将 php 数组传递给 javascript 数组

问题描述：

我正在寻找将 php 数组传递给 javascript 的最佳方法.

I'm looking for the best way to pass php array to javascript.

我正在制作一个 RPG，当他们登录时，我想从数据库返回他们保存的数据并存储在 javascript 端的数组中:

I'm making a RPG, and when they login, I'd like to return their saved data from database and store in an array on javascript side:

为了获取数据，我这样做:

To get data, I do:

$.getJSON("php/CRUD.php", {"_functionToRun" : ""},
    function (returned_data) {
        game.data.dataArray.push(returned_data.split(" "));
        log("1: " + game.data.dataArray); //output: `1: 120,mymap2` 
        log("2: " + game.data.dataArray[0]); //output: `2: 120,mymap2`
    }
);

PHP:

    $qry = 
        'SELECT * 
        FROM userstats
        WHERE id_user_fk ="' . $_SESSION['userid'] . '"
        LIMIT 1';

    $result = $mysqli->query($qry) or die(mysqli_error($mysqli));

    while ($row = $result->fetch_assoc()) {
        $message =  $row['experience'] . $row['levelname'];
    }   

 echo json_encode($message);

1) 这是将一组值放入 js 数组的最佳方法吗?

1) Is this the best way to get a set of values into a js array?

2) 为什么我无法使用game.data.dataArray[0] 访问game.data.dataArray 的某个数据元素.

2) Why can't I access a certain data element of game.data.dataArray using game.data.dataArray[0].

这些给出与 1: 120,mymap2

        log("1: " + game.data.dataArray);
        log("2: " + game.data.dataArray[0]);

是否应该returned_data.split(" ")将返回的字符串拆分成两个数组元素?

Should returned_data.split(" ") split the returned string into two array elements?

好的，我已经完成了 echo json_encode($message); 并且它返回了引号，但是对于 game.data.dataArray 和 仍然返回相同的结果代码>game.data.dataArray[0]

Okay I've done echo json_encode($message); and it returns with quotes, but still returns same results for game.data.dataArray and game.data.dataArray[0]

1: "120,mymap2" 
2: "120,mymap2"

我也将函数更改为 $.getJSON

又改成

    $qry = 
        'SELECT * 
        FROM userstats
        WHERE id_user_fk ="' . $_SESSION['userid'] . '"';

    $result = $mysqli->query($qry) or die(mysqli_error($mysqli));

    while ($row = $result->fetch_assoc()) {
        $array[] =  $row['experience'] . $row['levelname'];
        die(json_encode($array[]));
    }

和JS是:

$.getJSON("php/CRUD.php", {"_functionToRun" : ""},
    function (returned_data) {
        game.data.dataArray.push(returned_data.split(" "));
        log("1: " + game.data.dataArray);
        log("2: " + game.data.dataArray[0]);
    }
);

输出:

Uncaught melonJS: level <br />
<font size='1'><table class='xdebug-error' dir='ltr' border='1' cellspacing='0' cellpadding='1'>
<tr><th align='left' bgcolor='#f57900' colspan="5"><span style='background-color: #cc0000; color: #fce94f; font-size: x-...<omitted>...nd

答

如果你使用 $.getJSON 并且你提供了有效的 json，你不需要解析任何东西，所以你可以改变你的php到:

If you use $.getJSON and you provide valid json, you don't need to parse anything, so you can change your php to:

if ($row = $result->fetch_assoc()) {
    echo json_encode($row);
    exit;    // You're limiting the number of rows to 1, so there is no need to continue
}

还有 javascript:

And the javascript:

$.getJSON("php/CRUD.php", {"_functionToRun" : ""},
    function (returned_data) {
        // you can push `returned_data` to an array / add it to an object,
        // but then you need to keep track of its index
        log("1: " + returned_data.experience); //output: `1: 120` 
        log("2: " + returned_data.levelname); //output: `2: mymap2`
    }
);

将 php 数组传递给 javascript 数组

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