我有一個HTML表單,我通過jquery提交$ ajax();jQuery ajax表單上傳圖像&提交表單和保存值在MySQL分貝
形式有: 1.上傳圖像,到一個目錄,具有錯誤的檢查 2.保存圖像路徑到MySQL DB 3.將其它2個形態值:輸入+選擇值到DB 4。警告 - 致勝,在表單提交如果兩個圖像被上傳提交到DB成功
累了嘗試使用JSON的&值:
$.ajax({
url: 'submit.php',
type: 'POST',
data: data,
cache: false,
dataType: 'json',
processData: false, // Don't process the files
contentType: false, // Set content type to false as jQuery will tell the server its a query string request
success: function (data, textStatus, jqXHR) {
// lots of bullshit code
}
});
任何新鮮完整的工作代碼?哪些可以完成這些任務? 我已經發布了兩個尋求JSON幫助的問題,但似乎沒有人同意,與JSON作戰。
//////////////////////// HTML FORM CODE /////////////////// ////////
<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>Untitled Document</title>
</head>
<body>
<form name="fb_form" id="fb_form" method="post" action="" enctype="multipart/form-data">
<table class="sm" border="1px solid #CCCCCC">
<tr>
<td><strong>Facebook</strong></td>
<td><input type="file" name="fb_icon" style="border: solid 1px #90D8CF;"
placeholder="Upload Image"></td>
<td><input type="text" name="fb_link" style="border: solid 1px #90D8CF;"
placeholder="Paste redirect link"></td>
<td>
<select name="show_fb" class="myselect">
<option value="">---Select---</option>
<option value="1">Show</option>
<option value="0">Hide</option>
</select>
</td>
<td>
<button class="myButton" type="submit" id="fb_submit">UPDATE</button>
</td>
</tr>
</table>
</form><!--------FB Form ends here------------>
</body>
</html>
///////////////////// BASIC MySQL數據庫結構CODE IN submit.php ///// /////////////////
$filename_wpath = $imagepath;
$fb_link = $_POST['fb_link'];
$show_fb = $_POST['show_fb'];
$sql = "INSERT INTO `tblbasicheader`(fldFB_image, fldFB_link, fldHideShow)
VALUES('$filename_wpath','$fb_link','$show_fb')";
mysqli_query($db_conx, $sql);
任何猜測,什麼應該是好的代碼裏面:$ .ajax(); 感謝提前噸,作爲一個代碼戰鬥機:)