将json字符串转换为php关联数组

Gre*_*der 0 php json

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);
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任何帮助,将不胜感激!

Pra*_*man 6

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());
?>
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上面的代码返回a 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:

prevoew

你需要纠正它看起来像:

[{
    "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"]
    ]
}]
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你现在拥有的是一个JavaScript对象,而不是一个有效的JSON!

  • 我即将发布这个:) +1 (2认同)