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我的網站上有一個畫廊,當用戶從下拉菜單中選擇一個選項(選項代表位置)時需要更改。AJAX php gallery
目前我有以下兩種選擇:
<select class="form-control" id="photomenu">
<option value="Beckmouth Area">Beckmouth Area</option>
<option value="Barras Square area">Barras Square area</option>
</select>
我的jQuery的選擇選擇的值,並將其發送到我的AJAX請求,像這樣:
<script>
$(document).ready(function() {
$("#photomenu").change(function(){
var option = $("#photomenu option:selected").val();
var url = "access_database.php";
$.post(url, option, function(data)
{
$("#staithes_gallery").html(data)
});
});
});
</script>
我access_database.php
樣子:
<?php
$folder_location = $_GET["option"];
$path = "photos_staithes/Thumbnails/";
//connect to staithesbooks table
$servername = "localhost";
$username = "***********";
$password = "***********";
$dbname = "*********";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql_select_photo = "SELECT * FROM `staithesphotos` WHERE `folder` = '$folder_location'";
$results = $conn->query($sql_select_photo);
if ($results->num_rows > 0) {
// output data of each row
while($row = $results->fetch_assoc()) {
//create a list of all names and folder locations
$name[] = $row["name"];
}
//create the photo gallery
for ($x=0; $x<count($name); $x++)
{
$photo_gallery .= "<div class='col-md-4 work-grid animated wow slideInLeft'>
<a href='$path$folder_lcation$name[$x]' data-lightbox='featured' data-title='' style='cursor:pointer'><div class='item'>
<img src='$path$folder_location/$name[$x]' height='431' width='594' title='name' />
<div class='caption' style='display: none;'>
<h2>Some Title</h2>
<p>This is a caption to end all captions</p>
</div>
</div>
</a></div>";
}
};
echo $photo_gallery;
?>
我已經測試PHP
的工作原理是爲$folder_location
提供一個靜態文件位置。
在執行AJAX call
後,我希望用新庫更新#staithes_gallery
中的html。
通過測試並將結果輸出到警告框,我只需接收對象Object。我不確定我做錯了什麼。
任何幫助將不勝感激!
等等,這裏你真正的問題是什麼?你無法得到PHP返回的請求或$'folder_location'值或查詢的任何問題? – DontVoteMeDown
我不確定問題出在哪裏。 jquery只是返回對象Object而不是新的HTML文本。我已經單獨測試了每個組件,並且似乎可行。我不確定這個值是不是正確的PHP或從PHP返回的數據是錯誤的 – Jimmybob
設置'$ sql_select_photo'後'把'echo $ sql_select_photo;退出;'並直接在瀏覽器中調用URL。然後在你的數據庫中檢查你的查詢。 – DontVoteMeDown