如何在删除数据后更新表中特定列的值

Question

我需要帮助这个问题的家伙 我有一个4列的表。 no, name, address, phone。这种情况是在我删除一个数据之后，我必须减少列no以及我删除的数据量。

上次我使用vb.net制作程序时，我可以这样做。我使用这个代码我可以做得很好

Dim ab As Integer
                ab = lvrak.Items(i).SubItems(4).Text
                Dim stok As New SqlClient.SqlCommand("Update tbl_barangluar set jumlah_barang=jumlah_barang - '" & ab & "' where kode_barang='" & _
        lvrak.Items(i).SubItems(0).Text & "'")
                stok.Connection = koneksi
                stok.ExecuteNonQuery()

但现在我需要在php函数中这样做。我真的不知道该怎么做。 我试着做ex. update temp5 set no='$no' - 1并且进展不顺利。有人请帮助我

这是我的试用PHP

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<?php
include("../../Connections/koneksi.php");


	$no= $_POST['no_check'];
   
// Attempt insert query execution
$sql = "DELETE FROM temp5 WHERE no='$no'";

if(mysqli_query($db, $sql)){
    echo "Records were deleted successfully.";
	
} else{
    echo "ERROR: Could not able to execute $sql. " . mysqli_error($db);
}
 
// Close connection
mysqli_close($db);
?>
<?php
include("../../Connections/koneksi.php");


	$no= $_POST['no_check'];
   
// Attempt insert query execution
$sql = "alter table temp5 auto_increment = 1";

if(mysqli_query($db, $sql)){
    echo "Records were deleted successfully.";
	
} else{
    echo "ERROR: Could not able to execute $sql. " . mysqli_error($db);
}
 
// Close connection
mysqli_close($db);
?>
<?php
include("../../Connections/koneksi.php");

	$no= $_POST['no_check'];
	$min = 1;
 
// Attempt insert query execution
$sql = "UPDATE temp5 SET  no=no -1 where no";
if(mysqli_query($db, $sql)){
    echo "Records inserted successfully.";	

} else{
    echo "Records inserted failed ";
}

mysqli_close($db);
?>

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我会在删除功能之前和之后添加一张图片。 问题出在那里。我试图限制他们，但更多的是恐怖。 在第一张图片中包含删除前的所有数据。我会用no 4删除数据。结果如图2所示。结果错误..结果必须是1,2,3,4

Answer 1

您需要在删除success

后更新计数

UPDATE temp5 SET no=no -1

PHP：

<?php include("../../Connections/koneksi.php");
       $no= $_POST['no_check']; 
       $sql = "DELETE FROM temp5 WHERE no='$no'"; 
      if(mysqli_query($db, $sql)){ 

         $sql1 = "UPDATE temp5 SET no=no -1"; 
         mysqli_query($db, $sql1);

      } else{

        echo "ERROR: Could not able to execute $sql. " . mysqli_error($db); 

       }