未捕获的SoapFault异常:[soapenv:Server]在结果集开始之前

时间:2019-01-06 10:50:58

标签: java php soap resultset soap-client

我正在开发一个用于电子商务网站的Web服务,并且我将基于某些Supplier_id的所有产品从Java返回到php客户端,但这给了我这个错误

Fatal error: Uncaught SoapFault exception: [soapenv:Server] Before start of result set in C:\wamp\www\eshop\viewProductsSupplier.php:194 Stack trace: #0 C:\wamp\www\eshop\viewProductsSupplier.php(194): SoapClient->__soapCall('viewProducts', Array) #1 {main} thrown in C:\wamp\www\eshop\viewProductsSupplier.php on line 194

我试图检查查询是否执行,但这一切都很好

// php代码

$soapclient = new SoapClient('http://localhost:8090/Eshop_ecommerce/services/Eshop_ecommerce?wsdl', ['trace' => 1]);
            session_start();
            $id=array('id'=>@$_SESSION['supplierId']);
            $response=$soapclient->__soapCall("viewProducts",$id);
            if (isset($response->Name))
            {
                print $response;
            }
            else
            {
                print "Data is null";
            }

// Java代码

ViewProductsResponse res = new ViewProductsResponse();
                Class.forName(driver);
                conn = DriverManager.getConnection(url,username,password);
                stmt = conn.createStatement();
                String Query=null;
                String in=viewProducts.getViewProductsRequest();
                if(!in.isEmpty())
                {
                    Query="SELECT * FROM product";
                }
                else
                {
                    Query="SELECT * FROM product WHERE product_supplier_id='"+viewProducts.getViewProductsRequest()+"'";
                }
                ProductArray pa=new ProductArray();
                ResultSet result= stmt.executeQuery(Query);
                //System.out.println(result);
                while(result.next())
                {
                    Product p= new Product();
                    //System.out.println(result.getString("product_id"));
                    p.setId(result.getString("product_id"));
                    p.setName(result.getString("product_name"));
                    p.setPrice(result.getString("product_price"));
                    p.setStock(result.getString("product_stock"));
                    String supplierNameQuery="SELECT supplier_id,supplier_name FROM supplier WHERE supplier_id='"+result.getString("product_supplier_id")+"'";
                    ResultSet resultSupplierName = stmt.executeQuery(supplierNameQuery);
                    p.setSupplierName(resultSupplierName.getString("supplier_name"));
                    String catQuery="SELECT * FROM product_category WHERE category_id='"+result.getString("product_category_id")+"'";
                    ResultSet resCat= stmt.executeQuery(catQuery);
                    p.setCategoryName(resCat.getString("category_name"));
                    pa.setProduct(p);
                    res.setProducts(pa);
                }
                //res.setViewProductsResponse("Data Entered");
                return res;

我希望它可以在其中打印所有最终产品

1 个答案:

答案 0 :(得分:1)

ResultSet返回的Statement#executeQuery位于第一行之前(请参阅链接的JavaDoc)。为了使此类ResultSet指向第一行,它需要重新定位到第一行,例如使用next()方法。这样做是针对外部SELECT,而不是针对两个内部next()。此外,需要检查ResultSet调用的返回值,以查看该 if(resultSupplierName.next()) { p.setSupplierName(resultSupplierName.getString("supplier_name")); } if(resCat.next()) { p.setCategoryName(resCat.getString("category_name")); } 中是否有任何行。

例如可以这样实现:

{{1}}

请注意,发布的代码还有其他未解决的问题,例如JDBC资源未关闭。我强烈建议您研究一下JEE或Spring(因为该代码显然在应用服务器中运行)或至少在this处进行研究。更容易产生SQL injection

的代码