibm.swg.im.dbclient.php.doc/doc/t0023505.html

// If the search record was found, populate it on the html table.
if (($resultSet !== false) && ($stmt->rowCount() > 0)) {

// Set JSON headers
header('Content-type:application/json;charset=utf-8');

// Encode entire $resultSet array to JSON.
$json_encoded = json_encode($resultSet);
// Ready to return as JSON to client side JQuery / Java Script...

echo $json_encoded;

}

?>

尝试使用php从json中获取数据不会返回任何内容

4 个答案: