如果值匹配，则选中复选框，并在提交时使用复选框值更新数据库
[英]If value match then checkbox checked and update database with checkbox values on submit
问题描述
我正在尝试如果值在数据库显示复选框中被选中。 我检查了其他值，然后单击“提交”按钮以更新数据库，其中更新值为复选框。
这是我的代码：
<?php
include "database_connection.php";
$cutomername = $_GET['username'];
$productid = $_GET['userid'];
$cust_id = $_GET['custid'];
$productid_arr = explode(',' , $productid);
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title> <?php echo $cutomername;?></title>
</head>
<body>
<form method="post" name="checkboxForm">
<?php
$query_flags = "select * from products";
$row_flags=mysql_query($query_flags);
$i=0;
while ($row = mysql_fetch_array($row_flags)){
$product_id = $row['product_id'];
$product= $row['product'];
if($productid_arr[$i] ==$row['product_id'])
{
$check = 'checked="checked"';
} else{
$check = '';
}
if(in_array($row['product_id'], $chvalues))
{
echo "<input type=\"checkbox\" name=\"checkbox[]\" id=\"$product_id\" value='" . $row['product_id'] . "' $check />";
} else{
echo "<input type=\"checkbox\" name=\"checkbox[]\" id=\"$product_id\" value='" . $row['product_id'] . "' $check/>";
}
echo $row['product'];
echo "<hr />";
$i++;
}
?>
<input class="submit" type="submit" value="Submit" name="submit">
<?php
$chvalues = array();
if(isset($_POST['checkbox']))
{
foreach($_POST['checkbox'] as $ch => $value)
{
$chvalues[] = $value;
}
}
$des_prod_id = implode(',' , $chvalues);
$query = mysql_query("UPDATE customer SET product_id = '$des_prod_id' where ID = '$cust_id'");
?>
</form>
</body>
</html>
2 个解决方案
解决方案1
0 2013-09-09 14:07:24
更改此行：
$query = mysql_query("UPDATE customer SET product_id = $products where customer_name = $cutomername");
至：
$query = mysql_query("UPDATE customer SET product_id = $value where customer_name = $cutomername");
$products应该是$value因为您要在foreach循环中迭代并将每个元素分配给此变量。
解决方案2
0 已采纳 2013-09-17 12:03:38
终于我解决了这个问题
这是我的代码：
<?php
include "database_connection.php";
$cutomername = $_GET['username'];
$productid = $_GET['userid'];
$cust_id = $_GET['custid'];
$productid_arr = explode(',' , $productid);
if(isset($_POST['submit']))
{
$productid_arr = array();
foreach($_POST['checkbox'] as $ch => $value)
{
$productid_arr[] = $value;
}
$des_prod_id = implode(',' , $productid_arr);
$query = mysql_query("UPDATE customer SET product_id = '$des_prod_id' where ID = '$cust_id'");
}
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title> <?php echo $cutomername;?></title>
</head>
<body>
<form method="post" name="checkboxForm">
<?php
$query_flags = "select * from products";
$row_flags=mysql_query($query_flags);
$i=0;
while ($row = mysql_fetch_array($row_flags)){
$product_id = $row['product_id'];
$product= $row['product'];
if(in_array($product_id, $productid_arr))
{
$check = 'checked';
} else{
$check = '';
}
echo "<input type=\"checkbox\" name=\"checkbox[]\" id=\"$product_id\" value='" . $row['product_id'] . "' $check />";
echo $row['product'];
echo "<hr />";
$i++;
}
?>
<input class="submit" type="submit" value="Submit" name="submit">
</form>
</body>
</html>
复选框值匹配，但现在显示已选中复选框
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