Question

 <?php 

    $DatabaseServer = "localhost";
    $DatabaseUsername = "root";
    $DatabasePassword = "root";
    $DatabaseName = "demo";

    $Connection = mysqli_connect($DatabaseServer, $DatabaseUsername, $DatabasePassword, $DatabaseName);

    if ($Connection === false) {
        die("ERROR: Could not connect. " . mysqli_connect_error());
    }

    $sqlusers = "select * from user";
    $result = mysqli_query($Connection, $sqlusers);
        echo "<form method='POST'>";
    while($rowuser = mysqli_fetch_array($result)){
        $user = $rowuser['FirstName'];


        echo "<input type='text' name='firstName' value='$user' readonly>";
        echo "<select name='attendanceType'>";

            $sqltype = "select * from attendancetype";
            $resultaType = mysqli_query($Connection, $sqltype);
            while($rowtype = mysqli_fetch_array($resultaType)){
                echo "<option>";
                echo $rowtype['name'];
                echo "</option>";
            }

        echo "</select>";
        echo "<br>";

    }
    echo "<input type='submit' name='submit' value='submit'>";
    echo "</form>";
    ?>

用户表。

INSERT INTO `user` (`UserID`, `FirstName`, `LastName`, `Email`, `Password`, `City`) VALUES
(7, 'Rahul', 'Rajshekaran', 'Rahul@zzz.xxx', 'Rahul@123', 'Pune'),
(8, 'Mahesh', 'Krishna', 'Mahesh@xxx.xxx', 'Mahesh@123', 'Delhi');

出勤类型表：

INSERT INTO `attendancetype` (`attendanceTypeID`, `name`) VALUES
(0001, 'Present'),
(0002, 'Absent');

如何在单次提交表单时将数据插入表格？

Answer 1

你的代码有一个错误。如果要循环表单元素，则必须实现每个输入都具有唯一名称。

     databaseReference.child("notifications/"+uuid+"/data_notification").orderByChild("status").startAt("read").endAt("unread").addChildEventListener(new ChildEventListener() {
        @Override
        public void onChildAdded(DataSnapshot dataSnapshot) {.....


        //add items to list (like you did)



         @Override
        public void onChildChanged(DataSnapshot dataSnapshot) {.....

        //set the items again to the list (update them here) and notify
        //data set changed (so values update in your list).

处理表格：

$i = 0; 
while($rowuser = mysqli_fetch_array($result)){
    $user = $rowuser['FirstName'];


    echo "<input type='text' name='firstName[".$i."]' value='$user' readonly>";
    echo "<select name='attendanceType[".$i."]'>";

        $sqltype = "select * from attendancetype";
        $resultaType = mysqli_query($Connection, $sqltype);
        while($rowtype = mysqli_fetch_array($resultaType)){
            echo "<option>";
            echo $rowtype['name'];
            echo "</option>";
        }

    echo "</select>";
    echo "<br>";
    $i++;
}

escapeFunction用作事实的参考者，你应该以某种方式逃避（有更多的方法）每个输入，并且需要替换或定义

while循环结果中的表单提交仅在最后一行中出现？

1 个答案: