[英]how should i put data fetched from ajax call in hidden div box
我正在一個項目上,遇到一個模塊。
第1頁
用戶必須從搜索欄進行搜索,這會將他帶到第2頁。
第2頁
在第2頁上,所有提取的結果都將以div的形式顯示給用戶。 每個結果都有一個與之關聯的復選框。
當我單擊“添加以進行比較”復選框時,將執行ajax調用,並且所提取的選定結果應顯示在隱藏的div中。
我的問題是,它僅在隱藏的div中顯示第一個結果,而不與另一個結果一起使用。
我的第2頁代碼
<script type="text/javascript">
$(document).ready(function()
{
var check = $('#compare').val();
$("#compare").change(function() {
if(this.checked) {
$.ajax({
type: 'POST',
url: 'compare.php',
dataType : 'JSON',
data:{value : check},
success: function(data)
{
console.log(data);
$('#compare_box').html(data);
}
});
$("#compare_box").show();
}
else
{
$("#compare_box").hide();
}
});
});
</script>
</head>
<body>
<?php
$query = $_GET['search_bar'];
$query = "call fetch_data('$query')"or die(mysqli_error($conn));
$result = mysqli_query($conn,$query);
while($row = mysqli_fetch_array($result))
{
$id = $row['course_id'];
$title = $row['course_title'];
$description = $row['course_description'];
$course_url = $row['course_url'];
$video_url = $row['course_video_url'];
$fee = $row['course_fee'];
$duration = $row['course_duration'];
$start_date = $row['course_start_date'];
$university = $row['university_name'];
$course_provider = $row['course_provider_name'];
$instructor = $row['instructor_name'];
$_SESSION['result'][$id] = Array('id'=> $id,'course_title' => $title,'course_description'=> $description,'course_url' => $course_url,'video_url' => $video_url,'fee' => $fee,'course_duration'=>$duration,'start_date'=>$start_date,'university' => $university,'course_provider'=>$course_provider,'instructor'=>$instructor);
?>
<div id='compare_box'>
</div>
<div class="col-md-3 photo-grid " style="float:left">
<div class="well well-sm">
<a href="final.php?id=<?php echo $id;?>&name=<?php echo $title;?>" target="_blank">
<h4><small><?php echo $title; ?></small></h4>
</a>
<br>
<input type ='checkbox' name="compare" id="compare" value="<?php echo $id;?>">add to compare
</div>
</div>
<?php
}
?>
page3 compare.php
<?php
session_start();
include 'includes/dbconfig.php';
$check = $_POST['value'];
$sql = "SELECT * from course_info_table where course_id = '$check' " or die(mysqli_error($conn));
$result = mysqli_query($conn,$sql);
$index = 0;
while($row = mysqli_fetch_array($result))
{
$title = $row['course_title'];
?>
<?php
}
echo json_encode($title);
?>
你可以改變
<input type ='checkbox' name="compare" id="compare" value="<?php echo $id;?>">
至
<input type ='checkbox' name="compare" class="compare" value="<?php echo $id;?>"> ^you can only have one unique 'id' value in your html doc, which means your first id="compare" will work fine and others with id="compare" will be ignored by the DOM tree
參考:
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