Question

我在这里有点问题。没有显示错误通知，但数据未插入/更新到数据库中。谁能告诉我为什么？

这是我的代码：

$perbandingan = mysql_query("SELECT * FROM ar_produk_detail WHERE id_produk = '$_GET[id]'");
    $rows = mysql_num_rows($perbandingan);
    while($w = mysql_fetch_array($perbandingan)){
        $w_id[] = $w['id_subkategori'];
        $p_id[] = $w['id_produk_det'];
    }

    $data = $_POST['checkbox'];
    $sum = count($data);

    for ($r = 0; $r < $sum; $r++){
            if($w_id[$r] != $data[$r]){
                $do = "UPDATE FROM ar_produk_detail SET id_subkategori = '".$data[$r]."' WHERE id_produk_det = '".$p_id[$r]."'";
                mysql_query($do); //Hapus yang tidak sama / tidak terpilih
                echo $do;
        }
    }

Answer 1

不应该是

UPDATE FROM ar_produk_detail SET id_subkateg

但是

"UPDATE  ar_produk_detail SET id_subkateg = '".$data[$r]."' WHERE id_produk_det = '".$p_id[$r]."'";

在UPDATE查询中不需要FROM。

Answer 2

$perbandingan = mysql_query("SELECT * FROM ar_produk_detail WHERE id_produk = '$_GET[id]'");
    $rows = mysql_num_rows($perbandingan);
    while($w = mysql_fetch_array($perbandingan)){
        $w_id[] = $w['id_subkategori'];
        $p_id[] = $w['id_produk_det'];
    }

    $data = $_POST['checkbox'];
    $sum = count($data);

    for ($r = 0; $r < $sum; $r++){
            if($w_id[$r] != $data[$r]){
                $do = mysql_query("UPDATE  ar_produk_detail SET id_subkateg = '$data[$r]' WHERE id_produk_det = '$p_id[$r]'");
        }
    }

Answer 3

 $do = "UPDATE ar_produk_detail SET id_subkategori = '".$data[$r]."' WHERE id_produk_det = '".$p_id[$r]."'";

// Remove from in  the query

Answer 4

$ do =“UPDATE FROM ar_produk_detail SET id_subkategori ='”。$ data [$ r]。“'WHERE id_produk_det ='”。$ p_id [$ r]。“'”;

使用“FORM”查看更新查询，这是不正确的。

正确的语法 UPDATE table_name SET column1 = value1 WHERE some_column = some_value;

因此您必须从查询中删除“FROM” $ do =“UPDATE ar_produk_detail SET id_subkategori ='”。$ data [$ r]。“'WHERE id_produk_det ='”。$ p_id [$ r]。“'”;

MySQL数据无法更新

4 个答案: