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我的形式我失去了一些它不能將數據插入到表
ITEM FORM
<form method="POST" action="index.php">
<div class="col-xs-4">ITEM ID<input type="text" name="itemid" class="form-control"/></div>
<div class="col-xs-4">ITEM NAME<input type="text" name="itemname" class="form-control"/></div>
<div class="col-xs-4">ITEM DETAIL<input type="text" name="itemdetail" class="form-control"/></div>
ITEM DESCRIPTION<input type="text" name="itemdescription" class="form- control"/>
<div class="col-xs-4">MANUFACTURER ID <input type="text" name="manufacturerid" class="form-control"/></div>
<div class="col-xs-4">TYPE ID <input type="text" name="typeid" value="4001" class="form-control"/></div>
<div class="col-xs-4">CATEGORY ID <input type="text" value="1003" name="categoryid" class="form-control"/></div>
<div class="col-xs-4">MODULE ID <input type="text" name="moduleid" class="form-control"/></div>
<input type="submit" name="itemSubmit" class="btn btn-default"/>
</form>
<?php echo $_GET['$lastid'] ?>
<table class="table table-hover">
<tbody>
<?php while($row = mysqli_fetch_array($allresult)) { ?>
<tr>
<td><?php echo $row['itemid']?></td>
<td><?php echo $row['itemname']?></td>
<td><?php echo $row['itemdetail']?></td>
<td><?php echo $row['manufacturerid']?></td>
<td><?php echo $row['moduleid']?></td>
</tr>
<?php } ?>
</tbody>
</table>
PRICE FORM
<form method="POST" action="index.php">
<div class="col-xs-4">ITEM ID <input type="text" name="itemid" class="form-control"/></div>
<div class="col-xs-4">SHOP ID <input type="text" name="shopid" class="form-control"/></div>
<div class="col-xs-4">PRICE <input type="text" name="price" class="form-control"/></div>
ITEM URL <input type="text" name="itemurl" class="form-control"/>
ITEM IMAGE <input type="text" name="itemimage" class="form-control"/>
<input type="submit" name="priceSubmit" class="btn btn-default"/>
</form>
我的PHP
<?php
$servername = "localhost";
$username = "abc";
$password = "abc";
$database = "cd";
// Create connection
$conn = new mysqli($servername, $username, $password, $database);
// Check connection
if ($conn->mysqli_connect_error) {
die("Connection failed: " . $conn->mysqli_connect_error);
}
echo "Connected successfully";
$query1 = "
INSERT INTO items
(itemid
, itemname
, itemdetail
, itemdescription
, manufacturerid
, typeid
, categoryid
, moduleid
) VALUES
('".$_POST['itemid']."'
,'".$_POST['itemname']."'
,'".$_POST['itemdetail']."'
,'".$_POST['itemdescription']."'
,'".$_POST['manufacturerid']."'
,'".$_POST['typeid']."'
,'".$_POST['categoryid']."'
,'".$_POST['moduleid']."'
)";
$query2 = "SELECT COUNT(itemid) FROM products";
$query3 = "INSERT INTO prices (itemid, shopid, price, itemurl, itemimage VALUES ('".$_POST['itemid']."','".$_POST['shopid']."','".$_POST['price']."','".$_POST['itemurl']."','".$_POST['itemimage']."')";
$query4 = "SELECT * FROM items ORDER BY itemid DESC LIMIT 1";
if(isset($_POST['itemSubmit']))
{
mysqli_query($conn, $query1);
}
else if(isset($_POST['priceSubmit']))
{
mysqli_query($conn, $query3);
}
$lastid = mysqli_query($conn, $query2);
$allresult = mysqli_query($conn, $query4);
?>
請查看我失去了什麼？它不顯示任何錯誤，並且不會將數據插入到表中！我能否在一個php文件中實現兩種形式？
另外我試圖添加邏輯，如果我按itemSubmit或priceSubmit按鈕，那麼它應該運行相應的查詢。另一件事是我創建了一個表，它也不顯示。
順便說一句，mysqli的的美妙之處在於它提供了使用預處理語句，所以利用這一點。 – Strawberry
好的，但在這裏你可以檢查爲什麼表不迭代？ –