Question

首先，让我打个招呼，感谢所有帮助我找到过去一年左右答案的人，只需查看我的问题。我现在有一个问题，过去2天一直让我发疯。我已经找到了以前的答案，尝试了变化，但仍然卡住了。

无论如何，我修改了一些代码来从数据库表而不是CSV文件中获取输入。我首先得到一个错误，说'找不到文件＆＃39;，这是我的数据库连接文件（证明可以在所有其他使用它的脚本中工作）。它与执行文件的目录相同，并且debian系统的权限为755。我已将连接代码移动到StoreUpload.php文件中。原始代码调用函数文件来执行mysqli_query，我得到了'期望2，得到1＆＃39;错误。我将查询移动到StoreUpload文件中。我仍然得到了'期待2，得到1＆＃39;来自原始代码的第一个查询出错（我的输入工作正常，同一个DB）。我没有想法，它可能很简单，但我无法看到它。以下是一些代码段。输出屏幕截图：

SELECT
  p.products_id as v_products_id,
  p.products_model as v_products_model,
  p.products_sku as v_products_sku,
  p.products_upc as v_products_upc,
  p.products_image as v_products_image,
  p.products_price as v_products_price,
  p.products_weight as v_products_weight,
  p.products_date_added as v_date_added,
  p.products_date_available as v_date_avail,
  p.products_tax_class_id as v_tax_class_id,
  p.products_quantity as v_products_quantity,
  p.manufacturers_id as v_manufacturers_id,
  subc.categories_id as v_categories_id FROM
  `zc_products` as p,
  `zc_categories` as subc,
  `zc_products_to_categories` as ptoc
WHERE
  p.products_id = 'ptoc.products_id' AND
  p.products_model = 'EPPIRFA-9V' AND
  ptoc.categories_id = 'subc.categories_id'

PHP Warning: mysqli_query() expects at least 2 parameters, 1 given in /home/mark/development/epStoreUpload.php on line 375

和代码：

$sql = "SELECT
    p.products_id as v_products_id,
    p.products_model as v_products_model,
    p.products_sku as v_products_sku,
    p.products_upc as v_products_upc,
    p.products_image as v_products_image,
    p.products_price as v_products_price,
    p.products_weight as v_products_weight,
    p.products_date_added as v_date_added,
    p.products_date_available as v_date_avail,
    p.products_tax_class_id as v_tax_class_id,
    p.products_quantity as v_products_quantity,
    p.manufacturers_id as v_manufacturers_id,
    subc.categories_id as v_categories_id".
    $custom_filelayout_sql.
    " FROM
    `zc_products` as p,
    `zc_categories` as subc,
    `zc_products_to_categories` as ptoc
    WHERE
    p.products_id = 'ptoc.products_id' AND
    p.products_model = '$items[1]' AND
    ptoc.categories_id = 'subc.categories_id'
    ";
    echo $sql;
$result = mysqli_query($dbc,$sql) or trigger_error("Query Failed! SQL: $dbc -

错误：＆＃34; .mysqli_error（），E_USER_ERROR）; $ result = mysqli_query（$ sql）;

SELECT查询直接在PHPMyAdmin中工作。

Answer 1

我认为这会有所帮助

mysqli_query($con,$sql);

它有两个参数，一个是连接，另一个是SQL查询

Answer 2

ARRRGH！我有2个mysqli_query语句，第二个只有1个参数。我一直在阅读这段代码，我一直在阅读我的想法，而不是那里。在我离开一点之后找到它。无论如何，谢谢你的期待。

PHP警告：mysqli_query（）需要至少2个参数，给定1个

2 个答案: