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PHPでajaxとJSONによるエンコードとJSでのデコードを使用してphp配列をjsに取得しようとしていますが、目的の結果を得ることができません。

エンコードされた文字列をエコーする直前に var_dump を使用すると、次の出力が得られます。

$boardArrayEncoded = json_encode($boardArray);
var_dump($boardArrayEncoded);
echo $boardArrayEncoded;

JSON_encode の後のこの文字列は次のとおりです。

string '{"1x1":0,"1x2":0,"1x3":0,"1x4":0,"1x5":0,"1x6":0,"2x1":0,"2x2":0,"2x3":0,"2x4":0,"2x5":0,"2x6":0,"3x1":0,"3x2":0,"3x3":0,"3x4":0,"3x5":0,"3x6":0,"4x1":0,"4x2":0,"4x3":0,"4x4":0,"4x5":0,"4x6":0,"5x1":0,"5x2":0,"5x3":0,"5x4":0,"5x5":0,"5x6":0,"6x1":0,"6x2":0,"6x3":0,"6x4":0,"6x5":0,"6x6":0}'

JavaScriptで私はそれを解析しようとします:

if (xmlhttp.readyState == 4 && xmlhttp.status == 200)
{
   document.getElementById("logDiv1").innerHTML = xmlhttp.responseText;
   array = JSON.parse(xmlhttp.responseText);
}

これは xmlhttp.responseText です。 {"1x1":0,"1x2":0,"1x3":0,"1x4":0,"1x5":0,"1x6":0,"2x1":0,"2x2":0,"2x3":0,"2x4":0,"2x5":0,"2x6":0,"3x1":0,"3x2":0,"3x3":0,"3x4":0,"3x5":0,"3x6":0,"4x1":0,"4x2":0,"4x3":0,"4x4":0,"4x5":0,"4x6":0,"5x1":0,"5x2":0,"5x3":0,"5x4":0,"5x5":0,"5x6":0,"6x1":0,"6x2":0,"6x3":0,"6x4":0,"6x5":0,"6x6":0}

firebug では、構文エラーが発生します。

SyntaxError: JSON.parse: unexpected character
[Break On This Error]   

array1 = JSON.parse(xmlhttp.responseText);

私は何を間違っていますか？この配列を JS で使用する必要があります。どうすれば適切にエンコードできますか?

前もって感謝します

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4 に答える 4

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http://jsfiddle.net/ELc8L/

var responseText='{"1x1":0,"1x2":0,"1x3":0,"1x4":0,"1x5":0,"1x6":0,"2x1":0,"2x2":0,"2x3":0,"2x4":0,"2x5":0,"2x6":0,"3x1":0,"3x2":0,"3x3":0,"3x4":0,"3x5":0,"3x6":0,"4x1":0,"4x2":0,"4x3":0,"4x4":0,"4x5":0,"4x6":0,"5x1":0,"5x2":0,"5x3":0,"5x4":0,"5x5":0,"5x6":0,"6x1":0,"6x2":0,"6x3":0,"6x4":0,"6x5":0,"6x6":0}'; // Your JSON string.
parsedJSON=eval('('+responseText+')'); // Parsed JSON. Object.
alert(parsedJSON["1x1"]); // Access object element's like this, because you can't write parsedJSON.1x1;
for(element in parsedJSON){
    document.body.innerHTML+=element+" => "+parsedJSON[element]+"<br />";
} // Just a vardump to show everything's perfect
于 2013-08-03T12:28:01.317 に答える