如何在不同的 php 頁面上顯示輸出，非常感謝您的幫助，謝謝<?php$connect=mysqli_connect("localhost","root","","test"); $sql = "SELECT * FROM icheck "; $result = mysqli_query($connect, $sql); //$row = mysqli_fetch_array($result);?>//MY FIRST PAGE<form method="POST"><table> <thead> <tr> <th>ID</th> <th>Name</th> <th>Price</th> <th>Check</th> </tr> </thead> <?php if(mysqli_num_rows($result)>0){ $i=1; while($row=mysqli_fetch_array($result)) {echo'<tr> <td>'.$row['id'].'</td> <td>'.$row['name'].'</td> <td>'.$row['price'].'</td> <td><input type="checkbox" name="check[]" value='.$row['id'].'></td> '; $i++; } } ?></table><input type="submit" name="submit" Value="Submit"/></form>//OUTPUT TO BE DISPLAYED ON OTHER PAGE or 2nd PAGE<?php if(isset($_POST['submit'])){if(!empty($_POST['check'])) {// Counting number of checked checkboxes.$checked_count = count($_POST['check']);echo "You have selected following ".$checked_count." option(s): <br/>";if(isset($_POST['check'])){$sum=0;foreach($_POST['check'] as $selected) { $ah="SELECT * FROM icheckWHERE id=".$selected; $rowz = mysqli_query($connect, $ah); $row=mysqli_fetch_array($rowz); echo $row['name']."</br>"; $test=$row['price']; $sum+=(int)$test;}echo "Jay Total na ket: ".$sum;}}else{echo "<b>Please Select Atleast One Option.</b>";}}?>
2 回答
侃侃爾雅
TA貢獻1801條經驗 獲得超16個贊
您在第一頁中的表單不包含表單操作，因此您必須將表單操作放到第二頁，然后可以從第二頁捕獲發布的數據。
//我的第一頁
<form method="POST" action="mysecondpage.php">
<table>
<thead>
<tr>
<th>ID</th>
<th>Name</th>
<th>Price</th>
<th>Check</th>
</tr>
</thead>
<?php
if(mysqli_num_rows($result)>0){
$i=1;
while($row=mysqli_fetch_array($result))
{
echo'<tr>
<td>'.$row['id'].'</td>
<td>'.$row['name'].'</td>
<td>'.$row['price'].'</td>
<td><input type="checkbox" name="check[]" value='.$row['id'].'></td>
';
$i++;
}
}
?>
</table>
<input type="submit" name="submit" Value="Submit"/>
</form>
慕田峪4524236
TA貢獻1875條經驗 獲得超5個贊
起初 vardump 這個$_POST['check']。它是數組嗎，您的代碼可以工作。通過 vardump 變量，您可以找到解決方案。
您可以與我分享您的 vardump 結果。
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