我知道在SO中多次询问过这类问题。但我在下面的代码中遇到错误。
我在模态框中有一个表单。
<form method="post" action="<?php $_SERVER["PHP_SELF"];?>" role="form">
<div class="modal-body">
<div class="row">
<label class="col-xs-3 topjob_desc" for="title">Active list :</label>
<div class="col-xs-8">
<select class="form-control" id="title">
<?php
$total_view = $mysqli->prepare("SELECT id,title FROM posting where id= ?");
$total_view ->bind_param("i", $id);
$total_view->execute();
$total_view->store_result();
$total_view->bind_result($id,$name);
while ($total_view->fetch()) {
$name = $name;
$id = $id;
echo '<option value="'.$id.'">'.$name.'</option>';
}
?>
</select>
</div></div>
<br />
<div class="row">
<label class="col-xs-3 topjob_desc" for="title1">Name</label>
<div class="col-xs-8">
<textarea class="form-control" rows="5" id="comment" name="comment">Message</textarea>
</div>
</div>
</div>
<div class="modal-footer">
<input type="submit" class="btn btn-primary" id="submit" name="submit" value="Send" />
<button type="button" class="btn btn-default" data-dismiss="modal">Close</button>
</div>
</form>
以下是表单提交的代码。这是在同一页面上。
if (isset($_POST['submit'])) {
$messages = $_POST['comment'];
$id = $_POST['id'];
$total_viewss = $mysqli->prepare("SELECT title,company_name FROM posting where emid= ? and jid=?");
$total_viewss ->bind_param("si", $id,$jid);
$total_viewss->execute();
$total_viewss->store_result();
$total_viewss->bind_result($title,$company);
while ($total_viewss->fetch()) {
$company = $company;
$title=$title;
}
$result = $mysqli->prepare("INSERT INTO user_messages (uid, profile_pic, company, title, id, messages,emid) VALUES (?, ?, ?, ?, ?, ?, ?)");
$result ->bind_param("issssii", $uid, $profile_pic, $company, $title, $jid, $messages,$emid);
$result->execute();
$result->store_result();
}
当我提交表单时,数据未保存在数据库中。我不知道我做错了什么。请指教。
请忽略错别字。