使用mysqli更新记录的问题

Question

我想更新/编辑我的用户表单，但是当我点击“编辑”按钮时，我没有得到预期的结果。它应该显示新数据，但它会显示旧数据。

编码：

<?php 

include"errorReporting.php";
 include "conn.php";
 $conn = connect();
 $conndb = connectdb();

$wardID = $_REQUEST["wardID"];
$RequestName = $_REQUEST["RequestName"];
$Department = $_REQUEST["Department"];
$Position = $_REQUEST["Position"];
$Date= $_REQUEST["Date"];
$TypeOfRequest = $_REQUEST["TypeOfRequest"];
$PleaseSpecify = $_REQUEST["PleaseSpecify"];
$DateRequire = $_REQUEST["DateRequire"];
$DateReturn = $_REQUEST["DateReturn"];

mysqli_select_db($conn,"misadmin") or die ($conn->error ."\n");
$query = "select * from requestform";
$result2= $conn->query($query) or die ($conn->error ."\n");
$row_result =mysqli_fetch_assoc($result2);

mysqli_select_db($conn,"misadmin") or die ($conn->error ."\n");
$conn ->query("UPDATE requestform SET RequestName='$RequestName',Department='$Department',Position='$Position',Date='$Date',TypeOfRequest='$TypeOfRequest',PleaseSpecify='$PleaseSpecify',DateRequire='$DateRequire',DateReturn='$DateReturn' where wardID='$wardID'",$conn->affected_rows);

$result_update=mysqli_fetch_assoc($result);



header("Location:requestform3.php");
?>

输出：

Answer 1

where wardID='???'中的

必须在数据发生变化之前检索数据。我是说＆＃34; 4f＆＃34;不是＆＃34; med＆＃34;。

您可以在帖子表单中添加一个textbok：

<input type="hidden" id="original_wardID" value="<?php echo $wardID?>" />

在你的php代码中添加：

$ori_wardID=$_REQUEST['original_wardID'];

然后改变你的sql：

UPDATE requestform SET wardID='$wardID',RequestName='$RequestName',Department='$Department',Position='$Position',Date='$Date',TypeOfRequest='$TypeOfRequest',PleaseSpecify='$PleaseSpecify',DateRequire='$DateRequire',DateReturn='$DateReturn' where wardID='$ori_wardID'

Answer 2

试一试：

$conn ->query("UPDATE requestform SET RequestName='$RequestName',Department='$Department',Position='$Position',Date='$Date',TypeOfRequest='$TypeOfRequest',PleaseSpecify='$PleaseSpecify',DateRequire='$DateRequire',DateReturn='$DateReturn' where wardID=$wardID ");

Answer 3

"UPDATE requestform SET RequestName='$RequestName',Department='$Department',Position='$Position',Date='$Date',TypeOfRequest='$TypeOfRequest',PleaseSpecify='$PleaseSpecify',DateRequire='$DateRequire',DateReturn='$DateReturn' where wardID='$wardID'"

更改为

"UPDATE requestform SET RequestName='".$RequestName."',Department='".$Department."',Position='".$Position."',Date='".$Date."',TypeOfRequest='".$TypeOfRequest."',PleaseSpecify='".$PleaseSpecify."',DateRequire='".$DateRequire."',DateReturn='".$DateReturn."' where wardID=".$wardID

或者喜欢这个演示代码

$db_sql="select id ,uid,regdate from `newtable` where id=?";
$stmt=$mysqli->prepare($db_sql);//
$stmt->bind_param("i",$id);// i int d double s string b blob
$result = $stmt->execute();

但我想使用pdo会更好

Answer 4

您没有使用if ($_SERVER['REQUEST_METHOD'] == 'POST'){//save database}或(isset $_POST['edit']){//save database}作为编辑按钮，以便在点击编辑按钮时保存数据。

另外，您在UPDATE查询中使用日期（日期=）。 DATE也是MYSQL中使用的本机函数，您需要更改它。如果您将其更改为DateChanged使用预准备语句，它可以帮助您的代码更具可读性。

if ( $_SERVER['REQUEST_METHOD'] == 'POST' )
{
    $stmt = $conn->prepare("UPDATE requestform  SET RequestName=?, 
    Department=?, Position=?, DateChanged=?, TypeOfRequest=?, 
    PleaseSpecify=?, DateRequire=?, DateReturn=? WHERE wardID=?");

    $stmt->bind_param("sssssssss", $RequestName, $Department, 
    $Position, $DateChanged, $TypeOfRequest, $PleaseSpecify, 
    $DateRequire, $wardID);
    $stmt->execute();
    $stmt->close();
    $conn->close();

}

Answer 5

我找到了答案。在我使用旧模板重做我的编码之后

以下是代码。

{{1}}

感谢那些提出建议的人。我从你们那里学到了很多东西。