将json字符串转换为php关联数组

Question

不知道为什么这不会转换，我会假设它可能与字符串有关，但我得到np输出。

$string = '[{title : "Comp 1 Product",columns : ["Our Vehicle","Features","Their Vehicle"], items : [["dcq","adv","asdvasdv"],["sdv","sdv","sdv"]]},{title : "qwefqw",columns : ["Section 1","Features","Section 2"],items : [["qqwec","qwe","qwegqwev"]]}]';

print_r(json_decode($string), true);

任何帮助将不胜感激！

Answer 1

如果你看到：

<?php
    header("Content-type: text/plain");
    $string = '[{title : "Comp 1 Product",columns : ["Our Vehicle","Features","Their Vehicle"], items : [["dcq","adv","asdvasdv"],["sdv","sdv","sdv"]]},{title : "qwefqw",columns : ["Section 1","Features","Section 2"],items : [["qqwec","qwe","qwegqwev"]]}]';
    print_r(json_decode($string), true);
    print_r(json_last_error());
?>

上面的代码返回4，这意味着JSON_ERROR_SYNTAX，这是JSON的语法错误。使用JSON Lint检查时，您的JSON将抛出：

您需要将其更正为：

[{
    "title": "Comp 1 Product",
    "columns": ["Our Vehicle", "Features", "Their Vehicle"],
    "items": [
        ["dcq", "adv", "asdvasdv"],
        ["sdv", "sdv", "sdv"]
    ]
}, {
    "title": "qwefqw",
    "columns": ["Section 1", "Features", "Section 2"],
    "items": [
        ["qqwec", "qwe", "qwegqwev"]
    ]
}]

您现在拥有的是 JavaScript对象，而不是有效的JSON ！

Answer 2

除了无效的json之外，print_r(json_decode($string), true);不打印任何内容，但返回值。要将结果打印到输出，您需要：

print_r(json_decode($string));

或

echo print_r(json_decode($string), true);

前者更好。