[英]Using mod_scorm_insert_scorm_tracks
我正在將我的App與Moodle mod_scorm_get_scorm_sco_tracks
並且mod_scorm_get_scorm_attempt_count
成功通過Ajax(XMLHttpRequest)為給定用戶( userid
)調用了mod_scorm_get_scorm_sco_tracks
和mod_scorm_get_scorm_attempt_count
。
現在,我希望我的應用將一些SCORM曲目推回Moodle。 所以我試圖使用mod_scorm_insert_scorm_tracks
但沒有成功。 問題在於該方法沒有使用userid
參數,因此我不理解如何使用它(如果嘗試將userid
添加到輸入參數中,則會得到invalid parameter exception
)。
通過發送以下命令,我獲得了一些成功(沒有錯誤消息):
scoid = 206&attempt = 2&tracks [0] [element] = cmi.completion_status&tracks [0] [value] = completed&tracks [1] [element] = cmi.interactions.0.id&tracks [1] [value] = multiplechoice_page_1_1&tracks [2] [element] ] = cmi.interactions.0.learner_response&tracks [2] [value] = White&tracks [3] [element] = cmi.interactions.0.result&tracks [3] [value] = correct&tracks [4] [element] = cmi.interactions。 0.description&tracks [4] [value] = Which%20color%20was%20Garibaldi's%20white%20horse%3F&tracks [5] [element] = cmi.interactions.1.id&tracks [5] [value] = hotobject_page_2_1&tracks [6] [element] ] = cmi.interactions.1.learner_response&tracks [6] [value] = butterfly&tracks [7] [element] = cmi.interactions.1.result&tracks [7] [value] = incorrect&tracks [8] [element] = cmi.interactions。 1.description&tracks [8] [value] =%20is%20the%20fish%3F&tracks [9] [element] = cmi.score.max&tracks [9] [value] = 2&tracks [10] [element] = cmi.score。 raw&tracks [10] [value] = 1&tracks [11] [element] = cmi.score.scaled&tracks [11] [value] = 0.5&tracks [12] [element] = cmi.session_time&tracks [12] [value] = PT0H0M15S&tracks [ 13] [element] = timemodified&tracks [13] [value] = 1480947821&tracks [14] [element] = userid&tracks [14] [value] = 26&tracks [15] [element] = scoid&tracks [15] [value] = 206&wstoken = 69f2471506c4c49ff47cd0de0c4c9f01 mod_scorm_insert_scorm_tracks&moodlewsrestformat = json
但是,由於我無法指定這些數據所屬的用戶,因此我的用戶的嘗試不會更新(可預測)。 這是Moodle的回應:
{“ trackids”:[44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59],“警告”:[]}
我嘗試將userid
信息插入到Traks中( tracks[14][element]=userid&tracks[14][value]=26
),但還是沒有運氣。
因此,問題是:
userid
? 用戶身份來自完全登錄Moodle的HTTP上下文:您不能代表任何用戶(實際登錄的用戶)提供SCORM跟蹤信息。
更多信息:
HTH,
馬泰奧
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.