Php 未在选定的特定单选按钮上显示其他结果
使用4单选按钮作为选项，选择付款列，将所有数字相加。我 如果select等于first，则写入第一个查询；如果select等于first，则写入第二个查询。即使我选择了第二次付款，它也会在第一次付款时显示。是因为我在代码末尾的回音吗？有人能帮忙吗？是否应将Php 未在选定的特定单选按钮上显示其他结果,php,mysql,database,mysqli,Php,Mysql,Database,Mysqli,使用4单选按钮作为选项，选择付款列，将所有数字相加。我 如果select等于first，则写入第一个查询；如果select等于first，则写入第二个查询。即使我选择了第二次付款，它也会在第一次付款时显示。是因为我在代码末尾的回音吗？有人能帮忙吗？是否应将用于 HTML编码（paycal.HTML）： <body> <form name="input" action="paycal.php" method="post"> <fieldset> <lege
用于<body>
<form name="input" action="paycal.php" method="post">
<fieldset>
<legend>Which payment do you want to SUM up:</legend>
<input type="radio" name="payment" value="first" /> First
<input type="radio" name="payment" value="second" /> Second
<input type="radio" name="payment" value="third" /> Third
<input type="radio" name="payment" value="fourth" /> Fourth
</fieldset>
<input type="submit" value="Submit" />
</form>
</body>
</html>
<?php
error_reporting(E_ALL);
ini_set("display_errors", 1);
$pay=$_REQUEST['payment'];
// MySQL database connection, username, password, database name
$con=mysqli_connect("localhost","username","password","database_name");
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
if ($pay=="first") $query="SELECT *, SUM(first_payment) FROM `Customer Information`";
else if ($pay=="second") $query="SELECT *, SUM(second_payment) FROM `Customer Information`";
else if ($pay=="third") $query="SELECT *, SUM(third_payment) FROM `Customer Information`";
else if ($pay=="fourth") $query="SELECT *, SUM(fourth_payment) FROM `Customer Information`";
$result = mysqli_query($query);
// Executing and error checking of query
if (!mysqli_query($con,$query)) {
die('Error: ' . mysqli_error($con));
}
while($row = mysqli_fetch_array($result)) {
echo $row['first_payment'];
}
// Close MySQL
mysqli_close($con);
?>
您希望汇总哪笔付款：
弗斯特
第二
第三
第四
<body>
<form name="input" action="paycal.php" method="post">
<fieldset>
<legend>Which payment do you want to SUM up:</legend>
<input type="radio" name="payment" value="first" /> First
<input type="radio" name="payment" value="second" /> Second
<input type="radio" name="payment" value="third" /> Third
<input type="radio" name="payment" value="fourth" /> Fourth
</fieldset>
<input type="submit" value="Submit" />
</form>
</body>
</html>
<?php
error_reporting(E_ALL);
ini_set("display_errors", 1);
$pay=$_REQUEST['payment'];
// MySQL database connection, username, password, database name
$con=mysqli_connect("localhost","username","password","database_name");
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
if ($pay=="first") $query="SELECT *, SUM(first_payment) FROM `Customer Information`";
else if ($pay=="second") $query="SELECT *, SUM(second_payment) FROM `Customer Information`";
else if ($pay=="third") $query="SELECT *, SUM(third_payment) FROM `Customer Information`";
else if ($pay=="fourth") $query="SELECT *, SUM(fourth_payment) FROM `Customer Information`";
$result = mysqli_query($query);
// Executing and error checking of query
if (!mysqli_query($con,$query)) {
die('Error: ' . mysqli_error($con));
}
while($row = mysqli_fetch_array($result)) {
echo $row['first_payment'];
}
// Close MySQL
mysqli_close($con);
?>
您想用什么显示：
echo $row['first_payment'];
switch($pay)
{
case "first":
echo $row['first_payment'];
break;
case "second":
echo $row['second_payment'];
break;
case "third":
echo $row['third_payment'];
break;
case "fourth":
echo $row['fourth_payment'];
break;
default:
echo "nothing selected";
}
SELECT SUM(first) as paymentSum From ...
echo $row['paymentSum'];
$pay换句话说，这是非常糟糕的桌子设计。如果你有一个更好的设计，你就不会陷入这样的问题。我建议你改一下。这张桌子的设计有点不规范。我会先解决这个问题。是的，我理解，但是如果我的html提供了选择显示哪一笔付款的选项，那么如果用户选择第一笔付款，我将如何进行选择，那么它将显示第一笔付款，如果从html端选择了第二笔付款选项，则显示第二笔付款。您已经使用sql查询进行了切换<代码>如果（$pay==xxx）是，我必须在代码中这样做，但当我单击“立即提交”时，它仅显示第一笔付款，即使我选择显示第二笔付款。是的，因为您回显$row[“第一笔付款”]！！！您必须从数据库中回显相应的列；）谢谢你，但是说你上面写的只是一次创建多个输出。我只想要一个结果，而不是所有的结果，这就是我为什么要写
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