1

我这里有个小问题。没有显示错误通知，但数据没有插入/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;
        }
    }
4

4 回答 4

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]'");
        }
    }
于 2013-05-23T09:42:26.550 回答
2

不应该

UPDATE FROM ar_produk_detail SET id_subkateg

但

"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。

于 2013-05-23T08:37:06.537 回答
1 
 $do = "UPDATE ar_produk_detail SET id_subkategori = '".$data[$r]."' WHERE id_produk_det = '".$p_id[$r]."'";

// Remove from in  the query
于 2013-05-23T08:37:32.250 回答
0

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

查看您使用“FORM”的更新查询，这是不正确的。

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

所以你必须从你的查询中删除“FROM” $do = "UPDATE ar_produk_detail SET id_subkategori = '".$data[$r]."' WHERE id_produk_det = '".$p_id[$r]."'";

于 2013-05-23T09:30:47.737 回答