Question

请检查我的下面的代码，请让我知道我正在做的错误，我只想解码json并在浏览器上打印为数组

<?php

$json = '{
"Token" : "xxx-xxx-xxx",
"ID": "1",
"Recipients": [
    {
        "Recipient_ID": "XX",
        "From_Name": "XXX",
        "From_Email": "XXX",
        "To_Name": "XXX",
        "To_Email": "XXX",
        "Subject": "XXX",
        "Message": "XXX",
        "Attachments": [
            {
                "File_Name": "XXX",
                "File_Path": "XXX",
            }
        ],
    }
],
}';

$input = $json;
print_r(json_decode(stripslashes($input)));

?>

我试过这个json字符串在线解码（http://jsonviewer.stack.hu/）并且它的工作正常，所以json字符串没有问题。任何帮助都会很感激。

Answer 1

<?php

$json = '{
    "Token": "xxx-xxx-xxx",
    "ID": "1",
    "Recipients": [{
        "Recipient_ID": "XX",
        "From_Name": "XXX",
        "From_Email": "XXX",
        "To_Name": "XXX",
        "To_Email": "XXX",
        "Subject": "XXX",
        "Message": "XXX",
        "Attachments": [{
            "File_Name": "XXX",
            "File_Path": "XXX"
        }]
    }]
}';

$input = $json;
var_dump(json_decode(stripslashes($input)));

?>

Answer 2

问题从你所拥有的json格式开始......你有一些'，'需要被删除，这就是为什么json_decode（）无法完成他的工作。此函数抛出错误异常，但您应该做一个小技巧来查看错误。您可以使用此代码查看错误。

for (x=0; x<5; x++){
    printf("loop %d\n",x);
.
.
.

你json应该是这样的。

switch (json_last_error()) {
    case JSON_ERROR_NONE:
        echo ' - No errors';
    break;
    case JSON_ERROR_DEPTH:
        echo ' - Maximum stack depth exceeded';
    break;
    case JSON_ERROR_STATE_MISMATCH:
        echo ' - Underflow or the modes mismatch';
    break;
    case JSON_ERROR_CTRL_CHAR:
        echo ' - Unexpected control character found';
    break;
    case JSON_ERROR_SYNTAX:
        echo ' - Syntax error, malformed JSON';
    break;
    case JSON_ERROR_UTF8:
        echo ' - Malformed UTF-8 characters, possibly incorrectly encoded';
    break;
    default:
        echo ' - Unknown error';
    break;
}

json_decode无法在php中运行

2 个答案: