Question

编辑：我重写了我的问题以包含更多细节：

以下是生成用户数据库结果页面的代码快照：

userdetails.php

$currenttechnicians = "SELECT * FROM ahsinfo ORDER BY FirstName";
$currenttechniciansresults = sqlsrv_query($conn,$currenttechnicians) or die("Couldn't     execut query");

while ($techniciandata=sqlsrv_fetch_array($currenttechniciansresults)){

echo "<tr align=center>";
echo "<td height=35 background=largebg.gif style=text-align: center valign=center>";

echo "<input type=hidden name=\"ID\" value=";
echo $techniciandata['ID'];
echo ">";
echo "<input type=textbox name=\"FirstName\" value=";
echo $techniciandata['FirstName'];
echo "> ";
echo "</td><td height=35 background=largebg.gif style=text-align: center valign=center>";
echo "<input type=textbox name=\"LastName\" value=";
echo $techniciandata['LastName'];
echo "> ";
echo "</td><td height=35 background=largebg.gif style=text-align: center valign=center>";
echo "<input type=textbox size=40 name=\"SIPAddress\" value=";
echo $techniciandata['SIPAddress'];
echo "> ";
echo "</td><td height=35 background=largebg.gif style=text-align: center valign=center>";
echo "<input type=textbox name=\"MobileNumber\" value=";
echo $techniciandata['MobileNumber'];
echo "> ";
echo "</td><td height=35 background=largebg.gif style=text-align: center valign=center>";

?>

这是处理表单提交的代码:(再次，只是更新数据库的SQL查询。我删除了数据库连接字符串。）

edituserdetails.php

<?PHP
//Add the new users details to the database
$edituser = " UPDATE ahsinfo SET FirstName='{$_POST['FirstName']}',     LastName='{$_POST['LastName']}', SIPAddress='{$_POST['SIPAddress']}',     MobileNumber='{$_POST['MobileNumber']}' WHERE ID='{$_POST['ID']}' ";
$editresult = sqlsrv_query($conn,$edituser) or die("Couldn't execute query to update     database");
?>

我正在尝试用userdetails.php页面上的任何更改来更新整个数据库（它是一个小型数据库）。

Answer 1

您的字符串值周围缺少引号，并在其中一个POST变量上使用大括号而不是括号：

UPDATE ahsinfo 
SET FirstName="{$_POST['FirstName']}", 
    LastName="{$_POST['LastName']}", 
    EmailAddress="{$_POST['EmailAddress']}", 
    MobileNumber="{$_POST['MobileNumber']}" 
WHERE ID={$_POST['ID']}

如果您检查了代码中的错误，就会发现这一点。

仅供参考，您对SQL injections

持开放态度

SQL使用不同的值更新多行

1 个答案: