Question

选择数据库值并更新到具有相同ID的另一个数据库：我得到的错误是哪里试图获得非对象的属性我怎么能实现它有没有解决方案??

   $servername = "localhost";
   $username = "root";
   $password = "pass";
   $dbname = "db1";
   $dbname2="db2";


   $conn = new mysqli($servername, $username, $password, $dbname);

   if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
   } 


   $conn2 = new mysqli($servername, $username, $password, $dbname2);

   if ($conn2->connect_error) {
die("Connection failed: " . $conn2->connect_error);
   } 

   $sql = "SELECT * FROM affiliates";
   $result = $conn->query($sql);

   if ($result->num_rows > 0) {
   //output data of each row

      while($row = $result->fetch_assoc()) {
            echo "from db1 id: " . $row["id"]. "publish " .$row["publishinsuppliercontants"]. "<br>";

      $sql2 = "UPDATE a1_affilates_cstm SET publish_in_supplier_contacts_c=".$row["publishinsuppliercontants"]." WHERE id_c=".$row["id"]."";
     $result2 = $conn2->query($sql2);

    $sql2 = "SELECT * FROM a1_affilates_cstm WHERE id_c = ". $row["id"]."";

    $result2=mysqli_query($conn2,$sql2) or die mysqli_error($conn2);
    $row1 = mysqli_fetch_array($result2);
   ///check whether inserted ...
           echo "from db2 id: " . $row1["id_c"]. " - publish: ".row1["publish_in_supplier_contacts_c"]. " <br>";

       }
   } else {
       echo "0 results";
   }

Answer 1

看起来你没有得到结果对象。 SELECT查询是否成功？

在使用结果对象进行make操作之前，需要检查select查询状态/结果。

例如在您的情况下：

$sql = "SELECT * FROM affiliates";
if ( $result = $conn->query($sql) ) {
  if ($result->num_rows > 0) {
     ... some logic ...

来自PHP mysqli manual：

“失败时返回FALSE。对于成功的SELECT，SHOW，DESCRIBE或EXPLAIN查询，mysqli_query（）将返回一个mysqli_result对象。对于其他成功的查询，mysqli_query（）将返回TRUE。”

从一个数据库中选择并使用匹配的行更新到另一个

1 个答案: