2016-08-11 2 views
0

내 양식나는 테이블

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> 

내 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); 

    ?> 

확인하시기 바랍니다 내가 놓친 게 아무것도에 데이터를 삽입하지 않는 뭔가를 놓치고 있습니까? 어떤 오류도 보이지 않으며 테이블에 데이터를 삽입하지 않습니다! 하나의 PHP 파일에 두 개의 양식을 구현할 수 있습니까?

또한 itemSubmit 또는 priceSubmit 버튼을 누르면 해당 쿼리를 실행해야한다는 논리를 추가하려고했습니다. 또 하나는 테이블을 만들었고 테이블도 표시되지 않습니다.

+0

을 대체 옆으로, mysqli의 장점은 준비된 명령문의 사용을 준다, 그래서 활용할 것입니다. – Strawberry

+0

그래,하지만 여기 왜 테이블이 반복되지 않는지 확인할 수 있니? –

답변

-1

변경 $의 $query2 = "SELECT * FROM items";에 쿼리 2와 같이 <?php echo $_GET['$lastid'] ?><?php echo mysqli_num_rows($lastid); ?>

+0

아직 레코드의 삽입 없음 테이블의 반복 없음! –

0
<?php 
$servername = "localhost"; 
$username = "root"; 
$password = ""; 
$database = "stack11"; 

// 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 * FROM items"; 

$query3 = "INSERT INTO prices (itemid, shopid, price, itemurl, itemimage) VALUES ('".$_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); 

?> 

ITEM FORM 
<form method="POST" action="#"> 
    <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 mysqli_num_rows($lastid); ?> 
    <table class="table table-hover"> 
    <tbody> 
    <?php while($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="#"> 
    <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"/> 

+0

이것은 전체 코드이며, 저에게 잘 작동합니다. 데이터베이스 필드 이름과 테이블 이름을 확인하십시오. –

+0

Yess 그것은 지금 삽입하지만, 나는 테이블의 마지막 행을 반복하지 않습니다. check4를 확인하십시오 –

+0

그것은 나에게 마지막 행을 제공하므로 코드를 다시 확인하십시오. –