我的代码是：（在回答建议后编辑）
UNKNOWN我希望从数据库中删除选中复选框的元素，包含在选中的[]数组中。在删除之前，我想要打开一个确认对话框，其中包含“提交”按钮。这将导致实际删除。但是，出于某种原因，上述代码不起作用。我甚至无法确定是否提交了帖子，因为行<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>SomuFinance - Personal Finance Manager</title>
<link rel="stylesheet" type="text/css" href="indexStyle.css">
<script src="scripts/jquery-3.1.0.min.js"></script>
</head>
<body>
<form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>">
<div id="container">
<input type="submit" class="button" name="edit" value="Edit" />
<input type="button" class="button" name="delete" value="Delete" />
<input type="text" id="action" name="action">
<table id="listDB">
<tr>
<th>Select</th>
<th>ID</th>
<th>Category ID</th>
<th>Shop</th>
<th>Item</th>
<th>Quantity</th>
<th>Unit</th>
<th>Price Based On</th>
<th>MRP</th>
<th>Seller's Price</th>
<th>Last Updated On</th>
</tr>
<?php
$dbc = mysqli_connect('localhost','root','atlantis2016','itemDB')
or die("Error Connecting to Database");
if(isset($_POST['submit']))
{
echo "Action Set to ".$_POST['action'];
if($_POST['action']=='confirmDelete')
{
echo "Now Deleting!!";
foreach ($_POST['selected'] as $delete_id)
{
$query = "DELETE FROM grocery WHERE id = $delete_id";
mysqli_query($dbc, $query)
or die('Error querying database.');
}
}
}
$query1 = "SELECT DISTINCT category FROM grocery";
$result1 = mysqli_query($dbc, $query1)
or die("Error Querying Database");
while($row = mysqli_fetch_array($result1))
{
$category = $row['category'];
$query2 = "SELECT * FROM grocery WHERE category='$category' ORDER BY item ASC";
$result2 = mysqli_query($dbc, $query2)
or die("Error Querying Database");
echo '<tr>';
echo '<td class="catHead" colspan=11>'.$category.'</td>';
echo '</tr>';
$catCount=1;
while($inRow = mysqli_fetch_array($result2))
{
$id = $inRow['id'];
$shop = $inRow['shop'];
$item = $inRow['item'];
$qnty = $inRow['quantity'];
$unit = $inRow['unit'];
$price_based_on = $inRow['price_based_on'];
$mrp = $inRow['MRP'];
$sellers_price = $inRow['sellers_price'];
$last_updated_on = $inRow['last_updated_on'];
echo '<tr>';
echo '<td><input type="checkbox" value="' . $id . '" name="selected[]" /></td>';
echo '<td>'.$id.'</td>';
echo '<td>'.$catCount.'</td>';
echo '<td>'.$shop.'</td>';
echo '<td class="leftAligned">'.$item.'</td>';
echo '<td>'.$qnty.'</td>';
echo '<td>'.$unit.'</td>';
echo '<td>'.$price_based_on.'</td>';
echo '<td class="pri">₹'.$mrp.'</td>';
echo '<td class="pri">₹'.$sellers_price.'</td>';
echo '<td>'.$last_updated_on.'</td>';
echo '</tr>';
$catCount++;
}
}
mysqli_close($dbc);
?>
<input type="submit" value="Submit">
</table>
</div>
<div class="dialogBG">
<div id="deleteConfirmDialog" class="dialog">
<div class="closeDialog"></div>
<p>Sure you want to delete the selected Data?</p>
<input type="submit" id="confirmDelete" class="dialogButton" name="edit" value="Delete" />
<input type="button" class="dialogButton cancelButton" name="delete" value="Cancel" />
</div>
</div>
</form>
<script type="text/javascript">
$(document).ready(function(){
$('.button').click(function(){
if($(this).val()=="Delete")
{
$(".dialogBG").fadeIn(200);
$("#deleteConfirmDialog").show(200);
$("#action").val('confirmDelete');
}
else if($(this).val()=="Edit")
{
}
});
$('#confirmDelete').click(function(){
$(".closeDialog").trigger("click");
});
$('#cancelDelete').click(function(){
});
$(".closeDialog").click(function (e){
$(this).parent(".dialog").hide('200').parent(".dialogBG").fadeOut('200');
});
$(".cancelButton").click(function (e){
$(this).parent(".dialog").hide('200').parent(".dialogBG").fadeOut('200');
});
$("form").submit(function(e){
alert("Form is being sumbitted!");
});
});
</script>
</body>
</html>
没有返回任何输出。请帮忙。
我相信整段代码都不起作用（来自人工测试）。
echo "Action Set to ".$_POST['action'];任何想法为什么？
答案 0 :(得分：1)
您需要将这些输入放在表单中并设置操作。如下所示：
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post" role="form">
<input></input>
<input></input>
</form>
这将是您的代码
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>SomuFinance - Personal Finance Manager</title>
<link rel="stylesheet" type="text/css" href="indexStyle.css">
<script src="scripts/jquery-3.1.0.min.js"></script>
</head>
<body>
<div id="container">
<form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>" role="form">
<input type="submit" class="button" name="edit" value="Edit" />
<input type="button" class="button" name="delete" value="Delete" />
<input type="text" id="action" name="action">
<table id="listDB">
<tr>
<th>Select</th>
<th>ID</th>
<th>Category ID</th>
<th>Shop</th>
<th>Item</th>
<th>Quantity</th>
<th>Unit</th>
<th>Price Based On</th>
<th>MRP</th>
<th>Seller's Price</th>
<th>Last Updated On</th>
</tr>
<?php
$dbc = mysqli_connect('localhost','root','atlantis2016','itemDB')
or die("Error Connecting to Database");
if(isset($_POST['submit']))
{
echo "Action Set to ".$_POST['action'];
if($_POST['action']=='confirmDelete')
{
echo "Now Deleting!!";
foreach ($_POST['selected'] as $delete_id)
{
$query = "DELETE FROM grocery WHERE id = $delete_id";
mysqli_query($dbc, $query)
or die('Error querying database.');
}
}
}
$query1 = "SELECT DISTINCT category FROM grocery";
$result1 = mysqli_query($dbc, $query1)
or die("Error Querying Database");
while($row = mysqli_fetch_array($result1))
{
$category = $row['category'];
$query2 = "SELECT * FROM grocery WHERE category='$category' ORDER BY item ASC";
$result2 = mysqli_query($dbc, $query2)
or die("Error Querying Database");
echo '<tr>';
echo '<td class="catHead" colspan=11>'.$category.'</td>';
echo '</tr>';
$catCount=1;
while($inRow = mysqli_fetch_array($result2))
{
$id = $inRow['id'];
$shop = $inRow['shop'];
$item = $inRow['item'];
$qnty = $inRow['quantity'];
$unit = $inRow['unit'];
$price_based_on = $inRow['price_based_on'];
$mrp = $inRow['MRP'];
$sellers_price = $inRow['sellers_price'];
$last_updated_on = $inRow['last_updated_on'];
echo '<tr>';
echo '<td><input type="checkbox" value="' . $id . '" name="selected[]" /></td>';
echo '<td>'.$id.'</td>';
echo '<td>'.$catCount.'</td>';
echo '<td>'.$shop.'</td>';
echo '<td class="leftAligned">'.$item.'</td>';
echo '<td>'.$qnty.'</td>';
echo '<td>'.$unit.'</td>';
echo '<td>'.$price_based_on.'</td>';
echo '<td class="pri">₹'.$mrp.'</td>';
echo '<td class="pri">₹'.$sellers_price.'</td>';
echo '<td>'.$last_updated_on.'</td>';
echo '</tr>';
$catCount++;
}
}
mysqli_close($dbc);
?>
</table>
</form>
</div>
<script type="text/javascript">
$(document).ready(function(){
$('.button').click(function(){
if($(this).val()=="Delete")
{
$(".dialogBG").fadeIn(200);
$("#deleteConfirmDialog").show(200);
$("#action").val('confirmDelete');
}
else if($(this).val()=="Edit")
{
}
});
$('#confirmDelete').click(function(){
$(".closeDialog").trigger("click");
});
$('#cancelDelete').click(function(){
});
$(".closeDialog").click(function (e){
$(this).parent(".dialog").hide('200').parent(".dialogBG").fadeOut('200');
});
$(".cancelButton").click(function (e){
$(this).parent(".dialog").hide('200').parent(".dialogBG").fadeOut('200');
});
});
</script>
<div class="dialogBG">
<div id="deleteConfirmDialog" class="dialog">
<div class="closeDialog"></div>
<p>Sure you want to delete the selected Data?</p>
<input type="submit" id="confirmDelete" class="dialogButton" name="edit" value="Delete" />
<input type="button" class="dialogButton cancelButton" name="delete" value="Cancel" />
</div>
</div>
</body>
</html>
答案 1 :(得分：1)
解决方案很简单。 if(isset($_POST['confirmDelete']))需要在问题的问题部分而不是if(isset($_POST['submit']))，因为没有名为＆＃34;提交＆＃34;的按钮。在整个形式。