我想在PHp上创建购物车,
代码很简单,当客户填写QTY并单击按钮添加到购物车时,代码会将产品ID和数量保存到购物车表。但问题是循环中的Form。如何仅从客户点击按钮获取ID和数量。
程序看起来像这样
和这样的剧本
<?php
if(isset($_POST[ADD]))
{
$qty = $_POST[QTY];
$harga = $_POST[HARGA_ASLI];
$id = $_POST[ID];
print_r($_POST);
}
$kolom = 3;
$sql = "SELECT *,FORMAT(harga,0)AS harga_digit FROM item";
$hasil = mysql_query($sql);
echo "<form method=POST action=index.php>";
echo "<table>
<tr>";
$i = 0;
while($data=mysql_fetch_array($hasil))
{
if($i >= $kolom)
{
echo "</tr><tr>";
$i = 0;
}
$i++;
echo "<td align='center'><br><a href='detailBarang.php?ID=$data[ID]'><img src='$data[img]' width='200' height='150'/><br>$data[nama_produk]</a><br>
Rp. $data[harga_digit]<br>
<input type='submit' name='ADD' id='ADD' value='Add to Cart'>
<input type='text' name='QTY' id='QTY' placeholder='Qty' /><br>
<input type='hidden' name='HARGA_ASLI' id='HARGA_ASLI' value='$data[harga]' /><br>
<input type='hidden' name='ID' id='ID' value='$data[ID]' />
<br></td>";
}//end of while
echo "<tr></table>";
echo "</form>";
?>
如果我填写数量并单击添加到购物车,则只有最后一个项目可以发布数据。
如何仅为客户选择发布数据?
我非常感谢您的回答。
由于
答案 0 :(得分:1)
首先,让我们将您的MySQL转换为MySQLi。注释/ * * /中的更多解释:
<?php
$connection=mysqli_connect("YourHost","YourUsername","YourPassword","NameofYourDatabase");
if(mysqli_connect_errno()){
echo "Error".mysqli_connect_error();
}
$res=mysqli_query($con,"SELECT * FROM item");
while($row=mysqli_fetch_array($res)){
$nameofsubmitbutton=$row['ID'];
if(isset($_POST[$nameofsubmitbutton])){
$nameofproduct=$row['namaproduk'];
$nameofnumbersubmitted=$nameofsubmitbutton."number";
$quantity=$_POST[$nameofnumbersubmitted];
if(empty($quantity)){
echo "You wanted to buy a ".$nameofproduct."?<br>Type in a number so you can add it to your cart.";
}
else {
mysqli_query($connection,"INSERT INTO yourTable ('','') VALUES ('$quantity','$nameofproduct')");
echo "You bought ".$quantity." of ".$nameofproduct;
}
} /* END OF IF ISSET */
} /* END OF WHILE LOOP $RES */
$kolom = 3;
$hasil = mysqli_query($connection,"SELECT *,FORMAT(harga,0) AS harga_digit FROM item"); /* YOU SURE WITH THIS QUERY? */
echo "<form method=POST action=''>"; /* SUBMIT ON ITSELF */
echo "<table><tr>";
$i = 0; /* THIS WOULD ALSO SET AS YOUR COUNTER */
while($data=mysqli_fetch_array($hasil))
{
$id=$data['ID'];
if($i >= $kolom){
echo "</tr><tr>";
$i = 0;
} /* END OF IF $i >= $KOLOM */
$i++;
echo "<td align='center'><br><a href='detailBarang.php?ID=$data[ID]'><img src='$data[img]' width='200' height='150'/><br>".$data[nama_produk]."</a><br>Rp. ".$data[harga_digit]."<br>"; /* IF TO ECHO VARIABLES, USE ".$variable." */
$numbername=$id."number";
echo "<input type='number' name='$numbername' id='QTY' placeholder='Qty' /><br>"; /* CHANGE YOUR INPUT TYPE TO NUMBER */
/* NO NEED FOR THE HIDDEN INPUT */
echo "<input type='submit' name='$id' id='ADD' value='Add to Cart'></td>"; /* CHANGE THE NAME OF SUBMIT BUTTON TO THE CORRESPONDING ID FROM YOUR TABLE */
} /* END OF WHILE LOOP */
echo "<tr></table>";
echo "</form>";
?>
我在我的本地计算机上试过了。你也应该。
这是一个示例屏幕截图。
答案 1 :(得分:0)
这可以解决问题:
<?php
$kolom = 3;
$sql = "SELECT *,FORMAT(harga,0)AS harga_digit FROM item";
$hasil = mysql_query($sql);
while($data=mysql_fetch_array($hasil))
{
if(isset($_POST['ADD'.$data[ID]]))
{
$qty = $_POST['QTY'.$data[ID]];
$harga = $_POST['HARGA_ASLI'.$data[ID]];
$id = $_POST['ID'.$data[ID]];
print_r($_POST);
}
}
echo "<form method=POST action=index.php>";
echo "<table>
<tr>";
$i = 0;
while($data=mysql_fetch_array($hasil))
{
if($i >= $kolom)
{
echo "</tr><tr>";
$i = 0;
}
$i++;
echo "<td align='center'><br><a href='detailBarang.php?ID=$data[ID]'><img src='$data[img]' width='200' height='150'/><br>$data[nama_produk]</a><br>
Rp. $data[harga_digit]<br>
<input type='submit' name='ADD'".$data[ID]." id='ADD' value='Add to Cart'>
<input type='text' name='QTY'".$data[ID]." id='QTY' placeholder='Qty' /><br>
<input type='hidden' name='HARGA_ASLI'".$data[ID]." id='HARGA_ASLI' value='$data[harga]' /><br>
<input type='hidden' name='ID' id='ID'".$data[ID]." value='$data[ID]' />
<br></td>";
}//end of while
echo "<tr></table>";
echo "</form>";
?>