如何使用 php 在更新查询中用多列更新特定的单列？

Question

我有一张名为Master 的表，它有四列名为id、A、B 和C。

当我更新表格时，我需要修改用户选择的特定单列。例如，如果用户选择列A，那么只有A 记录将在数据库中更新。如果用户选择C，那么只会更新C 记录。

以下是我尝试过的代码，但它正在更新所有列。你会在这方面帮助我吗？

$column_name=$row['column_name'];//A,B,C
if (isset($result->num_rows) > 0) {
    // output data of each row
    while($row = $result->fetch_assoc()) {
        $A=$row['A'];
        $B=$row['B'];
        $C=$row['C'];
      }}

 UPDATE Master SET $column_name='$A'+20, $column_name='$B'+30, $column_name='$C'+50 WHERE user_id='$id'";

Answer 1

虽然你的要求对我来说并不完全清楚，所以我认为 将帮助您满足您的要求。这是一个动态的解决方案。如果不是，您可以澄清您的要求，我必须尽力解决它。谢谢你的提问。

* table schema is stakcoverflow
* table name is master_tbl
* fields name 
id | a | b | c


<?php 
$myHost = "localhost"; // use your real host name ex. myDomainName.com
$myUserName = "root";   // use your real login user name ex. myUserName
$myPassword = "";   // use your real login password ex. myPassword
$myDataBaseName = "stakcoverflow"; // use your real database name ex. myDataBaseName

$con = mysqli_connect( "$myHost", "$myUserName", "$myPassword", "$myDataBaseName" );

if( !$con ) // == null if creation of connection object failed
{ 
    // report the error to the user, then exit program
    die("connection object not created: ".mysqli_error($con));
}

if( mysqli_connect_errno() )  // returns false if no error occurred
{ 
    // report the error to the user, then exit program
    die("Connect failed: ".mysqli_connect_errno()." : ". mysqli_connect_error());
} 

$sql="SELECT a,b,c,id FROM master_tbl ORDER BY id";
$affectedrowsno  = 0;
if ($result=mysqli_query($con,$sql))
  {
    // Get field information for all fields
    while ($fieldinfo=mysqli_fetch_field($result))
    {
        $column_name = $fieldinfo->name;
        // Get field information for all fields

        while ($row=mysqli_fetch_assoc($result))
        {
            $a = $row['a'];
            $b = $row['b'];
            $c = $row['c'];
            $id = $row['id'];
            $sql = "UPDATE master_tbl SET $column_name=$a+20, $column_name=$b+50, $column_name=$c+80 WHERE id=$id";
            mysqli_query($con,$sql);

            $affectedrowsno += count(mysqli_affected_rows($con));
        }

   }
   echo "Affected rows = " . $affectedrowsno;
   // Free result set
    mysqli_free_result($result);

}

mysqli_close($con);

?>

Answer 2

你可以使用if语句：

$column_name=$row['column_name'];//A,B,C
if (isset($result->num_rows) > 0) {
    // output data of each row
    while($row = $result->fetch_assoc()) {
        $A=$row['A'];
        $B=$row['B'];
        $C=$row['C'];
      }}

    if($column_name == 'A'){
       UPDATE Master SET A='$A'+20 WHERE user_id='$id'";
    } elseif($column_name == 'B'){
       UPDATE Master SET B ='$B'+30 WHERE user_id='$id'";
    }elseif($column_name == 'C'){
       UPDATE Master SET C ='$C'+50 WHERE user_id='$id'";
    }    
}