是否将选中项插入另一个sql表?(PHP和SQL)
大家好,我是新来的!我想知道当选中特定复选框时,如何插入打印行的值 这是我的密码:是否将选中项插入另一个sql表?(PHP和SQL),php,sql,Php,Sql,大家好,我是新来的!我想知道当选中特定复选框时,如何插入打印行的值 这是我的密码: <?php $query = "SELECT * FROM phpmyreservation_forvalidation"; $result = mysql_query($query); ?> <form action='insertrecords.php' method='post'> <?php echo "<table cellpadding='0' cellspac
<?php
$query = "SELECT * FROM phpmyreservation_forvalidation";
$result = mysql_query($query);
?>
<form action='insertrecords.php' method='post'>
<?php
echo "<table cellpadding='0' cellspacing='0' border='1'>";
echo "<tr> <th>reservation_id</th> <th>reservation_user_name</th> <th>reservation_day</th> <th>reservation_week</th>
<th> # </th></tr>";
while($row = mysql_fetch_array( $result )) {
echo "<tr><td>";
echo $row['reservation_id'];
echo "</td><td>";
echo $row['reservation_user_name'];
echo "</td><td>";
echo $row['reservation_day'];
echo "</td><td>";
echo $row['reservation_week'];
echo "</td><td align='center'>";
echo "</td>";
?>
<td><input name="checkbox[]" type="checkbox" id="checkbox[]" value="<?= $row['id'] ?>" ></td>
<?php
}
echo "</table>";
?>
<input type='submit' value='Submit' />
</form>
将“phpmyreservation\u for Validation”结果提取到变量中,并在提交表单时将其插入到“phpmyreservation\u Reservation”表中。在records.php中:
$row_id=$_POST['checkbox'];
foreach ($row_id as $key=>$val)
{
//do insert here with $val
}
如果我是您,我将使用AJAX调用从数据库中获取所需的所有数据,并将数据插入到新表中。在下面的示例中,我使用jQuery进行ajax调用
我的示例网页代码:
<html>
<head>
<meta charset="utf-8" />
<meta name="viewport" content="width=device-width" />
<title>Title of the page</title>
<script src="jquery-1.10.2.min.js"></script>
<script>
// Function beeing called when a checkbox is changed
function MyCheckboxChanged(sender)
{
// Send the data to be processed to the server if the checkbox is checked
if (sender.checked)
{
var MyObject = {};
MyObject.ValueToInsert = sender.value;
DoAJAXData("insert", MyObject);
}
}
// Function responsible for AJAX calls - requires jQuery
function DoAJAXData(command, MyObject)
{
$.post("phpReceiver.php",
{
command:command,
sendObject:JSON.stringify(MyObject)
})
.success(function(data)
{
// Process the data echoed from php
// In my case, should alert: "successfully inserted data: " + your selected checkbox value
alert(data);
})
.fail(function(error){
//alert("Unable to retrieve data from the server");
});
}
</script>
</head>
<body>
<?php
for ($i=0; $i < 5; $i++)
{
$var = <<<CHB
<input type="checkbox" onchange="MyCheckboxChanged(this);" name="$i" value="Bike number $i">I have a bike number $i</br>
CHB;
echo $var;
}
?>
</body>
</html>
书名
//更改复选框时调用函数
函数MyCheckboxChanged(发送方)
{
//如果选中该复选框,则将要处理的数据发送到服务器
如果(发件人已选中)
{
var MyObject={};
MyObject.ValueToInsert=sender.value;
数据(“插入”,MyObject);
}
}
//负责AJAX调用的函数-需要jQuery
函数DoAJAXData(命令,MyObject)
{
$.post(“phpreciver.php”,
{
命令:命令,,
sendObject:JSON.stringify(MyObject)
})
.成功(功能(数据)
{
//处理从php回显的数据
//在我的情况下,应该提醒:“成功插入数据:”+您选择的复选框值
警报(数据);
})
.失败(功能(错误){
//警报(“无法从服务器检索数据”);
});
}
以及phpreciver.php代码:
<?php
$command = $_POST['command'];
$object = json_decode($_POST['sendObject'], true);
if ($command == "insert")
{
$valueToInsert = $object['ValueToInsert'];
// Do you insertion to the database here
// Echo any results if wanted
echo "successfully inserted data: $valueToInsert";
}
?>
谢谢!插入值是这样的吗?mysql_查询(“插入到“'phpmyreservation_reservations.”(预订时间、预订年份、预订周、预订日、预订时间、预订价格、预订用户id、预订用户电子邮件、预订用户名称)值(“$val”)”;请详细说明或解释您的问题。事实上,我知道了,我需要做的是使每一行都有一个隐藏的输入类型,并将其值放入变量中,以将其插入数据库:DThanks!下次我做类似的操作时,我肯定会这样做