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我使用Laravel新數據,我有以下數據視圖顯示使用Ajax和jQuery
<div class="content-inside-main">
<div class="content-inside" id="content-inside-feedback">
<div class="row header-content space-div">
<div class="col-lg-1"><h5>#</h5></div>
<div class="col-lg-1"><h5>Member Id</h5></div>
<div class="col-lg-4"><h5>Question</h5></div>
<div class="col-lg-4"><h5>Reply</h5></div>
<div class="col-lg-1"><h5>Replied by</h5></div>
<div class="col-lg-1"><h5>Options</h5></div>
</div>
<div>
<hr class="line-div"/>
</div>
<?php
$questions= \App\Question::all();
?>
<?php
foreach ($questions as $question):
$id = $question->id;
$member_id = $question->user_id;
$body = $question->message;
$status=$question->replied;
$reply=$question->reply;
$user_id=$question->replied_id;
$member=\App\Member::find($member_id);
$m_id=$member->id;
$m_name=$member->nick_name;
$m_reg_time=$member->reg_time;
$m_unreg_time=$member->unreg_time;
$m_status=$member->unreg;
$m_group_id=$member->group;
$group=\App\Group::find($m_group_id);
$m_group_name=$group->name;
if($id != NULL) {
?>
<div class="row content-messages" >
<input type="hidden" id="count" value="{{$id}}"/>
<div class="col-lg-1"><?php echo $id; ?></div>
<div class="col-lg-1"><?php echo $member_id; ?></div>
<div class="col-lg-4"><?php echo $body; ?></div>
<div class="col-lg-4">
<?php
if($status == 0){
?>
<div class="according-form-container" id="reply-feedback-form_<?php echo $id; ?>">
<a class="btn-link show-add-form-div" data-toggle="collapse" data-parent="#reply-feedback-form_<?php echo $id; ?>" href="#reply-feedback-form_content_<?php echo $id; ?>" >
Reply
</a>
<div id="reply-feedback-form_content_<?php echo $id; ?>" class="collapse collapse-add-form">
<form class="form" id="reply-feedback_<?php echo $id; ?>" enctype="multipart/form-data" method="post" action="addreply">
{{csrf_field()}}
<div class="control-group">
<label class="control-label" for="description">Message: </label>
<div class="controls">
<input type="hidden" name="id" id="id" value="{{$id}}"/>
<input type="hidden" name="member_id" id="member_id" value="{{$member_id}}"/>
<input type="hidden" name="user_id" id="user_id" value="{{Auth::id()}}"/>
<textarea name="description" id="feedback-message_<?php echo $id; ?>" class="input-block-level" required></textarea>
<br/><br/>
<button id="submitfeedback_<?php echo $id . '_' . $member_id; ?>" type="submit" class="btn feedback-reply-submit-btn">Send</button>
</div>
</div>
</form>
<div id='preview_feedback_<?php echo $id; ?>'>
</div>
</div>
</div>
<?php
} else {?>
<div class="col-lg-4">{{$reply}}</div>
<?php
}
?>
</div>
<div class="col-lg-1">
<?php
if($user_id != null){
$user_name= DB::table('admin')->where('id',$user_id)->value('name');
echo $user_name;
}else {
echo 'None';
}
?>
</div>
<div class="col-lg-1">
<button id="view_member" name="view_member" class="btn btn-primary btn-sm" data-toggle="modal" data-target="#view_memeber"
>View
</button>
</div>
</div>
<hr class="line-div" />
<?php
}
endforeach;
?>
<div id="show"></div>
<div class="text-center navigation-paginator" id="paginator" >
</div>
</div>
我還有一個application.It填充問題表anytime.I想做的事。如果問題表有新的記錄,在不刷新的情況下在本頁面顯示它們。
像下面的截圖:
1.儘量避免使用''裏面model' view'你有控制器爲 2.您需要一個AJAX與間隔被調用,將檢查新或編輯記錄 其他選項是使用套接字+ Laravel + Redis所以套接字將推送到客戶端更新信息(但實施起來更復雜) – Froxz
如何做到這一點?我還沒有清楚的想法 – Lara
看看這裏https://laracasts.com/discuss/channels/general-discussion/using-ajax-for-real-time-update-in-laravel?page=1 – Froxz