为什么php json_解码只接受整数，而不接受字符串？

经过几天的努力，终于意识到我的json语法并没有被php json_解码所接受。它似乎只在值是整数而不是字符串时才起作用。我如何使用json（如以下内容）来工作

<?php
$json = '{"fname":2,"surname":2}'; //return No errors
// $json = '{"fname": SJ011MS,"surn": something}'; //return Syntax error, malformed JSON

$obj = json_decode($json, true);
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;
    }
?>

---

这绝对是格式错误的JSON：

$json = '{"fname": SJ011MS,"surn": something}'

您必须使用

“

（甚至不是单引号

”

）将字符串括起来。它实际上应该是：

$json = '{"fname": "SJ011MS","surn": "something"}'
//-----------------^-------^---------^---------^

JSON只能有基本的JavaScript类型、数组和对象

（来源：）

这不是有效的JSON:

{"fname": SJ011MS,"surn": something}

属性值也需要在引号中：

{"fname": "SJ011MS","surn": "something"}

---

考虑一下您的示例json

{"widget": {
    "text": {
        "data": "Click Here",
        "size": 36,
        "style": "bold",
        "name": "text1",
        "hOffset": 250,
        "vOffset": 100,
        "alignment": "center",
        "onMouseUp": "sun1.opacity = (sun1.opacity / 100) * 90;"
    }
}}

我已经编写了以下代码，似乎工作得很好：-

<?php
$json = '{"widget": {
    "text": {
        "data": "Click Here",
        "size": 36,
        "style": "bold",
        "name": "text1",
        "hOffset": 250,
        "vOffset": 100,
        "alignment": "center",
        "onMouseUp": "sun1.opacity = (sun1.opacity / 100) * 90;"
    }
}}';
$decodedJson = json_decode($json,TRUE);
echo $json;
print_r($decodedJson);
?>

---

这里要学习的是，如果JSON属性包含任何字符串，那么它应该在引号内“

我真的不明白问题出在哪里；您的最终json很好，格式错误的json缺少引号。但你似乎已经知道了。这应该作为一个打字错误来结束吗？谢谢你们的帮助，伙计们，我正在绞尽脑汁去理解this@Diego你知道答案了吗？是的，我的错误是没有在字符串上使用“on string”，ruckie错了，谢谢again@Diego通过点击我的答案旁边的嘀嗒按钮来接受答案。

<?php
$json = '{"widget": {
    "text": {
        "data": "Click Here",
        "size": 36,
        "style": "bold",
        "name": "text1",
        "hOffset": 250,
        "vOffset": 100,
        "alignment": "center",
        "onMouseUp": "sun1.opacity = (sun1.opacity / 100) * 90;"
    }
}}';
$decodedJson = json_decode($json,TRUE);
echo $json;
print_r($decodedJson);
?>

{"widget": {
    "text": {
        "data": "Click Here",
        "size": 36,
        "style": "bold",
        "name": "text1",
        "hOffset": 250,
        "vOffset": 100,
        "alignment": "center",
        "onMouseUp": "sun1.opacity = (sun1.opacity / 100) * 90;"
    }
}}Array
(
    [widget] => Array
        (
            [text] => Array
                (
                    [data] => Click Here
                    [size] => 36
                    [style] => bold
                    [name] => text1
                    [hOffset] => 250
                    [vOffset] => 100
                    [alignment] => center
                    [onMouseUp] => sun1.opacity = (sun1.opacity / 100) * 90;
                )

        )

)