doucu9677 2013-08-01 01:08
浏览 6
已采纳

php json_decode,数据由[]包装

with true option, it works fine.

sorry and thanks guys.

=============================================

I encoded php array to json with this code

$rows = array();
if ($result = $mysqli->query($query)) {
    while ($row = $result->fetch_assoc()) {
        $rows[] = $row;
    }
    echo json_encode($rows);
    /* free result set */
    $result->free();
}

end decode with

$array = json_decode($server_output)

the $server_output is like this

[{"userid":"96679","userinfor":"xxxxxxxxx","userlocation":"CA"}]
[{"userid":"153795","userinfor":"xxxxxxxxx","userlocation":"CA"}]
[{"userid":"131878","userinfor":"xxxxxxxxx","userlocation":"CA"}]

but, $array is NULL :(

Thanks in advence,

  • 写回答

1条回答 默认 最新

  • dongleibeng5602 2013-08-01 01:28
    关注

    This is not valid json together:

    [{"userid":"96679","userinfor":"xxxxxxxxx","userlocation":"CA"}]
    [{"userid":"153795","userinfor":"xxxxxxxxx","userlocation":"CA"}]
    [{"userid":"131878","userinfor":"xxxxxxxxx","userlocation":"CA"}]
    

    While, this part is valid:

    [{"userid":"96679","userinfor":"xxxxxxxxx","userlocation":"CA"}]
    

    or any one of them separately

    $json = '[{"userid":"96679","userinfor":"xxxxxxxxx","userlocation":"CA"}]';
    print_r(json_decode($json, true));
    

    Output:

    Array
    (
        [0] => Array
            (
                [userid] => 96679
                [userinfor] => xxxxxxxxx
                [userlocation] => CA
            )
    
    )
    

    The valid format for all data is like this:

    [
        {
            "userid": "96679",
            "userinfor": "xxxxxxxxx",
            "userlocation": "CA"
        },
        {
            "userid": "153795",
            "userinfor": "xxxxxxxxx",
            "userlocation": "CA"
        },
        {
            "userid": "131878",
            "userinfor": "xxxxxxxxx",
            "userlocation": "CA"
        }
    ]
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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