我一直在這個小時裏撓了撓頭;新來的PHP,請親切。將PHP表格連接到兩個表中的鏈接
我試過多種連接方法和一些工作有的沒有,但我想在一次從一個單一的形式保存到兩個表,但第二個表具有第一表的主鍵作爲外鍵。
對於原因,我無法找到,它只是不想要工作,並顯示第二個錯誤消息(失敗)。任何幫助將大大接近。
HTML片段 - 除非您是使用BULK INSERT忽略了垃圾HTML
<form action="savedata.php" method="POST" >
<div class="form-group">
<label>Full Company Name</label>
<input class="form-control" name="company_name" placeholder="For example - Aviva UK Zone PLC">
</div>
<div class="form-group">
<label>Company URL(s)</label>
<textarea class="form-control" rows="3" name="company_url" placeholder="Place a new domain on each line without http or www."></textarea>
</div>
<hr>
<div class="form-group">
<label>POC Name</label>
<input class="form-control" name="company_poc_name"placeholder="Point of contact name">
</div>
<div class="form-group">
<label>POC Telephone</label>
<input class="form-control" name="company_poc_telephone" placeholder="Point of contact telephone">
</div>
<div class="form-group">
<label>POC Email</label>
<input class="form-control" name="company_poc_email" placeholder="Point of contact email">
</div>
</div>
</div>
</div>
<div class="panel panel-info">
<div class="panel-heading">
New User Information
</div>
<div class="panel-body">
<p>This user will be hard-paired with the company above; the user will not be emailed until you complete a dataset to the company.</p> <div class="form-group">
<label>Full User Name</label>
<input class="form-control" name="user_name" placeholder="Full User Name i.e. Joe Blogs">
</div>
<div class="form-group">
<label>Temp Password</label>
<input class="form-control" name="user_password" placeholder="User Work Email Address">
</div>
<div class="form-group">
<label>User Email</label>
<input class="form-control" name="user_email" placeholder="User Work Email Address">
</div>
</div>
</div>
<div class="panel panel-danger">
<div class="panel-heading">
Complete New Company + User
</div>
<div class="panel-body">
<p>The company and user will be created, you must then add a data set to pair with the company you have just created.</p> <div class="form-group">
<input type="submit" class="btn btn-default btn-lg btn-block" value="Save new Company"></input>
</form>
PHP
<?php mysql_connect("localhost","user","password");
mysql_select_db("corporat_account");
$order = "INSERT INTO corp_companies (company_name, company_url, company_poc_name, company_poc_telephone, company_poc_email)
VALUES ('$_POST[company_name]','$_POST[company_url]','$_POST[company_poc_name]','$_POST[company_poc_telephone]','$_POST[company_poc_email]')
INSERT INTO corp_users (comp_id, user_name, user_password, user_email)
VALUES ('1','$_POST[user_name]','$_POST[user_password]','$_POST[user_email]')";
$result = mysql_query($order);
if($result){
echo("<br>Input data is succeed");
} else{
echo("<br>Input data is fail");
}
?>
你不想使用MySQL的** _ ***爲您查詢了。這是貶值。檢查MySQLi或PDO。另外,您不應該像用戶那樣直接將用戶數據放入查詢中。您需要使用[real_escape_string](http://uk3.php.net/mysqli_real_escape_string)或使用[prepared statements](http://php.net/manual/en/mysqli.prepare.php)將其轉義防止SQL注入。 – Styphon