JSON.parse不起作用

-1

我正在使用AJAX從數據庫中獲取值。我正在使用JSON.parse不起作用

echo json_encode($writers);

in php。

在我的呼喚

document.getElementById("writer").innerHTML = xmlhttp.responseText;

我得到結果的JScript

[{"name":"demo-1","user_id":"13","writing_level":"","writing_category":"","pri":3},{"name":"Atif Rauf 
Alvi","user_id":"12","writing_level":"High 
School","writing_category":"Social 
Sciences,History,Mathematics and 
Economics,Nature,Health and 
Medicine,Creative 
writing","pri":3},{"name":"ffffo","user_id":"14","writing_level":"High 
School,College,Masters","writing_category":"Literature 
and Language,Social 
Sciences,Mathematics and 
Economics","pri":3},{"name":"mariam","user_id":"15","writing_level":"High 
School","writing_category":"Communications 
and Media,Religion and Theology,Life 
Sciences,Creative 
writing","pri":3},{"name":"ddd","user_id":"17","writing_level":"High 
School,College","writing_category":"Literature 
and Language,Art,Social 
Sciences,History,Law","pri":3},{"name":"maria","user_id":"16","writing_level":"High 
School","writing_category":"Art,Social 
Sciences,History,Law,Mathematics and 
Economics","pri":3},{"name":"Muhammad 
Zoyeb","user_id":"11","writing_level":"High 
School,College","writing_category":"Education,Tourism","pri":3},{"name":"wewe","user_id":"10","writing_level":"","writing_category":"","pri":3},{"name":"janea","user_id":"5","writing_level":"","writing_category":"","pri":3},{"name":"shazia","user_id":"4","writing_level":"","writing_category":"","pri":3},{"name":"s","user_id":"6","writing_level":"","writing_category":"","pri":3},{"name":"iuiui","user_id":"8","writing_level":"","writing_category":"","pri":3},{"name":"demo","user_id":"9","writing_level":"","writing_category":"","pri":3},{"name":"arsalan","user_id":"3","writing_level":"","writing_category":"Nature,Education,Health and Medicine,Communications and Media","pri":3}]

這是一個有效的JSON對象

但是當我使用

var writer=JSON.parse(xmlhttp.responseText);

我得到錯誤誰能plz幫助，並解釋如何，我可以解決這個

感謝

我使用Firefox和螢火蟲它只是顯示錯誤的JSON.parse

來源

2011-03-25 shazia

**你得到了什麼**錯誤？ – Quentin 2011-03-25 07:18:49

「我得到錯誤」不是診斷。 – 2011-03-25 07:19:02

什麼瀏覽器會出錯？ – 2011-03-25 07:22:26

設置適當的內容類型，在PHP中：

header('Content-type: application/json'); 
echo json_encode($writers);

來源

2011-03-25 07:20:28

儘管應該完成這個工作，但並不清楚它並沒有什麼區別，因爲瀏覽器最終會以xhr.responseText中的一個未分類的字符串結束。 – Quentin 2011-03-25 07:22:17

爲什麼選擇投票？ – 2011-03-25 07:34:12

我第二... – 2011-03-25 09:04:08

仔細檢查使用jsonLint

我快速複製並粘貼您的JSON到它，並得到了一些錯誤，你的JSON - 確保日ere在json字符串中沒有換行符，因爲這會使json無效。

來源

2011-03-25 15:53:22 Matt27

JSON.parse不起作用

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