我有一個用戶註冊表單 ,然後插入到數據庫但其拋出了一個綁定PARAM錯誤bind_param問題PHP編寫插入語句
if(isset($_POST['submit'])){
// Login Query
// Connection
$mysqli = new mysqli("localhost", "root", "", "pg");
if ($mysqli->connect_errno) {
echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error;
}
// Query
$query = "INSERT INTO `affiliates` (aid,affname,title,fname,lname,add1,add2,add3,city,county,country,pcode,email,password,paymethod) VALUES (,?,?,?,?,?,?,?,?,?,?,?,?,?,?)";
$stmt = $mysqli->prepare($query);
$stmt->bind_param('ssssssssssssss',$_POST['affname'],$_POST['title'],$_POST['fname'],$_POST['lname'],$_POST['add1'],$_POST['add2'],$_POST['add3'],$_POST['city'],$_POST['county'],$_POST['country'],$_POST['pcode'],$_POST['email'],$_POST['password'],$_POST['paymethod']); // If 2x Variables are used etc 's' would become 'ss',$_GET['VAR'],$_GET['VAR']
// Bind Results
$stmt->execute();
//$result = $stmt->get_result()->fetch_all();
}
我不能工作了什麼是錯誰能幫助 感謝
更新的表單&查詢代碼
<form id="form1" name="form1" method="post" action="<?php echo $_SERVER['PHP_SELF'];?>">
<p>
<label for="textfield"></label>
<input type="text" name="affname" id="textfield" />
</p>
<p>
<label for="textfield"></label>
<input type="text" name="title" id="textfield" />
</p> <p>
<label for="textfield"></label>
<input type="text" name="fname" id="textfield" />
</p> <p>
<label for="textfield"></label>
<input type="text" name="lname" id="textfield" />
</p> <p>
<label for="textfield"></label>
<input type="text" name="add1" id="textfield" />
</p> <p>
<label for="textfield"></label>
<input type="text" name="add2" id="textfield" />
</p> <p>
<label for="textfield"></label>
<input type="text" name="add3" id="textfield" />
</p> <p>
<label for="textfield"></label>
<input type="text" name="city" id="textfield" />
</p> <p>
<label for="textfield"></label>
<input type="text" name="county" id="textfield" />
</p> <p>
<label for="textfield"></label>
<input type="text" name="country" id="textfield" />
</p> <p>
<label for="textfield"></label>
<input type="text" name="pcode" id="textfield" />
</p> <p>
<label for="textfield"></label>
<input type="text" name="email" id="textfield" />
</p>
<p>
<label for="textfield"></label>
<input type="text" name="password" id="textfield" />
</p>
<p>
<label for="textfield"></label>
<input type="text" name="paymethod" id="textfield" />
<input type="hidden" value="" name="aid" />
</p>
<p>
<input type="submit" name="submit" id="submit" value="Submit" />
</p>
</form>
<?php
if(isset($_POST['submit'])){
// Login Query
// Connection
$mysqli = new mysqli("localhost", "root", "", "pg");
if ($mysqli->connect_errno) {
echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error;
}
// Query
$query = "INSERT INTO `affiliates` (aid,affname,title,fname,lname,add1,add2,add3,city,county,country,pcode,email,password,paymethod) VALUES (?,?,?,?,?,?,?,?,?,?,?,?,?,?,?)";
$stmt = $mysqli->prepare($query);
$stmt->bind_param('sssssssssssssss',$_POST['aid'],$_POST['affname'],$_POST['title'],$_POST['fname'],$_POST['lname'],$_POST['add1'],$_POST['add2'],$_POST['add3'],$_POST['city'],$_POST['county'],$_POST['country'],$_POST['pcode'],$_POST['email'],$_POST['password'],$_POST['paymethod']); // If 2x Variables are used etc 's' would become 'ss',$_GET['VAR'],$_GET['VAR']
// Bind Results
$stmt->execute();
//$result = $stmt->get_result()->fetch_all();
}
我修改了查詢,現在已經被傳遞 瓦爾正確的金額不過現在沒有顯示出錯誤,但也可以作爲你的援助autocremented只是查詢改變這一點,沒有必要不插入數據
錯誤信息是? –
該SQL無效,因此您的準備工作可能會失敗。我可以猜出錯誤。 '(,?,?,?,?,?,?,?,?,?,?,?,?'缺少結尾括號和開頭後的內容。 –