Question

感谢大家这个伟大的网站，这是我第二次提问，我希望它不是太多^^ 我正在处理我的项目，我的更新页面出现了一些奇怪的问题，它基本上显示了mysql中一个表中的所有数据，当我勾选两行以使用新信息更新它们时，它会复制所有新信息来自＆＃34;（仅最后一行）＆＃34;并将其全部复制到其他行上，结果是，所有行都变得相同!!!!难道我做错了什么？请帮帮我们这是我的代码..

我也尝试更新任何其他行时不会更新，但只更新最后一行......

<?php
session_start();
if( isset($_SESSION['username']) ){
 include('../CIEcon.php');


echo "<form action= 'adminCleaning.php'  method = 'post'>" ; 

if(isset($_POST['update'])){

        if( isset($_POST['id']) ){
                if( empty($_POST['id']) || $_POST['id'] == 0 ){
                    echo"<h4>  please choose something to delete   </h4>";
                  }else{

            echo $implid = implode("' , '", $_POST['id']);

            $sqlUpdate = "UPDATE Cleaning  SET  JobName= '$_POST[jobname]',Description= '$_POST[description]',NoStudent='$_POST[nostudent]',DueDate='$_POST[duedate]' WHERE Id IN('" . $implid . "')";
            $resultUpdate = mysqli_query($dbCIE,$sqlUpdate )or die(mysqli_error($dbCIE));


        if (mysqli_affected_rows($dbCIE) > 0) {
            echo "You have successfully updated your data.<br><br>";
        }
        else {
            echo "The data you submitted matched the current data so nothing was changed.<br><br>";
        }


        } // end of else..


        } // end of  if isset($_POST['id']) ... 
} // end of   if isset($_POST['update']) ... 


    $sql = "SELECT * FROM Cleaning  ";
    $result = mysqli_query($dbCIE, $sql) or die(mysqli_error($dbCIE));

                  /// NOW DISPLAY ALL INFO FROM CHOSEN DATABASE...
                            echo "

                            <table cellpadding ='4' border='1' width='80%' align='center'>
                            <tr>

                            <th class='tt' >Check </th>
                            <th class='tt'> Job's Name</th>
                            <th class='tt' >Description</th>
                            <th class='tt' > No Students needed</th>
                            <th class='tt' >Due Date</th>
                            </tr>";

                            while($row = mysqli_fetch_array($result))
                            {
                                echo "<br>";
                            echo "<tr>";
                            echo "<td> <input type='checkbox'  name='id[]'  value='". $row['Id'] ."' /> </td>";  // array[] cause to edit more than one record... 
                            echo "<td><input type='text' name='jobname' value='" . $row['JobName']  . "'> </td>";
                            echo "<td><input type='text' name='description' value='" . $row['Description']  . "'> </td>";
                            echo "<td><input type='text' name='nostudent' value='" . $row['NoStudent']  . "'> </td>";
                            echo "<td><input type='text' name='duedate' value='" . $row['DueDate']  . "'> </td>";
                            echo "</tr>";
                            }
                            echo "</table>";
                   /// END THE SEARCH HERE...........

                            echo " <br>
                                <div align='center'>
                                <input type='reset' value='clear' /> 

                                <input type='submit' name='update' value='update' />

                                </div> ";

                          mysqli_close($dbCIE);

echo "</form>";
}
else{echo "must logout to see this page..!!";}

?>

<html>

<head><title> ..Cleanding.... </title></head>

<style type="text/css">

body{
    margin-top: 70px;    /*space above the table....*/
    background-color: #23438e; 
}
table{
    background-color: white; 
}


.tt{
    background: #f26822;
    color: white ;
}
</style>


<body>

<!-- <a href= "../AdminIndex.php" > <button> Main Page </button></a>     -->

</body>
</html>

Answer 1

问题是您的所有输入都具有相同的名称，并且您使用 IN 进行更新，而不是 =

更改输入标记以遵循此模式：

<input type='text' name='jobname[".$row['Id']."]' value='" . $row['JobName']  . "'>

以及使用此模式的SQL：

JobName= '$_POST[jobname][$id]' ... WHERE Id = $id

如果一次发送多个ID，您需要循环遍历所有输入并运行查询。

应将HTML表的每一行发送到具有唯一ID的服务器，并且应使用适当的数据和id将每个更新发送到数据库。

更新页面不会起作用

1 个答案: